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• why Cannot invoke super constructor from enum constructor ?

  - by hilal

  public enum A { A(1); private A(int i){ } private A(){ super(); // compile - error // Cannot invoke super constructor from enum constructor A() } } and here is the hierarchy of enum A extends from abstract java.lang.Enum extends java.lang.Object Class c = Class.forName("/*path*/.A"); System.out.println(c.getSuperclass().getName()); System.out.println(Modifier.toString(c.getSuperclass().getModifiers()).contains("abstract")); System.out.println(c.getSuperclass().getSuperclass().getName());

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• minimum L sum in a mxn matrix - 2

  - by hilal

  Here is my first question about maximum L sum and here is different and hard version of it. Problem : Given a mxn *positive* integer matrix find the minimum L sum from 0th row to the m'th row . L(4 item) likes chess horse move Example : M = 3x3 0 1 2 1 3 2 4 2 1 Possible L moves are : (0 1 2 2), (0 1 3 2) (0 1 4 2) We should go from 0th row to the 3th row with minimum sum I solved this with dynamic-programming and here is my algorithm : 1. Take a mxn another Minimum L Moves Sum array and copy the first row of main matrix. I call it (MLMS) 2. start from first cell and look the up L moves and calculate it 3. insert it in MLMS if it is less than exists value 4. Do step 2. until m'th row 5. Choose the minimum sum in the m'th row Let me explain on my example step by step: M[ 0 ][ 0 ] sum(L1 = (0, 1, 2, 2)) = 5 ; sum(L2 = (0,1,3,2)) = 6; so MLMS[ 0 ][ 1 ] = 6 sum(L3 = (0, 1, 3, 2)) = 6 ; sum(L4 = (0,1,4,2)) = 7; so MLMS[ 2 ][ 1 ] = 6 M[ 0 ][ 1 ] sum(L5 = (1, 0, 1, 4)) = 6; sum(L6 = (1,3,2,4)) = 10; so MLMS[ 2 ][ 2 ] = 6 ... the last MSLS is : 0 1 2 4 3 6 6 6 6 Which means 6 is the minimum L sum that can be reach from 0 to the m. I think it is O(8*(m-1)*n) = O(m*n). Is there any optimal solution or dynamic-programming algorithms fit this problem? Thanks, sorry for long question

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• computes the number of possible orderings of n objects under the relations < and =

  - by hilal

  Here is the problem : Give a algorithm that takes a positive integer n as input, and computes the number of possible orderings of n objects under the relations < and =. For example, if n = 3 the 13 possible orderings are as follows: a = b = c, a = b < c, a < b = c, a < b < c, a < c < b, a = c < b, b < a = c, b < a < c, b < c < a, b = c < a, c < a = b, c < a < b, c < b < a. Your algorithm should run in time polynomial in n. I'm null to this problem. Can you find any solution to this dynamic-programming problem?

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• in R, question about generate table

  - by alex

  a = matrix(1:25,5,5) B = capture.output(for (X in 1:5){ A = c(min(a[,X]),quantile(a[,X],0.25),median(a[,X]),quantile(a[,X],0.75),max(a[,X]),mean(a[,X]),sd(a[,X])/m^(1/2),var(a[,X])) cat(A,"\n") }) matrix(B,8,5) what i was trying to do is to generate a table which each column has those element in A and in that order. i try to use the matrix, but seems like it dont reli work here...can anyone help 1 2 3 4 5 min 1st quartile median SEM VAR THIS IS WHAT I WANT THE TABLE LOOKS LIKE ..

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• Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

  - by hilal

  Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

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• Interview question : What is the fastest way to generate prime number recursively ?

  - by hilal

  Generation of prime number is simple but what is the fastest way to find it and generate( prime numbers) it recursively ? Here is my solution. However, it is not the best way. I think it is O(N*sqrt(N)). Please correct me, if I am wrong. public static boolean isPrime(int n) { if (n < 2) { return false; } else if (n % 2 == 0 & n != 2) { return false; } else { return isPrime(n, (int) Math.sqrt(n)); } } private static boolean isPrime(int n, int i) { if (i < 2) { return true; } else if (n % i == 0) { return false; } else { return isPrime(n, --i); } } public static void generatePrimes(int n){ if(n < 2) { return ; } else if(isPrime(n)) { System.out.println(n); } generatePrimes(--n); } public static void main(String[] args) { generatePrimes(200); }

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• Stata Nearest neighbor of percentile

  - by Kyle Billings

  This has probably already been answered, but I must just be searching for the wrong terms. Suppose I am using the built in Stata data set auto: sysuse auto, clear and say for example I am working with 1 independent and 1 dependent variable and I want to essentially compress down to the IQR elements, min, p(25), median, p(75), max... so I use command, keep weight mpg sum weight, detail return list local min=r(min) local lqr=r(p25) local med = r(p50) local uqr = r(p75) local max = r(max) keep if weight==`min' | weight==`max' | weight==`med' | weight==`lqr' | weight==`uqr' Hence, I want to compress the data set down to only those 5 observations, and for example in this situation the median is not actually an element of the weight vector. there is an observation above and an observation below (due to the definition of median this is no surprise). is there a way that I can tell stata to look for the nearest neighbor above the percentile. ie. if r(p50) is not an element of weight then search above that value for the next observation? The end result is I am trying to get the data down to 2 vectors, say weight and mpg such that for each of the 5 elements of weight in the IQR have their matching response in mpg. Any thoughts?

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• SPARC T4-4 Delivers World Record Performance on Oracle OLAP Perf Version 2 Benchmark

  - by Brian

  Oracle's SPARC T4-4 server delivered world record performance with subsecond response time on the Oracle OLAP Perf Version 2 benchmark using Oracle Database 11g Release 2 running on Oracle Solaris 11. The SPARC T4-4 server achieved throughput of 430,000 cube-queries/hour with an average response time of 0.85 seconds and the median response time of 0.43 seconds. This was achieved by using only 60% of the available CPU resources leaving plenty of headroom for future growth. The SPARC T4-4 server operated on an Oracle OLAP cube with a 4 billion row fact table of sales data containing 4 dimensions. This represents as many as 90 quintillion aggregate rows (90 followed by 18 zeros). Performance Landscape Oracle OLAP Perf Version 2 Benchmark 4 Billion Fact Table Rows System Queries/hour Users* Response Time (sec) Average Median SPARC T4-4 430,000 7,300 0.85 0.43 * Users - the supported number of users with a given think time of 60 seconds Configuration Summary and Results Hardware Configuration: SPARC T4-4 server with 4 x SPARC T4 processors, 3.0 GHz 1 TB memory Data Storage 1 x Sun Fire X4275 (using COMSTAR) 2 x Sun Storage F5100 Flash Array (each with 80 FMODs) Redo Storage 1 x Sun Fire X4275 (using COMSTAR with 8 HDD) Software Configuration: Oracle Solaris 11 11/11 Oracle Database 11g Release 2 (11.2.0.3) with Oracle OLAP option Benchmark Description The Oracle OLAP Perf Version 2 benchmark is a workload designed to demonstrate and stress the Oracle OLAP product's core features of fast query, fast update, and rich calculations on a multi-dimensional model to support enhanced Data Warehousing. The bulk of the benchmark entails running a number of concurrent users, each issuing typical multidimensional queries against an Oracle OLAP cube consisting of a number of years of sales data with fully pre-computed aggregations. The cube has four dimensions: time, product, customer, and channel. Each query user issues approximately 150 different queries. One query chain may ask for total sales in a particular region (e.g South America) for a particular time period (e.g. Q4 of 2010) followed by additional queries which drill down into sales for individual countries (e.g. Chile, Peru, etc.) with further queries drilling down into individual stores, etc. Another query chain may ask for yearly comparisons of total sales for some product category (e.g. major household appliances) and then issue further queries drilling down into particular products (e.g. refrigerators, stoves. etc.), particular regions, particular customers, etc. Results from version 2 of the benchmark are not comparable with version 1. The primary difference is the type of queries along with the query mix. Key Points and Best Practices Since typical BI users are often likely to issue similar queries, with different constants in the where clauses, setting the init.ora prameter "cursor_sharing" to "force" will provide for additional query throughput and a larger number of potential users. Except for this setting, together with making full use of available memory, out of the box performance for the OLAP Perf workload should provide results similar to what is reported here. For a given number of query users with zero think time, the main measured metrics are the average query response time, the median query response time, and the query throughput. A derived metric is the maximum number of users the system can support achieving the measured response time assuming some non-zero think time. The calculation of the maximum number of users follows from the well-known response-time law N = (rt + tt) * tp where rt is the average response time, tt is the think time and tp is the measured throughput. Setting tt to 60 seconds, rt to 0.85 seconds and tp to 119.44 queries/sec (430,000 queries/hour), the above formula shows that the T4-4 server will support 7,300 concurrent users with a think time of 60 seconds and an average response time of 0.85 seconds. For more information see chapter 3 from the book "Quantitative System Performance" cited below. -- See Also Quantitative System Performance Computer System Analysis Using Queueing Network Models Edward D. Lazowska, John Zahorjan, G. Scott Graham, Kenneth C. Sevcik external local Oracle Database 11g – Oracle OLAP oracle.com OTN SPARC T4-4 Server oracle.com OTN Oracle Solaris oracle.com OTN Oracle Database 11g Release 2 oracle.com OTN Disclosure Statement Copyright 2012, Oracle and/or its affiliates. All rights reserved. Oracle and Java are registered trademarks of Oracle and/or its affiliates. Other names may be trademarks of their respective owners. Results as of 11/2/2012.

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• Accelerated C++, problem 5-6 (copying values from inside a vector to the front)

  - by Darel

  Hello, I'm working through the exercises in Accelerated C++ and I'm stuck on question 5-6. Here's the problem description: (somewhat abbreviated, I've removed extraneous info.) 5-6. Write the extract_fails function so that it copies the records for the passing students to the beginning of students, and then uses the resize function to remove the extra elements from the end of students. (students is a vector of student structures. student structures contain an individual student's name and grades.) More specifically, I'm having trouble getting the vector.insert function to properly copy the passing student structures to the start of the vector students. Here's the extract_fails function as I have it so far (note it doesn't resize the vector yet, as directed by the problem description; that should be trivial once I get past my current issue.) // Extract the students who failed from the "students" vector. void extract_fails(vector<Student_info>& students) { typedef vector<Student_info>::size_type str_sz; typedef vector<Student_info>::iterator iter; iter it = students.begin(); str_sz i = 0, count = 0; while (it != students.end()) { // fgrade tests wether or not the student failed if (!fgrade(*it)) { // if student passed, copy to front of vector students.insert(students.begin(), it, it); // tracks of the number of passing students(so we can properly resize the array) count++; } cout << it->name << endl; // output to verify that each student is iterated to it++; } } The code compiles and runs, but the students vector isn't adding any student structures to its front. My program's output displays that the students vector is unchanged. Here's my complete source code, followed by a sample input file (I redirect input from the console by typing " < grades" after the compiled program name at the command prompt.) #include <iostream> #include <string> #include <algorithm> // to get the declaration of `sort' #include <stdexcept> // to get the declaration of `domain_error' #include <vector> // to get the declaration of `vector' //driver program for grade partitioning examples using std::cin; using std::cout; using std::endl; using std::string; using std::domain_error; using std::sort; using std::vector; using std::max; using std::istream; struct Student_info { std::string name; double midterm, final; std::vector<double> homework; }; bool compare(const Student_info&, const Student_info&); std::istream& read(std::istream&, Student_info&); std::istream& read_hw(std::istream&, std::vector<double>&); double median(std::vector<double>); double grade(double, double, double); double grade(double, double, const std::vector<double>&); double grade(const Student_info&); bool fgrade(const Student_info&); void extract_fails(vector<Student_info>& v); int main() { vector<Student_info> vs; Student_info s; string::size_type maxlen = 0; while (read(cin, s)) { maxlen = max(maxlen, s.name.size()); vs.push_back(s); } sort(vs.begin(), vs.end(), compare); extract_fails(vs); // display the new, modified vector - it should be larger than // the input vector, due to some student structures being // added to the front of the vector. cout << "count: " << vs.size() << endl << endl; vector<Student_info>::iterator it = vs.begin(); while (it != vs.end()) cout << it++->name << endl; return 0; } // Extract the students who failed from the "students" vector. void extract_fails(vector<Student_info>& students) { typedef vector<Student_info>::size_type str_sz; typedef vector<Student_info>::iterator iter; iter it = students.begin(); str_sz i = 0, count = 0; while (it != students.end()) { // fgrade tests wether or not the student failed if (!fgrade(*it)) { // if student passed, copy to front of vector students.insert(students.begin(), it, it); // tracks of the number of passing students(so we can properly resize the array) count++; } cout << it->name << endl; // output to verify that each student is iterated to it++; } } bool compare(const Student_info& x, const Student_info& y) { return x.name < y.name; } istream& read(istream& is, Student_info& s) { // read and store the student's name and midterm and final exam grades is >> s.name >> s.midterm >> s.final; read_hw(is, s.homework); // read and store all the student's homework grades return is; } // read homework grades from an input stream into a `vector<double>' istream& read_hw(istream& in, vector<double>& hw) { if (in) { // get rid of previous contents hw.clear(); // read homework grades double x; while (in >> x) hw.push_back(x); // clear the stream so that input will work for the next student in.clear(); } return in; } // compute the median of a `vector<double>' // note that calling this function copies the entire argument `vector' double median(vector<double> vec) { typedef vector<double>::size_type vec_sz; vec_sz size = vec.size(); if (size == 0) throw domain_error("median of an empty vector"); sort(vec.begin(), vec.end()); vec_sz mid = size/2; return size % 2 == 0 ? (vec[mid] + vec[mid-1]) / 2 : vec[mid]; } // compute a student's overall grade from midterm and final exam grades and homework grade double grade(double midterm, double final, double homework) { return 0.2 * midterm + 0.4 * final + 0.4 * homework; } // compute a student's overall grade from midterm and final exam grades // and vector of homework grades. // this function does not copy its argument, because `median' does so for us. double grade(double midterm, double final, const vector<double>& hw) { if (hw.size() == 0) throw domain_error("student has done no homework"); return grade(midterm, final, median(hw)); } double grade(const Student_info& s) { return grade(s.midterm, s.final, s.homework); } // predicate to determine whether a student failed bool fgrade(const Student_info& s) { return grade(s) < 60; } Sample input file: Moo 100 100 100 100 100 100 100 100 Fail1 45 55 65 80 90 70 65 60 Moore 75 85 77 59 0 85 75 89 Norman 57 78 73 66 78 70 88 89 Olson 89 86 70 90 55 73 80 84 Peerson 47 70 82 73 50 87 73 71 Baker 67 72 73 40 0 78 55 70 Davis 77 70 82 65 70 77 83 81 Edwards 77 72 73 80 90 93 75 90 Fail2 55 55 65 50 55 60 65 60 Thanks to anyone who takes the time to look at this!

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• Difference in performance: local machine VS amazon medium instance

  - by user644745

  I see a drastic difference in performance matrix when i run it with apache benchmark (ab) in my local machine VS production hosted in amazon medium instance. Same concurrent requests (5) and same total number of requests (111) has been run against both. Amazon has better memory than my local machine. But there are 2 CPUs in my local machine vs 1 CPU in m1.medium. My internet speed is very low at the moment, I am getting Transfer rate as 25.29KBps. How can I improve the performance ? Do not know how to interpret Connect, Processing, Waiting and total in ab output. Here is Localhost: Server Hostname: localhost Server Port: 9999 Document Path: / Document Length: 7631 bytes Concurrency Level: 5 Time taken for tests: 1.424 seconds Complete requests: 111 Failed requests: 102 (Connect: 0, Receive: 0, Length: 102, Exceptions: 0) Write errors: 0 Total transferred: 860808 bytes HTML transferred: 847155 bytes Requests per second: 77.95 [#/sec] (mean) Time per request: 64.148 [ms] (mean) Time per request: 12.830 [ms] (mean, across all concurrent requests) Transfer rate: 590.30 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 0 0 0.5 0 1 Processing: 14 63 99.9 43 562 Waiting: 14 60 96.7 39 560 Total: 14 63 99.9 43 563 And this is production: Document Path: / Document Length: 7783 bytes Concurrency Level: 5 Time taken for tests: 33.883 seconds Complete requests: 111 Failed requests: 0 Write errors: 0 Total transferred: 877566 bytes HTML transferred: 863913 bytes Requests per second: 3.28 [#/sec] (mean) Time per request: 1526.258 [ms] (mean) Time per request: 305.252 [ms] (mean, across all concurrent requests) Transfer rate: 25.29 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 290 297 14.0 293 413 Processing: 897 1178 63.4 1176 1391 Waiting: 296 606 135.6 588 1171 Total: 1191 1475 66.0 1471 1684

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• Nginx Slower than Apache??

  - by ichilton

  Hi, I've just setup 2x identical Rackspace Cloud instances and am doing some comparisons and benchmarks to compare Apache and Nginx. I'm testing with a 3.4k png file and initially 512MB server instances but have now moved to 1024MB server instances. I'm very surprised to see that whatever I try, Apache seems to consistently outperform Nginx....what am I doing wrong? Nginx: Server Software: nginx/0.8.54 Server Port: 80 Document Length: 3400 bytes Concurrency Level: 100 Time taken for tests: 2.320 seconds Complete requests: 1000 Failed requests: 0 Write errors: 0 Total transferred: 3612000 bytes HTML transferred: 3400000 bytes Requests per second: 431.01 [#/sec] (mean) Time per request: 232.014 [ms] (mean) Time per request: 2.320 [ms] (mean, across all concurrent requests) Transfer rate: 1520.31 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 0 11 15.7 3 120 Processing: 1 35 76.9 20 1674 Waiting: 1 31 73.0 19 1674 Total: 1 46 79.1 21 1693 Percentage of the requests served within a certain time (ms) 50% 21 66% 39 75% 40 80% 40 90% 98 95% 136 98% 269 99% 334 100% 1693 (longest request) And Apache: Server Software: Apache/2.2.16 Server Port: 80 Document Length: 3400 bytes Concurrency Level: 100 Time taken for tests: 1.346 seconds Complete requests: 1000 Failed requests: 0 Write errors: 0 Total transferred: 3647000 bytes HTML transferred: 3400000 bytes Requests per second: 742.90 [#/sec] (mean) Time per request: 134.608 [ms] (mean) Time per request: 1.346 [ms] (mean, across all concurrent requests) Transfer rate: 2645.85 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 0 1 3.7 0 27 Processing: 0 3 6.2 1 29 Waiting: 0 2 5.0 1 29 Total: 1 4 7.0 1 29 Percentage of the requests served within a certain time (ms) 50% 1 66% 1 75% 1 80% 1 90% 17 95% 19 98% 26 99% 27 100% 29 (longest request) I'm currently using worker_processes 4; and worker_connections 1024; but i've tried and benchmarked different values and see the same behaviour on all - I just can't get it to perform as well as Apache and from what i've read previously, i'm shocked about this! Can anyone give any advice? Thanks, Ian

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• AWS elastic load balancer basic issues

  - by Jones

  I have an array of EC2 t1.micro instances behind a load balancer and each node can manage ~100 concurrent users before it starts to get wonky. i would THINK if i have 2 such instances it would allow my network to manage 200 concurrent users... apparently not. When i really slam the server (blitz.io) with a full 275 concurrents, it behaves the same as if there is just one node. it goes from 400ms response time to 1.6 seconds (which for a single t1.micro is expected, but not 6). So the question is, am i simply not doing something right or is ELB effectively worthless? Anyone have some wisdom on this? AB logs: Loadbalancer (3x m1.medium) Document Path: /ping/index.html Document Length: 185 bytes Concurrency Level: 100 Time taken for tests: 11.668 seconds Complete requests: 50000 Failed requests: 0 Write errors: 0 Non-2xx responses: 50001 Total transferred: 19850397 bytes HTML transferred: 9250185 bytes Requests per second: 4285.10 [#/sec] (mean) Time per request: 23.337 [ms] (mean) Time per request: 0.233 [ms] (mean, across all concurrent requests) Transfer rate: 1661.35 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 1 2 4.3 2 63 Processing: 2 21 15.1 19 302 Waiting: 2 21 15.0 19 261 Total: 3 23 15.7 21 304 Single instance (1x m1.medium direct connection) Document Path: /ping/index.html Document Length: 185 bytes Concurrency Level: 100 Time taken for tests: 9.597 seconds Complete requests: 50000 Failed requests: 0 Write errors: 0 Non-2xx responses: 50001 Total transferred: 19850397 bytes HTML transferred: 9250185 bytes Requests per second: 5210.19 [#/sec] (mean) Time per request: 19.193 [ms] (mean) Time per request: 0.192 [ms] (mean, across all concurrent requests) Transfer rate: 2020.01 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 1 9 128.9 3 3010 Processing: 1 10 8.7 9 141 Waiting: 1 9 8.7 8 140 Total: 2 19 129.0 12 3020

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• subset in geom_point SOMETIMES returns full dataset, instead of none.

  - by Andreas

  I ask the following in the hope that someone might come up with a generic description about the problem.Basically I have no idea whats wrong with my code. When I run the code below, plot nr. 8 turns out wrong. Specifically the subset in geom_point does not work the way it should. (update: With plot nr. 8 the whole dataset is plottet, instead of only the subset). If somebody can tell me what the problem is, I'll update this post. SOdata <- structure(list(id = 10:55, one = c(7L, 8L, 7L, NA, 7L, 8L, 5L, 7L, 7L, 8L, NA, 10L, 8L, NA, NA, NA, NA, 6L, 5L, 6L, 8L, 4L, 7L, 6L, 9L, 7L, 5L, 6L, 7L, 6L, 5L, 8L, 8L, 7L, 7L, 6L, 6L, 8L, 6L, 8L, 8L, 7L, 7L, 5L, 5L, 8L), two = c(7L, NA, 8L, NA, 10L, 10L, 8L, 9L, 4L, 10L, NA, 10L, 9L, NA, NA, NA, NA, 7L, 8L, 9L, 10L, 9L, 8L, 8L, 8L, 8L, 8L, 9L, 10L, 8L, 8L, 8L, 10L, 9L, 10L, 8L, 9L, 10L, 8L, 8L, 7L, 10L, 8L, 9L, 7L, 9L), three = c(7L, 10L, 7L, NA, 10L, 10L, NA, 10L, NA, NA, NA, NA, 10L, NA, NA, 4L, NA, 7L, 7L, 4L, 10L, 10L, 7L, 4L, 7L, NA, 10L, 4L, 7L, 7L, 7L, 10L, 10L, 7L, 10L, 4L, 10L, 10L, 10L, 4L, 10L, 10L, 10L, 10L, 7L, 10L), four = c(7L, 10L, 4L, NA, 10L, 7L, NA, 7L, NA, NA, NA, NA, 10L, NA, NA, 4L, NA, 10L, 10L, 7L, 10L, 10L, 7L, 7L, 7L, NA, 10L, 7L, 4L, 10L, 4L, 7L, 10L, 2L, 10L, 4L, 12L, 4L, 7L, 10L, 10L, 12L, 12L, 4L, 7L, 10L), five = c(7L, NA, 6L, NA, 8L, 8L, 7L, NA, 9L, NA, NA, NA, 9L, NA, NA, NA, NA, 7L, 8L, NA, NA, 7L, 7L, 4L, NA, NA, NA, NA, 5L, 6L, 5L, 7L, 7L, 6L, 9L, NA, 10L, 7L, 8L, 5L, 7L, 10L, 7L, 4L, 5L, 10L), six = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("2010-05-25", "2010-05-27", "2010-06-07"), class = "factor"), seven = c(0.777777777777778, 0.833333333333333, 0.333333333333333, 0.888888888888889, 0.5, 0.888888888888889, 0.777777777777778, 0.722222222222222, 0.277777777777778, 0.611111111111111, 0.722222222222222, 1, 0.888888888888889, 0.722222222222222, 0.555555555555556, NA, 0, 0.666666666666667, 0.666666666666667, 0.833333333333333, 0.833333333333333, 0.833333333333333, 0.833333333333333, 0.722222222222222, 0.833333333333333, 0.888888888888889, 0.666666666666667, 1, 0.777777777777778, 0.722222222222222, 0.5, 0.833333333333333, 0.722222222222222, 0.388888888888889, 0.722222222222222, 1, 0.611111111111111, 0.777777777777778, 0.722222222222222, 0.944444444444444, 0.555555555555556, 0.666666666666667, 0.722222222222222, 0.444444444444444, 0.333333333333333, 0.777777777777778), eight = c(0.666666666666667, 0.333333333333333, 0.833333333333333, 0.666666666666667, 1, 1, 0.833333333333333, 0.166666666666667, 0.833333333333333, 0.833333333333333, 1, 1, 0.666666666666667, 0.666666666666667, 0.333333333333333, 0.5, 0, 0.666666666666667, 0.5, 1, 0.666666666666667, 0.5, 0.666666666666667, 0.666666666666667, 0.666666666666667, 0.333333333333333, 0.333333333333333, 1, 0.666666666666667, 0.833333333333333, 0.666666666666667, 0.666666666666667, 0.5, 0, 0.833333333333333, 1, 0.666666666666667, 0.5, 0.666666666666667, 0.666666666666667, 0.5, 1, 0.833333333333333, 0.666666666666667, 0.833333333333333, 0.666666666666667), nine = c(0.307692307692308, NA, 0.461538461538462, 0.538461538461538, 1, 0.769230769230769, 0.538461538461538, 0.692307692307692, 0, 0.153846153846154, 0.769230769230769, NA, 0.461538461538462, NA, NA, NA, NA, 0, 0.615384615384615, 0.615384615384615, 0.769230769230769, 0.384615384615385, 0.846153846153846, 0.923076923076923, 0.615384615384615, 0.692307692307692, 0.0769230769230769, 0.846153846153846, 0.384615384615385, 0.384615384615385, 0.461538461538462, 0.384615384615385, 0.461538461538462, NA, 0.923076923076923, 0.692307692307692, 0.615384615384615, 0.615384615384615, 0.769230769230769, 0.0769230769230769, 0.230769230769231, 0.692307692307692, 0.769230769230769, 0.230769230769231, 0.769230769230769, 0.615384615384615), ten = c(0.875, 0.625, 0.375, 0.75, 0.75, 0.75, 0.625, 0.875, 1, 0.125, 1, NA, 0.625, 0.75, 0.75, 0.375, NA, 0.625, 0.5, 0.75, 0.875, 0.625, 0.875, 0.75, 0.625, 0.875, 0.5, 0.75, 0, 0.5, 0.875, 1, 0.75, 0.125, 0.5, 0.5, 0.5, 0.625, 0.375, 0.625, 0.625, 0.75, 0.875, 0.375, 0, 0.875), elleven = c(1, 0.8, 0.7, 0.9, 0, 1, 0.9, 0.5, 0, 0.8, 0.8, NA, 0.8, NA, NA, 0.8, NA, 0.4, 0.8, 0.5, 1, 0.4, 0.5, 0.9, 0.8, 1, 0.8, 0.5, 0.3, 0.9, 0.2, 1, 0.8, 0.1, 1, 0.8, 0.5, 0.2, 0.7, 0.8, 1, 0.9, 0.6, 0.8, 0.2, 1), twelve = c(0.666666666666667, NA, 0.133333333333333, 1, 1, 0.8, 0.4, 0.733333333333333, NA, 0.933333333333333, NA, NA, 0.6, 0.533333333333333, NA, 0.533333333333333, NA, 0, 0.6, 0.533333333333333, 0.733333333333333, 0.6, 0.733333333333333, 0.666666666666667, 0.533333333333333, 0.733333333333333, 0.466666666666667, 0.733333333333333, 1, 0.733333333333333, 0.666666666666667, 0.533333333333333, NA, 0.533333333333333, 0.6, 0.866666666666667, 0.466666666666667, 0.533333333333333, 0.333333333333333, 0.6, 0.6, 0.866666666666667, 0.666666666666667, 0.6, 0.6, 0.533333333333333)), .Names = c("id", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "elleven", "twelve"), class = "data.frame", row.names = c(NA, -46L)) iqr <- function(x, ...) { qs <- quantile(as.numeric(x), c(0.25, 0.5, 0.75), na.rm = T) names(qs) <- c("ymin", "y", "ymax") qs } magic <- function(y, ...) { high <- median(SOdata[[y]], na.rm=T)+1.5*sd(SOdata[[y]],na.rm=T) low <- median(SOdata[[y]], na.rm=T)-1.5*sd(SOdata[[y]],na.rm=T) ggplot(SOdata, aes_string(x="six", y=y))+ stat_summary(fun.data="iqr", geom="crossbar", fill="grey", alpha=0.3)+ geom_point(data = SOdata[SOdata[[y]] > high,], position=position_jitter(w=0.1, h=0),col="green", alpha=0.5)+ geom_point(data = SOdata[SOdata[[y]] < low,], position=position_jitter(w=0.1, h=0),col="red", alpha=0.5)+ stat_summary(fun.y=median, geom="point",shape=18 ,size=4, col="orange") } for (i in names(SOdata)[-c(1,7)]) { p<- magic(i) ggsave(paste("magig_plot_",i,".png",sep=""), plot=p, height=3.5, width=5.5) }

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• how to use ggplot conditional on data

  - by Andreas

  I asked this question and it seams ggplot2 currently has a bug with empty data.frames. Therefore I am trying to check if the dataframe is empty, before I make the plot. But what ever I come up with, it gets really ugly, and doesn't work. So I am asking for your help. example data: SOdata <- structure(list(id = 10:55, one = c(7L, 8L, 7L, NA, 7L, 8L, 5L, 7L, 7L, 8L, NA, 10L, 8L, NA, NA, NA, NA, 6L, 5L, 6L, 8L, 4L, 7L, 6L, 9L, 7L, 5L, 6L, 7L, 6L, 5L, 8L, 8L, 7L, 7L, 6L, 6L, 8L, 6L, 8L, 8L, 7L, 7L, 5L, 5L, 8L), two = c(7L, NA, 8L, NA, 10L, 10L, 8L, 9L, 4L, 10L, NA, 10L, 9L, NA, NA, NA, NA, 7L, 8L, 9L, 10L, 9L, 8L, 8L, 8L, 8L, 8L, 9L, 10L, 8L, 8L, 8L, 10L, 9L, 10L, 8L, 9L, 10L, 8L, 8L, 7L, 10L, 8L, 9L, 7L, 9L), three = c(7L, 10L, 7L, NA, 10L, 10L, NA, 10L, NA, NA, NA, NA, 10L, NA, NA, 4L, NA, 7L, 7L, 4L, 10L, 10L, 7L, 4L, 7L, NA, 10L, 4L, 7L, 7L, 7L, 10L, 10L, 7L, 10L, 4L, 10L, 10L, 10L, 4L, 10L, 10L, 10L, 10L, 7L, 10L), four = c(7L, 10L, 4L, NA, 10L, 7L, NA, 7L, NA, NA, NA, NA, 10L, NA, NA, 4L, NA, 10L, 10L, 7L, 10L, 10L, 7L, 7L, 7L, NA, 10L, 7L, 4L, 10L, 4L, 7L, 10L, 2L, 10L, 4L, 12L, 4L, 7L, 10L, 10L, 12L, 12L, 4L, 7L, 10L), five = c(7L, NA, 6L, NA, 8L, 8L, 7L, NA, 9L, NA, NA, NA, 9L, NA, NA, NA, NA, 7L, 8L, NA, NA, 7L, 7L, 4L, NA, NA, NA, NA, 5L, 6L, 5L, 7L, 7L, 6L, 9L, NA, 10L, 7L, 8L, 5L, 7L, 10L, 7L, 4L, 5L, 10L), six = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("2010-05-25", "2010-05-27", "2010-06-07"), class = "factor"), seven = c(0.777777777777778, 0.833333333333333, 0.333333333333333, 0.888888888888889, 0.5, 0.888888888888889, 0.777777777777778, 0.722222222222222, 0.277777777777778, 0.611111111111111, 0.722222222222222, 1, 0.888888888888889, 0.722222222222222, 0.555555555555556, NA, 0, 0.666666666666667, 0.666666666666667, 0.833333333333333, 0.833333333333333, 0.833333333333333, 0.833333333333333, 0.722222222222222, 0.833333333333333, 0.888888888888889, 0.666666666666667, 1, 0.777777777777778, 0.722222222222222, 0.5, 0.833333333333333, 0.722222222222222, 0.388888888888889, 0.722222222222222, 1, 0.611111111111111, 0.777777777777778, 0.722222222222222, 0.944444444444444, 0.555555555555556, 0.666666666666667, 0.722222222222222, 0.444444444444444, 0.333333333333333, 0.777777777777778), eight = c(0.666666666666667, 0.333333333333333, 0.833333333333333, 0.666666666666667, 1, 1, 0.833333333333333, 0.166666666666667, 0.833333333333333, 0.833333333333333, 1, 1, 0.666666666666667, 0.666666666666667, 0.333333333333333, 0.5, 0, 0.666666666666667, 0.5, 1, 0.666666666666667, 0.5, 0.666666666666667, 0.666666666666667, 0.666666666666667, 0.333333333333333, 0.333333333333333, 1, 0.666666666666667, 0.833333333333333, 0.666666666666667, 0.666666666666667, 0.5, 0, 0.833333333333333, 1, 0.666666666666667, 0.5, 0.666666666666667, 0.666666666666667, 0.5, 1, 0.833333333333333, 0.666666666666667, 0.833333333333333, 0.666666666666667), nine = c(0.307692307692308, NA, 0.461538461538462, 0.538461538461538, 1, 0.769230769230769, 0.538461538461538, 0.692307692307692, 0, 0.153846153846154, 0.769230769230769, NA, 0.461538461538462, NA, NA, NA, NA, 0, 0.615384615384615, 0.615384615384615, 0.769230769230769, 0.384615384615385, 0.846153846153846, 0.923076923076923, 0.615384615384615, 0.692307692307692, 0.0769230769230769, 0.846153846153846, 0.384615384615385, 0.384615384615385, 0.461538461538462, 0.384615384615385, 0.461538461538462, NA, 0.923076923076923, 0.692307692307692, 0.615384615384615, 0.615384615384615, 0.769230769230769, 0.0769230769230769, 0.230769230769231, 0.692307692307692, 0.769230769230769, 0.230769230769231, 0.769230769230769, 0.615384615384615), ten = c(0.875, 0.625, 0.375, 0.75, 0.75, 0.75, 0.625, 0.875, 1, 0.125, 1, NA, 0.625, 0.75, 0.75, 0.375, NA, 0.625, 0.5, 0.75, 0.875, 0.625, 0.875, 0.75, 0.625, 0.875, 0.5, 0.75, 0, 0.5, 0.875, 1, 0.75, 0.125, 0.5, 0.5, 0.5, 0.625, 0.375, 0.625, 0.625, 0.75, 0.875, 0.375, 0, 0.875), elleven = c(1, 0.8, 0.7, 0.9, 0, 1, 0.9, 0.5, 0, 0.8, 0.8, NA, 0.8, NA, NA, 0.8, NA, 0.4, 0.8, 0.5, 1, 0.4, 0.5, 0.9, 0.8, 1, 0.8, 0.5, 0.3, 0.9, 0.2, 1, 0.8, 0.1, 1, 0.8, 0.5, 0.2, 0.7, 0.8, 1, 0.9, 0.6, 0.8, 0.2, 1), twelve = c(0.666666666666667, NA, 0.133333333333333, 1, 1, 0.8, 0.4, 0.733333333333333, NA, 0.933333333333333, NA, NA, 0.6, 0.533333333333333, NA, 0.533333333333333, NA, 0, 0.6, 0.533333333333333, 0.733333333333333, 0.6, 0.733333333333333, 0.666666666666667, 0.533333333333333, 0.733333333333333, 0.466666666666667, 0.733333333333333, 1, 0.733333333333333, 0.666666666666667, 0.533333333333333, NA, 0.533333333333333, 0.6, 0.866666666666667, 0.466666666666667, 0.533333333333333, 0.333333333333333, 0.6, 0.6, 0.866666666666667, 0.666666666666667, 0.6, 0.6, 0.533333333333333)), .Names = c("id", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "elleven", "twelve"), class = "data.frame", row.names = c(NA, -46L)) And the plot iqr <- function(x, ...) { qs <- quantile(as.numeric(x), c(0.25, 0.5, 0.75), na.rm = T) names(qs) <- c("ymin", "y", "ymax") qs } magic <- function(y, ...) { high <- median(SOdata[[y]], na.rm=T)+1.5*sd(SOdata[[y]],na.rm=T) low <- median(SOdata[[y]], na.rm=T)-1.5*sd(SOdata[[y]],na.rm=T) ggplot(SOdata, aes_string(x="six", y=y))+ stat_summary(fun.data="iqr", geom="crossbar", fill="grey", alpha=0.3)+ geom_point(data = SOdata[SOdata[[y]] > high,], position=position_jitter(w=0.1, h=0),col="green", alpha=0.5)+ geom_point(data = SOdata[SOdata[[y]] < low,], position=position_jitter(w=0.1, h=0),col="red", alpha=0.5)+ stat_summary(fun.y=median, geom="point",shape=18 ,size=4, col="orange") } for (i in names(SOdata)[-c(1,7)]) { p<- magic(i) ggsave(paste("magig_plot_",i,".png",sep=""), plot=p, height=3.5, width=5.5) } The problem is that sometimes in the call to geom_point the subset returns an empty dataframe, which sometimes (!) causes ggplot2 to plot all the data instead of none of the data. geom_point(data = SOdata[SOdata[[y]] > high,], position=position_jitter(w=0.1, h=0),col="green", alpha=0.5)+ This is kindda of important to me, and I am really stuck trying to find a solution. Any help that will get me started is much appreciated. Thanks in advance.

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• What is wrong here (will update): subset in geom_point does not work as expected

  - by Andreas

  I ask the following in the hope that someone might come up with a generic description about the problem.Basically I have no idea whats wrong with my code. When I run the code below, plot nr. 8 turns out wrong. Specifically the subset in geom_point does not work the way it should. If somebody can tell me what the problem is, I'll update this post. SOdata <- structure(list(id = 10:55, one = c(7L, 8L, 7L, NA, 7L, 8L, 5L, 7L, 7L, 8L, NA, 10L, 8L, NA, NA, NA, NA, 6L, 5L, 6L, 8L, 4L, 7L, 6L, 9L, 7L, 5L, 6L, 7L, 6L, 5L, 8L, 8L, 7L, 7L, 6L, 6L, 8L, 6L, 8L, 8L, 7L, 7L, 5L, 5L, 8L), two = c(7L, NA, 8L, NA, 10L, 10L, 8L, 9L, 4L, 10L, NA, 10L, 9L, NA, NA, NA, NA, 7L, 8L, 9L, 10L, 9L, 8L, 8L, 8L, 8L, 8L, 9L, 10L, 8L, 8L, 8L, 10L, 9L, 10L, 8L, 9L, 10L, 8L, 8L, 7L, 10L, 8L, 9L, 7L, 9L), three = c(7L, 10L, 7L, NA, 10L, 10L, NA, 10L, NA, NA, NA, NA, 10L, NA, NA, 4L, NA, 7L, 7L, 4L, 10L, 10L, 7L, 4L, 7L, NA, 10L, 4L, 7L, 7L, 7L, 10L, 10L, 7L, 10L, 4L, 10L, 10L, 10L, 4L, 10L, 10L, 10L, 10L, 7L, 10L), four = c(7L, 10L, 4L, NA, 10L, 7L, NA, 7L, NA, NA, NA, NA, 10L, NA, NA, 4L, NA, 10L, 10L, 7L, 10L, 10L, 7L, 7L, 7L, NA, 10L, 7L, 4L, 10L, 4L, 7L, 10L, 2L, 10L, 4L, 12L, 4L, 7L, 10L, 10L, 12L, 12L, 4L, 7L, 10L), five = c(7L, NA, 6L, NA, 8L, 8L, 7L, NA, 9L, NA, NA, NA, 9L, NA, NA, NA, NA, 7L, 8L, NA, NA, 7L, 7L, 4L, NA, NA, NA, NA, 5L, 6L, 5L, 7L, 7L, 6L, 9L, NA, 10L, 7L, 8L, 5L, 7L, 10L, 7L, 4L, 5L, 10L), six = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("2010-05-25", "2010-05-27", "2010-06-07"), class = "factor"), seven = c(0.777777777777778, 0.833333333333333, 0.333333333333333, 0.888888888888889, 0.5, 0.888888888888889, 0.777777777777778, 0.722222222222222, 0.277777777777778, 0.611111111111111, 0.722222222222222, 1, 0.888888888888889, 0.722222222222222, 0.555555555555556, NA, 0, 0.666666666666667, 0.666666666666667, 0.833333333333333, 0.833333333333333, 0.833333333333333, 0.833333333333333, 0.722222222222222, 0.833333333333333, 0.888888888888889, 0.666666666666667, 1, 0.777777777777778, 0.722222222222222, 0.5, 0.833333333333333, 0.722222222222222, 0.388888888888889, 0.722222222222222, 1, 0.611111111111111, 0.777777777777778, 0.722222222222222, 0.944444444444444, 0.555555555555556, 0.666666666666667, 0.722222222222222, 0.444444444444444, 0.333333333333333, 0.777777777777778), eight = c(0.666666666666667, 0.333333333333333, 0.833333333333333, 0.666666666666667, 1, 1, 0.833333333333333, 0.166666666666667, 0.833333333333333, 0.833333333333333, 1, 1, 0.666666666666667, 0.666666666666667, 0.333333333333333, 0.5, 0, 0.666666666666667, 0.5, 1, 0.666666666666667, 0.5, 0.666666666666667, 0.666666666666667, 0.666666666666667, 0.333333333333333, 0.333333333333333, 1, 0.666666666666667, 0.833333333333333, 0.666666666666667, 0.666666666666667, 0.5, 0, 0.833333333333333, 1, 0.666666666666667, 0.5, 0.666666666666667, 0.666666666666667, 0.5, 1, 0.833333333333333, 0.666666666666667, 0.833333333333333, 0.666666666666667), nine = c(0.307692307692308, NA, 0.461538461538462, 0.538461538461538, 1, 0.769230769230769, 0.538461538461538, 0.692307692307692, 0, 0.153846153846154, 0.769230769230769, NA, 0.461538461538462, NA, NA, NA, NA, 0, 0.615384615384615, 0.615384615384615, 0.769230769230769, 0.384615384615385, 0.846153846153846, 0.923076923076923, 0.615384615384615, 0.692307692307692, 0.0769230769230769, 0.846153846153846, 0.384615384615385, 0.384615384615385, 0.461538461538462, 0.384615384615385, 0.461538461538462, NA, 0.923076923076923, 0.692307692307692, 0.615384615384615, 0.615384615384615, 0.769230769230769, 0.0769230769230769, 0.230769230769231, 0.692307692307692, 0.769230769230769, 0.230769230769231, 0.769230769230769, 0.615384615384615), ten = c(0.875, 0.625, 0.375, 0.75, 0.75, 0.75, 0.625, 0.875, 1, 0.125, 1, NA, 0.625, 0.75, 0.75, 0.375, NA, 0.625, 0.5, 0.75, 0.875, 0.625, 0.875, 0.75, 0.625, 0.875, 0.5, 0.75, 0, 0.5, 0.875, 1, 0.75, 0.125, 0.5, 0.5, 0.5, 0.625, 0.375, 0.625, 0.625, 0.75, 0.875, 0.375, 0, 0.875), elleven = c(1, 0.8, 0.7, 0.9, 0, 1, 0.9, 0.5, 0, 0.8, 0.8, NA, 0.8, NA, NA, 0.8, NA, 0.4, 0.8, 0.5, 1, 0.4, 0.5, 0.9, 0.8, 1, 0.8, 0.5, 0.3, 0.9, 0.2, 1, 0.8, 0.1, 1, 0.8, 0.5, 0.2, 0.7, 0.8, 1, 0.9, 0.6, 0.8, 0.2, 1), twelve = c(0.666666666666667, NA, 0.133333333333333, 1, 1, 0.8, 0.4, 0.733333333333333, NA, 0.933333333333333, NA, NA, 0.6, 0.533333333333333, NA, 0.533333333333333, NA, 0, 0.6, 0.533333333333333, 0.733333333333333, 0.6, 0.733333333333333, 0.666666666666667, 0.533333333333333, 0.733333333333333, 0.466666666666667, 0.733333333333333, 1, 0.733333333333333, 0.666666666666667, 0.533333333333333, NA, 0.533333333333333, 0.6, 0.866666666666667, 0.466666666666667, 0.533333333333333, 0.333333333333333, 0.6, 0.6, 0.866666666666667, 0.666666666666667, 0.6, 0.6, 0.533333333333333)), .Names = c("id", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "elleven", "twelve"), class = "data.frame", row.names = c(NA, -46L)) iqr <- function(x, ...) { qs <- quantile(as.numeric(x), c(0.25, 0.5, 0.75), na.rm = T) names(qs) <- c("ymin", "y", "ymax") qs } magic <- function(y, ...) { high <- median(SOdata[[y]], na.rm=T)+1.5*sd(SOdata[[y]],na.rm=T) low <- median(SOdata[[y]], na.rm=T)-1.5*sd(SOdata[[y]],na.rm=T) ggplot(SOdata, aes_string(x="six", y=y))+ stat_summary(fun.data="iqr", geom="crossbar", fill="grey", alpha=0.3)+ geom_point(data = SOdata[SOdata[[y]] > high,], position=position_jitter(w=0.1, h=0),col="green", alpha=0.5)+ geom_point(data = SOdata[SOdata[[y]] < low,], position=position_jitter(w=0.1, h=0),col="red", alpha=0.5)+ stat_summary(fun.y=median, geom="point",shape=18 ,size=4, col="orange") } for (i in names(SOdata)[-c(1,7)]) { p<- magic(i) ggsave(paste("magig_plot_",i,".png",sep=""), plot=p, height=3.5, width=5.5) }

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• How&rsquo;s your Momma an&rsquo; them?

  - by Bill Jones Jr.

  When a Southern “boy” like me sees somebody that used to be, or should be, a close friend or relative that they haven’t seen in a long time, that’s a typical greeting.  Come to think of it, we were often related to close friends. So “back in the day”, we not only knew people but everybody close to them.  When I started driving, my Dad told me to always drive carefully in Polk county.  He said if I ran into anybody there, it was likely they would be related or close family friends. Not so much any more… the cities have gotten bigger and more people come south and stay.  One of the curses of air conditioning I guess. Anyway, it’s been a while.  So “How’s your Momma and them”?  Have you been waiting for me to blog again?  Too bad, I’m back anyway <smile>. Here in Charlotte we just had another great code camp.  The Enterprise Developers Guild is going strong, thanks to the help of a lot of dedicated people.  Mark Wilson, Brian Gough, Syl Walker, Ghayth Hilal, Alberto Botero, Dan Thyer, Jean Doiron, Matt Duffield all come to mind.  Plus all the regulars who volunteer for every special event we have. Brian Gough put on a successful SharePoint Saturday.  Rafael Salas and our friends at the local Pass SQL group had a great SQL Saturday.  Brian Hitney and Glen Gordon keep on doing their usual great job for developers in the southeast as our local Microsoft reps. Since my last post, I have the honor of being designated the INetA Membership Mentor for Georgia in addition to mentoring the groups in the Carolinas for the past several years.  Georgia could be a really good thing since my wife likes shopping in Atlanta, not to mention how much we both like Georgia in general.  As I recall, my Momma had people in Georgia.  Wonder how their “Mommas an’ them” are doing?   Bill J

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• jQuery tooltip: Trouble with remove()

  - by Rosarch

  I'm using a jQuery tooltip plugin. I have HTML like this: <li class="term ui-droppable"> <strong>Fall 2011</strong> <li class="course ui-draggable">Biological Statistics I<a class="remove-course-button" href="">[X]</a></li> <div class="term-meta-data"> <p class="total-credits too-few-credits">Total credits: 3</p> <p class="median-GPA low-GPA">Median Historical GPA: 2.00</p> </div> </li> I want to remove the .course element. So, I attach a click handler to the <a>: function _addDeleteButton(course, term) { var delete_button = $('<a href="" class="remove-course-button" title="Remove this course">[X]</a>'); course.append(delete_button); $(delete_button).click(function() { course.remove(); return false; }).tooltip(); } This all works fine, in terms of attaching the click handler. However, when course.remove() is called, Firebug reports an error in tooltip.js: Line 282 tsettings is null if ((!IE || !$.fn.bgiframe) && tsettings.fade) { What am I doing wrong? If the link has a tooltip attached, do I need to remove it specially? UPDATE: Removing .tooltip() solve the problem. I'd like to keep it in, but that makes me suspect that my use of .tooltip() is incorrect here.

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• Performing calculations by subsets of data in R

  - by Vivi

  I want to perform calculations for each company number in the column PERMNO of my data frame, the summary of which can be seen here: > summary(companydataRETS) PERMNO RET Min. :10000 Min. :-0.971698 1st Qu.:32716 1st Qu.:-0.011905 Median :61735 Median : 0.000000 Mean :56788 Mean : 0.000799 3rd Qu.:80280 3rd Qu.: 0.010989 Max. :93436 Max. :19.000000 My solution so far was to create a variable with all possible company numbers compns <- companydataRETS[!duplicated(companydataRETS[,"PERMNO"]),"PERMNO"] And then use a foreach loop using parallel computing which calls my function get.rho() which in turn perform the desired calculations rhos <- foreach (i=1:length(compns), .combine=rbind) %dopar% get.rho(subset(companydataRETS[,"RET"],companydataRETS$PERMNO == compns[i])) I tested it for a subset of my data and it all works. The problem is that I have 72 million observations, and even after leaving the computer working overnight, it still didn't finish. I am new in R, so I imagine my code structure can be improved upon and there is a better (quicker, less computationally intensive) way to perform this same task (perhaps using apply or with, both of which I don't understand). Any suggestions?

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• subset complete or balance dataset in r

  - by SHRram

  I have a dataset that unequal number of repetition. I want to subset a data by removing those entries that are incomplete (i.e. replication less than maximum). Just small example: set.seed(123) mydt <- data.frame (name= rep ( c("A", "B", "C", "D", "E"), c(1,2,4,4, 3)), var1 = rnorm (14, 3,1), var2 = rnorm (14, 4,1)) mydt name var1 var2 1 A 2.439524 3.444159 2 B 2.769823 5.786913 3 B 4.558708 4.497850 4 C 3.070508 2.033383 5 C 3.129288 4.701356 6 C 4.715065 3.527209 7 C 3.460916 2.932176 8 D 1.734939 3.782025 9 D 2.313147 2.973996 10 D 2.554338 3.271109 11 D 4.224082 3.374961 12 E 3.359814 2.313307 13 E 3.400771 4.837787 14 E 3.110683 4.153373 summary(mydt) name var1 var2 A:1 Min. :1.735 Min. :2.033 B:2 1st Qu.:2.608 1st Qu.:3.048 C:4 Median :3.120 Median :3.486 D:4 Mean :3.203 Mean :3.688 E:3 3rd Qu.:3.446 3rd Qu.:4.412 Max. :4.715 Max. :5.787 I want to get rid of A, B, E from the data as they are incomplete. Thus expected output: name var1 var2 4 C 3.070508 2.033383 5 C 3.129288 4.701356 6 C 4.715065 3.527209 7 C 3.460916 2.932176 8 D 1.734939 3.782025 9 D 2.313147 2.973996 10 D 2.554338 3.271109 11 D 4.224082 3.374961 Please note the dataset is big, the following may not a option: mydt[mydt$name == "C",] mydt[mydt$name == "D", ]

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• Foreign-key-like merge in R

  - by skyl

  I'm merging a bunch of csv with 1 row per id/pk/seqn. > full = merge(demo, lab13am, by="seqn", all=TRUE) > full = merge(full, cdq, by="seqn", all=TRUE) > full = merge(full, mcq, by="seqn", all=TRUE) > full = merge(full, cfq, by="seqn", all=TRUE) > full = merge(full, diq, by="seqn", all=TRUE) > print(length(full$ridageyr)) [1] 9965 > print(summary(full$ridageyr)) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.00 11.00 19.00 29.73 48.00 85.00 Everything is great. But, I have another file which has multiple rows per id like: "seqn","rxd030","rxd240b","nhcode","rxq250" 56,2,"","",NA,NA,"" 57,1,"ACETAMINOPHEN","01200",2 57,1,"BUDESONIDE","08800",1 58,1,"99999","",NA 57 has two rows. So, if I naively try to merge this file, I have a ton more rows and my data gets all skewed up. > full = merge(full, rxq, by="seqn", all=TRUE) > print(length(full$ridageyr)) [1] 15643 > print(summary(full$ridageyr)) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.00 14.00 41.00 40.28 66.00 85.00 Is there a normal idiomatic way to deal with data like this? Suppose I want a way to make a simple model like MYSPECIAL_FACTOR <- somehow() glm(MYSPECIAL_FACTOR ~ full$ridageyr, family=binomial) where MYSPECIAL_FACTOR is, say, whether or not rxd240b == "ACETAMINOPHEN" for the observations which are unique by seqn. You can reproduce by running the first bit of this.

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• Apache2 benchmarks - very poor performance

  - by andrzejp

  I have two servers on which I test the configuration of apache2. The first server: 4GB of RAM, AMD Athlon (tm) 64 X2 Dual Core Processor 5600 + Apache 2.2.3, mod_php, mpm prefork: Settings: Timeout 100 KeepAlive On MaxKeepAliveRequests 150 KeepAliveTimeout 4 <IfModule Mpm_prefork_module> StartServers 7 MinSpareServers 15 MaxSpareServers 30 MaxClients 250 MaxRequestsPerChild 2000 </ IfModule> Compiled in modules: core.c mod_log_config.c mod_logio.c prefork.c http_core.c mod_so.c Second server: 8GB of RAM, Intel (R) Core (TM) i7 CPU [email protected] Apache 2.2.9, **fcgid, mpm worker, suexec** PHP scripts are running via fcgi-wrapper Settings: Timeout 100 KeepAlive On MaxKeepAliveRequests 100 KeepAliveTimeout 4 <IfModule Mpm_worker_module> StartServers 10 MaxClients 200 MinSpareThreads 25 MaxSpareThreads 75 ThreadsPerChild 25 MaxRequestsPerChild 1000 </ IfModule> Compiled in modules: core.c mod_log_config.c mod_logio.c worker.c http_core.c mod_so.c The following test results, which are very strange! New server (dynamic content - php via fcgid+suexec): Server Software: Apache/2.2.9 Server Hostname: XXXXXXXX Server Port: 80 Document Path: XXXXXXX Document Length: 179512 bytes Concurrency Level: 10 Time taken for tests: 0.26276 seconds Complete requests: 1000 Failed requests: 0 Total transferred: 179935000 bytes HTML transferred: 179512000 bytes Requests per second: 38.06 Transfer rate: 6847.88 kb/s received Connnection Times (ms) min avg max Connect: 2 4 54 Processing: 161 257 449 Total: 163 261 503 Old server (dynamic content - mod_php): Server Software: Apache/2.2.3 Server Hostname: XXXXXX Server Port: 80 Document Path: XXXXXX Document Length: 187537 bytes Concurrency Level: 10 Time taken for tests: 173.073 seconds Complete requests: 1000 Failed requests: 22 (Connect: 0, Length: 22, Exceptions: 0) Total transferred: 188003372 bytes HTML transferred: 187546372 bytes Requests per second: 5777.91 Transfer rate: 1086267.40 kb/s received Connnection Times (ms) min avg max Connect: 3 3 28 Processing: 298 1724 26615 Total: 301 1727 26643 Old server: Static content (jpg file) Server Software: Apache/2.2.3 Server Hostname: xxxxxxxxx Server Port: 80 Document Path: /images/top2.gif Document Length: 40486 bytes Concurrency Level: 100 Time taken for tests: 3.558 seconds Complete requests: 1000 Failed requests: 0 Write errors: 0 Total transferred: 40864400 bytes HTML transferred: 40557482 bytes Requests per second: 281.09 [#/sec] (mean) Time per request: 355.753 [ms] (mean) Time per request: 3.558 [ms] (mean, across all concurrent requests) Transfer rate: 11217.51 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 3 11 4.5 12 23 Processing: 40 329 61.4 339 1009 Waiting: 6 282 55.2 293 737 Total: 43 340 63.0 351 1020 New server - static content (jpg file) Server Software: Apache/2.2.9 Server Hostname: XXXXX Server Port: 80 Document Path: /images/top2.gif Document Length: 40486 bytes Concurrency Level: 100 Time taken for tests: 3.571531 seconds Complete requests: 1000 Failed requests: 0 Write errors: 0 Total transferred: 41282792 bytes HTML transferred: 41030080 bytes Requests per second: 279.99 [#/sec] (mean) Time per request: 357.153 [ms] (mean) Time per request: 3.572 [ms] (mean, across all concurrent requests) Transfer rate: 11287.88 [Kbytes/sec] received Connection Times (ms) min mean[+/-sd] median max Connect: 2 63 24.8 66 119 Processing: 124 278 31.8 282 391 Waiting: 3 70 28.5 66 164 Total: 126 341 35.9 350 443 I noticed that in the apache error.log is a lot of entries: [notice] mod_fcgid: call /www/XXXXX/public_html/forum/index.php with wrapper /www/php-fcgi-scripts/XXXXXX/php-fcgi-starter What I have omitted, or do not understand? Such a difference in requests per second? Is it possible? What could be the cause?

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• Latency issues over internet

  - by Stevo

  I have a Media Temple server running http://www.popsapp.com which I am having latency issues with. If I run ab -n 100 -c 10 http://www.popsapp.com/ from my local machine I get very bad stats e.g.: Connection Times (ms) min mean[+/-sd] median max Connect: 179 3375 2185.4 2837 12525 Processing: 0 505 693.3 229 4564 Waiting: 0 50 115.4 0 415 Total: 964 3880 2094.5 3159 12608 Whereas if I run it from a rackspace server I have I get this: Connection Times (ms) min mean[+/-sd] median max Connect: 75 76 3.3 75 84 Processing: 235 339 81.4 315 579 Waiting: 159 249 61.7 234 411 Total: 311 415 82.0 390 663 To me this looks like intermediate network issues, but I wouldn't have thought it could be this bad! Any ideas how I can improve it? Here's the trace route traceroute to www.popsapp.com (216.70.105.183), 64 hops max, 52 byte packets 1 192.168.2.1 (192.168.2.1) 3.738 ms 0.953 ms 1.418 ms 2 host-92-22-112-1.as13285.net (92.22.112.1) 27.409 ms 97.093 ms 78.858 ms 3 host-78-151-225-141.static.as13285.net (78.151.225.141) 61.830 ms 170.484 ms 113.288 ms 4 host-78-151-225-80.static.as13285.net (78.151.225.80) 101.513 ms host-78-151-225-22.static.as13285.net (78.151.225.22) 64.718 ms 47.309 ms 5 xe-11-1-0-rt001.sov.as13285.net (62.24.240.14) 98.381 ms 114.424 ms xe-11-1-0-rt001.the.as13285.net (62.24.240.6) 96.592 ms 6 host-78-144-1-59.as13285.net (78.144.1.59) 36.799 ms host-78-144-1-63.as13285.net (78.144.1.63) 178.426 ms host-78-144-1-61.as13285.net (78.144.1.61) 85.516 ms 7 xe-10-0-0-scr010.thn.as13285.net (78.144.0.224) 88.158 ms host-78-144-0-207.as13285.net (78.144.0.207) 35.132 ms host-78-144-0-153.as13285.net (78.144.0.153) 121.464 ms 8 limelight-pp-thn.as13285.net (78.144.3.6) 46.987 ms limelight-pp-sov.as13285.net (78.144.5.18) 108.025 ms 40.169 ms 9 tge11-1.fr4.lga.llnw.net (69.28.172.149) 109.603 ms ve6.fr4.lon.llnw.net (68.142.88.221) 121.681 ms 38.609 ms 10 tge11-1.fr4.lga.llnw.net (69.28.172.149) 111.981 ms 113.744 ms 111.711 ms 11 tge8-2.fr4.iad.llnw.net (69.28.189.34) 117.102 ms ve5.fr4.iad.llnw.net (69.28.171.214) 184.372 ms 146.178 ms 12 cr02-1-1.iad1.net2ez.com (65.97.48.254) 182.880 ms net2ez.tge2-2.fr4.iad.llnw.net (69.28.156.170) 150.489 ms 121.862 ms 13 65.97.50.26 (65.97.50.26) 184.620 ms cr02-1-1.iad1.net2ez.com (65.97.48.254) 156.136 ms 131.963 ms 14 65.97.50.26 (65.97.50.26) 124.899 ms 126.537 ms 123.322 ms 15 e1.4.as02.iad01.mtsvc.net (70.32.64.246) 134.647 ms 186.307 ms 211.059 ms 16 popsapp.com (216.70.105.183) 118.876 ms 113.189 ms vzx258.mediatemple.net (216.70.104.17) 131.012 ms Looks to me like there is significant delay across the limelight network. This would explain why the traceroute via my rackspace server doesn't suffer from the same delay as they will be using their own trunk.

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• How should a new programmers behave at their first job? [on hold]

  - by Dimension

  What are programmers expected to know at their first job and how old will they typically be? I'm not going to school so I'm not around other programmers, therefore I have no idea what kind of programmers they are when they first get hired. I just want to get an idea what the median programmer's knowledge looks like. Will they already have had experience with version control? Are they writing good maintainable code? And what are they expected to do do? I'm programming my own software at home and because of the complexities of it and how new I am to programming I'm sometimes throwing all the code out and starting again with a better design. Aren't new programmers going to write terribly structured software for their employer, or is someone else going to be making the decisions on how everything is laid out?

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• Looking for a dynamic programming solution

  - by krammer

  Given a sequence of integers in range 1 to n. Each number can appear at most once. Let there be a symbol X in the sequence which means remove the minimum element from the list. There can be an arbitrarily number of X in the sequence. Example: 1,3,4,X,5,2,X The output is 1,2. We need to find the best way to perform this operation. The solution I have been thinking is: Scan the sequence from left to right and count number of X which takes O(n) time. Perform partial sorting and find the k smallest elements (k = number of X) which takes O(n+klogk) time using median of medians. Is there a better way to solve this problem using dynamic programming or any other way ?

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• Android: OutofMemoryError: bitmap size exceeds VM budget with no reason I can see.

  - by Meymann

  Hi. I am having an OutOfMemory exception with a gallery over 600x800 pixels JPEG's. The environment I've been using Gallery with JPG images around 600x800 pixels. Since my content may be a bit more complex than just images, I have set each view to be a RelativeLayout that wraps ImageView with the JPG. In order to "speed up" the user experience I have a simple cache of 4 slots that prefetches (in a looper) about 1 image left and 1 image right to the displayed image and keeps them in a 4 slot HashMap. The platform I am using AVD of 256 RAM and 128 Heap Size, with a 600x800 screen. It also happens on an Entourage Edge target, except that with the device it's harder to debug. The problem I have been getting an exception: OutofMemoryError: bitmap size exceeds VM budget And it happens when fetching the fifth image. I have tried to change the size of my image cache, and it is still the same. The strange thing: There should not be a memory problem In order to make sure the heap limit is very far away from what I need, I have defined a dummy 8MB array in the beginning, and left it unreferenced so it's immediately dispatched. It is a member of the activity thread and is defined as following static { @SuppressWarnings("unused") byte dummy[] = new byte[ 8*1024*1024 ]; } The result is that the heap size is nearly 11MB and it's all free. Note I have added that trick after it began to crash. It makes OutOfMemory less frequent. Now, I am using DDMS. Just before the crash (does not change much after the crash), DDMS shows: ID Heap Size Allocated Free %Used #Objects 1 11.195 MB 2.428 MB 8.767 MB 21.69% 47,156 And in the detail table it shows: Type Count Total Size Smallest Largest Median Average free 1,536 8.739MB 16B 7.750MB 24B 5.825KB The largest block is 7.7MB. And yet the LogCat says: ERROR/dalvikvm-heap(1923): 925200-byte external allocation too large for this process. If you mind the relation of the median and the average, it is plausible to assume that most of the available blocks are very small. However, there is a block large enough for the bitmap, it's 7.7M. How come it is still not enough? Note: I recorded a heap trace. When looking at the amount of data allocated, it does not feel like more than 2M is allocated. It does match the free memory report by DDMS. Could it be that I experience some problem like heap-fragmentation? How do I solve/workaround the problem? Is the heap shared to all threads? Could it be that I interpret the DDMS readout in a wrong way, and there is really no 900K block to allocate? If so, can anybody please tell me where I can see that? Thanks a lot Meymann

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