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  • Code Contracts: validating arrays and collections

    - by DigiMortal
    Validating collections before using them is very common task when we use built-in generic types for our collections. In this posting I will show you how to validate collections using code contracts. It is cool how much awful looking code you can avoid using code contracts. Failing code Let’s suppose we have method that calculates sum of all invoices in collection. We have class Invoice and one of properties it has is Sum. I don’t introduce here any complex calculations on invoices because we have another problem to solve in this topic. Here is our code. public static decimal CalculateTotal(IList<Invoice> invoices) {     var sum = invoices.Sum(p => p.Sum);     return sum; } This method is very simple but it fails when invoices list contains at least one null. Of course, we can test if invoice is null but having nulls in lists like this is not good idea – it opens green way for different coding bugs in system. Our goal is to react to bugs ASAP at the nearest place they occur. There is one more way how to make our method fail. It happens when invoices is null. I thing it is also one common bugs during development and it even happens in production environments under some conditions that are usually hardly met. Now let’s protect our little calculation method with code contracts. We need two contracts: invoices cannot be null invoices cannot contain any nulls Our first contract is easy but how to write the second one? Solution: Contract.ForAll Preconditions in code are checked using Contract.Ensures method. This method takes boolean value as argument that sais if contract holds or not. There is also method Contract.ForAll that takes collection and predicate that must hold for that collection. Nice thing is ForAll returns boolean. So, we have very simple solution. public static decimal CalculateTotal(IList<Invoice> invoices) {     Contract.Requires(invoices != null);     Contract.Requires(Contract.ForAll<Invoice>(invoices, p => p != null));       var sum = invoices.Sum(p => p.Sum);     return sum; } And here are some lines of code you can use to test the contracts quickly. var invoices = new List<Invoice>(); invoices.Add(new Invoice()); invoices.Add(null); invoices.Add(new Invoice()); //CalculateTotal(null); CalculateTotal(invoices); If your code is covered with unit tests then I suggest you to write tests to check that these contracts hold for every code run. Conclusion Although it seemed at first place that checking all elements in collection may end up with for-loops that does not look so nice we were able to solve our problem nicely. ForAll method of contract class offered us simple mechanism to check collections and it does it smoothly the code-contracts-way. P.S. I suggest you also read devlicio.us blog posting Validating Collections with Code Contracts by Derik Whittaker.

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  • F# Objects &ndash; Integration with the other .Net Languages &ndash; Part 2

    - by MarkPearl
    So in part one of my posting I covered the real basics of object creation. Today I will hopefully dig a little deeper… My expert F# book brings up an interesting point – properties in F# are just syntactic sugar for method calls. This makes sense… for instance assume I had the following object with the property exposed called Firstname. type Person(Firstname : string, Lastname : string) = member v.Firstname = Firstname I could extend the Firstname property with the following code and everything would be hunky dory… type Person(Firstname : string, Lastname : string) = member v.Firstname = Console.WriteLine("Side Effect") Firstname   All that this would do is each time I use the property Firstname, I would see the side effect printed to the screen saying “Side Effect”. Member methods have a very similar look & feel to properties, in fact the only difference really is that you declare that parameters are being passed in. type Person(Firstname : string, Lastname : string) = member v.FullName(middleName) = Firstname + " " + middleName + " " + Lastname   In the code above, FullName requires the parameter middleName, and if viewed from another project in C# would show as a method and not a property. Precomputation Optimizations Okay, so something that is obvious once you think of it but that poses an interesting side effect of mutable value holders is pre-computation of results. All it is, is a slight difference in code but can result in quite a huge saving in performance. Basically pre-computation means you would not need to compute a value every time a method is called – but could perform the computation at the creation of the object (I hope I have got it right). In a way I battle to differentiate this from lazy evaluation but I will show an example to explain the principle. Let me try and show an example to illustrate the principle… assume the following F# module namespace myNamespace open System module myMod = let Add val1 val2 = Console.WriteLine("Compute") val1 + val2 type MathPrecompute(val1 : int, val2 : int) = let precomputedsum = Add val1 val2 member v.Sum = precomputedsum type MathNormalCompute(val1 : int, val2 : int) = member v.Sum = Add val1 val2 Now assume you have a C# console app that makes use of the objects with code similar to the following… using System; using myNamespace; namespace CSharpTest { class Program { static void Main(string[] args) { Console.WriteLine("Constructing Objects"); var myObj1 = new myMod.MathNormalCompute(10, 11); var myObj2 = new myMod.MathPrecompute(10, 11); Console.WriteLine(""); Console.WriteLine("Normal Compute Sum..."); Console.WriteLine(myObj1.Sum); Console.WriteLine(myObj1.Sum); Console.WriteLine(myObj1.Sum); Console.WriteLine(""); Console.WriteLine("Pre Compute Sum..."); Console.WriteLine(myObj2.Sum); Console.WriteLine(myObj2.Sum); Console.WriteLine(myObj2.Sum); Console.ReadKey(); } } } The output when running the console application would be as follows…. You will notice with the normal compute object that the system would call the Add function every time the method was called. With the Precompute object it only called the compute method when the object was created. Subtle, but something that could lead to major performance benefits. So… this post has gone off in a slight tangent but still related to F# objects.

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  • Beginner to RUBY - require_relative problem

    - by WANNABE
    Hi, I'm learning Ruby (using version 1.8.6) on Windows 7. When I try to run the stock_stats.rb program below, I get the following error: C:\Users\Will\Desktop\ruby>ruby stock_stats.rb stock_stats.rb:1: undefined method `require_relative' for main:Object (NoMethodE rror) I have three v.small code files: stock_stats.rb require_relative 'csv_reader' reader = CsvReader.new ARGV.each do |csv_file_name| STDERR.puts "Processing #{csv_file_name}" reader.read_in_csv_data(csv_file_name) end puts "Total value = #{reader.total_value_in_stock}" csv_reader.rb require 'csv' require_relative 'book_in_stock' class CsvReader def initialize @books_in_stock = [] end def read_in_csv_data(csv_file_name) CSV.foreach(csv_file_name, headers: true) do |row| @books_in_stock << BookInStock.new(row["ISBN"], row["Amount"]) end end # later we'll see how to use inject to sum a collection def total_value_in_stock sum = 0.0 @books_in_stock.each {|book| sum += book.price} sum end def number_of_each_isbn # ... end end book_in_stock.rb require 'csv' require_relative 'book_in_stock' class CsvReader def initialize @books_in_stock = [] end def read_in_csv_data(csv_file_name) CSV.foreach(csv_file_name, headers: true) do |row| @books_in_stock << BookInStock.new(row["ISBN"], row["Amount"]) end end # later we'll see how to use inject to sum a collection def total_value_in_stock sum = 0.0 @books_in_stock.each {|book| sum += book.price} sum end def number_of_each_isbn # ... end end Thanks in advance for any help.

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  • How to write a subquery using DB

    - by Rita
    SELECT SUM( amount_disbursed ) disbusedamount, (select SUM( loaninstallmentpaid_amount ) FROM ourbank_loan_repayment) paidamount, SUM( amount_disbursed )- (select SUM( loaninstallmentpaid_amount ) FROM ourbank_loan_repayment) differenceamount FROM ourbank_loan_disbursement Any Help would be appreciated

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  • Why did i get this error?

    - by David
    here's the code: class Acount { int sum ; String owner ; //these seem to make sense //a constructor or two public Acount () { this.sum = 0 ; this.owner = "John Doe" ; } public Acount (String name) {this.sum = 0 ; this.owner = name ; } public Acount (String name, int sum) {this.sum = sum ; this.owner = name ; } //prints an acount in the format "owner" "sum" public static void printAcount (Acount Acount) {System.out.print (Acount.owner) ; System.out.print (" ") ; System.out.println (Acount.sum) ; } public static void main (String[]arg) { Acount Acount1 = new Acount ("david", 100) ; System.out.println ("heres the first acount as it was created:") ; printAcount (Acount1) ; System.out.println ("now i changed one of its instance varaibles with a static method") ; upOne (Acount1) ; printAcount (Acount1) ; } public static Acount upOne (Acount Acount) { Acount.sum = Acount.sum + 1 ; return Acount ; } } here's the error: Exception in thread "main" java.lang.NoClassDefFoundError: Acount/java What went wrong?

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  • round off and displaying the values

    - by S.PRATHIBA
    Hi all, I have the following code: import java.io.; import java.sql.; import java.math.; import java.lang.; public class Testd1{ public static void main(String[] args) { System.out.println("Sum of the specific column!"); Connection con = null; int m=1; double sum,sum1,sum2; int e[]; e=new int[100]; int p; int decimalPlaces = 5; for( int i=0;i< e.length;i++) { e[i]=0; } double b2,c2,d2,u2,v2; int i,j,k,x,y; double mat[][]=new double[10][10]; try { Class.forName("com.mysql.jdbc.Driver"); con = DriverManager.getConnection ("jdbc:mysql://localhost:3306/prathi","root","mysql"); try{ Statement st = con.createStatement(); ResultSet res = st.executeQuery("SELECT Service_ID,SUM(consumer_feedback) FROM consumer1 group by Service_ID"); while (res.next()){ int data=res.getInt(1); System.out.println(data); System.out.println("\n\n"); int c1 = res.getInt(2); e[m]=res.getInt(2); if(e[m]<0) e[m]=0; m++; System.out.print(c1); System.out.println("\t\t"); } sum=e[1]+e[2]+e[3]+e[4]+e[5]; System.out.println("\n \n The sum is" +sum); for( p=21; p<=25; p++) { if(e[p] != 0) e[p]=e[p]/(int)sum; //I have type casted sum to get output BigDecimal bd1 = new BigDecimal(e[p]); bd1 = bd1.setScale(decimalPlaces, BigDecimal.ROUND_HALF_UP); // setScale is immutable e[p] = bd1.intValue(); System.out.println("\n\n The normalized value is" +e[p]); mat[4][p-21]=e[p]; } } catch (SQLException s){ System.out.println("SQL statement is not executed!"); } } catch (Exception e1){ e1.printStackTrace(); } } } I have a table named consumer1.After calculating the sum i am getting the values as follows mysql select Service_ID,sum(consumer_feedback) from consumer1 group by Service_ ID; Service_ID sum(consumer_feedback) 31 17 32 0 33 60 34 38 35 | 38 In my program I am getting the sum for each Service_ID correctly.But,after normalization ie while I am calculating 17/153=0.111 I am getting the normalized value is 0.I want the normalized values to be displayed correctly after rounding off.My output is as follows C:javac Testd1.java C:java Testd1 Sum of the specific column! 31 17 32 0 33 60 34 38 35 38 The sum is153.0 The normalized value is0 The normalized value is0 The normalized value is0 The normalized value is0 The normalized value is0 But,after normalization i want to get 17/153=0.111 I am getting the normalized value is 0.I want these values to be rounded off.

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  • Linear regression confidence intervals in SQL

    - by Matt Howells
    I'm using some fairly straight-forward SQL code to calculate the coefficients of regression (intercept and slope) of some (x,y) data points, using least-squares. This gives me a nice best-fit line through the data. However we would like to be able to see the 95% and 5% confidence intervals for the line of best-fit (the curves below). What these mean is that the true line has 95% probability of being below the upper curve and 95% probability of being above the lower curve. How can I calculate these curves? I have already read wikipedia etc. and done some googling but I haven't found understandable mathematical equations to be able to calculate this. Edit: here is the essence of what I have right now. --sample data create table #lr (x real not null, y real not null) insert into #lr values (0,1) insert into #lr values (4,9) insert into #lr values (2,5) insert into #lr values (3,7) declare @slope real declare @intercept real --calculate slope and intercept select @slope = ((count(*) * sum(x*y)) - (sum(x)*sum(y)))/ ((count(*) * sum(Power(x,2)))-Power(Sum(x),2)), @intercept = avg(y) - ((count(*) * sum(x*y)) - (sum(x)*sum(y)))/ ((count(*) * sum(Power(x,2)))-Power(Sum(x),2)) * avg(x) from #lr Thank you in advance.

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  • Summation loop program in Pascal

    - by user2526598
    I am having a bit of an issue with this problem. I am taking a Pascal programming class and this problem was in my logic book. I am required to have the user enter a series of (+) numbers and once he/she enters a (-) number, the program should find the sum of all the (+) numbers. I accomplished this, but now I am attempting part two of this problem, which requires me to utilize a nested loop to run the program x amount of times based on the user's input. The following code is what I have so far and honestly I am stumped: program summation; //Define main program's variables var num, sum, numRun : integer; //Design procedure that will promt user for number of runs procedure numRunLoop ( var numRun : integer ); begin writeln('How many times shall I run this program?'); readln(numRun); end; //Design procedure that will sum a series of numbers //based on user input procedure numPromptLoop( numRun : integer; var num : integer ); var count : integer; begin //Utilize for to establish run limit for count := 1 to numRun do begin //Use repeat to prompt user for numbers repeat writeln('Enter a number: '); readln(num); //Tells program when to sum if num >= 0 then sum := sum + num; until num < 0; end; end; //Design procedure that will display procedure addItion( sum : integer ); begin writeln('The sum is; ', sum); end; begin numRunLoop(numRun); numPromptloop(numRun, num); addItion(sum); readln(); end.

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  • In Ruby, why is a method invocation not be able to be treated as a unit when "do" and "end" is used?

    - by Jian Lin
    The following question is related to the question "Ruby Print Inject Do Syntax". My question is, can we insist on using do and end and make it work with puts or p? This works: a = [1,2,3,4] b = a.inject do |sum, x| sum + x end puts b # prints out 10 so, is it correct to say, inject is a class method of the Array class, which takes a block of code, and then returns a number. If so, then it should be no different from calling a function and getting back a return value: b = foo(3) puts b or b = circle.getRadius() puts b In the above two cases, we can directly say puts foo(3) puts circle.getRadius() so, there is no way to make it work directly by using the following 2 ways: a = [1,2,3,4] puts a.inject do |sum, x| sum + x end but it gives ch01q2.rb:7:in `inject': no block given (LocalJumpError) from ch01q2.rb:4:in `each' from ch01q2.rb:4:in `inject' from ch01q2.rb:4 grouping the method call using ( ) doesn't work either: a = [1,2,3,4] puts (a.inject do |sum, x| sum + x end) and this gives: ch01q3.rb:4: syntax error, unexpected kDO_BLOCK, expecting ')' puts (a.inject do |sum, x| ^ ch01q3.rb:4: syntax error, unexpected '|', expecting '=' puts (a.inject do |sum, x| ^ ch01q3.rb:6: syntax error, unexpected kEND, expecting $end end) ^ finally, the following version works: a = [1,2,3,4] puts a.inject { |sum, x| sum + x } but why doesn't the grouping of the method invocation using ( ) work in the earlier example? What if a programmer insist that he uses do and end, can it be made to work?

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  • In Ruby, why does a method invocation not be able to be treated as a unit when "do" and "end" is use

    - by Jian Lin
    The following question is related to http://stackoverflow.com/questions/2127836/ruby-print-inject-do-syntax The question is, can we insist on using DO and END and make it work with puts or p? This works: a = [1,2,3,4] b = a.inject do |sum, x| sum + x end puts b # prints out 10 so, is it correct to say, inject is a class method of the Array class, which takes a block of code, and then returns a number. If so, then it should be no different from calling a function and getting back a return value: b = foo(3) puts b or b = circle.getRadius() puts b In the above two cases, we can directly say puts foo(3) puts circle.getRadius() so, there is no way to make it work directly by using the following 2 ways: a = [1,2,3,4] puts a.inject do |sum, x| sum + x end but it gives ch01q2.rb:7:in `inject': no block given (LocalJumpError) from ch01q2.rb:4:in `each' from ch01q2.rb:4:in `inject' from ch01q2.rb:4 grouping the method call using ( ) doesn't work either: a = [1,2,3,4] puts (a.inject do |sum, x| sum + x end) and this gives: ch01q3.rb:4: syntax error, unexpected kDO_BLOCK, expecting ')' puts (a.inject do |sum, x| ^ ch01q3.rb:4: syntax error, unexpected '|', expecting '=' puts (a.inject do |sum, x| ^ ch01q3.rb:6: syntax error, unexpected kEND, expecting $end end) ^ finally, the following version works: a = [1,2,3,4] puts a.inject { |sum, x| sum + x } but why doesn't the grouping of the method invocation using ( ) work? What if a programmer insists that he uses do and end, can it be made to work directly with p or puts, without an extra temporary variable?

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  • MySql selecting default value if there are no result?

    - by Kenan
    i'm having 2 tables: members and comments. I select all members, and then join comments. But in comments I'm selecting some SUM of points, and if user never commented, I can't get that user in listing?! So how to select default value for SUM, or some other solution: SELECT c.comment_id AS item_id, m.member_id AS member_id, m.avatar, SUM(c.vote_value) AS vote_value, SUM(c.best) AS best, SUM(c.vote_value) + SUM(c.najbolji)*10 AS total FROM members m LEFT JOIN comments c ON m.member_id = c.author_id GROUP BY c.author_id ORDER BY m.member_id DESC LIMIT {$sql_start}, {$sql_pokazi}

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  • How do I write recursive anonymous functions?

    - by James T Kirk
    In my continued effort to learn scala, I'm working through 'Scala by example' by Odersky and on the chapter on first class functions, the section on anonymous function avoids a situation of recursive anonymous function. I have a solution that seems to work. I'm curious if there is a better answer out there. From the pdf: Code to showcase higher order functions def sum(f: Int => Int, a: Int, b: Int): Int = if (a > b) 0 else f(a) + sum(f, a + 1, b) def id(x: Int): Int = x def square(x: Int): Int = x * x def powerOfTwo(x: Int): Int = if (x == 0) 1 else 2 * powerOfTwo(x-1) def sumInts(a: Int, b: Int): Int = sum(id, a, b) def sumSquares(a: Int, b: Int): Int = sum(square, a, b) def sumPowersOfTwo(a: Int, b: Int): Int = sum(powerOfTwo, a, b) scala> sumPowersOfTwo(2,3) res0: Int = 12 from the pdf: Code to showcase anonymous functions def sum(f: Int => Int, a: Int, b: Int): Int = if (a > b) 0 else f(a) + sum(f, a + 1, b) def sumInts(a: Int, b: Int): Int = sum((x: Int) => x, a, b) def sumSquares(a: Int, b: Int): Int = sum((x: Int) => x * x, a, b) // no sumPowersOfTwo My code: def sumPowersOfTwo(a: Int, b: Int): Int = sum((x: Int) => { def f(y:Int):Int = if (y==0) 1 else 2 * f(y-1); f(x) }, a, b) scala> sumPowersOfTwo(2,3) res0: Int = 12

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  • In Ruby, why is a method invocation not able to be treated as a unit when "do" and "end" is used?

    - by Jian Lin
    The following question is related to the question "Ruby Print Inject Do Syntax". My question is, can we insist on using do and end and make it work with puts or p? This works: a = [1,2,3,4] b = a.inject do |sum, x| sum + x end puts b # prints out 10 so, is it correct to say, inject is an instance method of the Array object, and this instance method takes a block of code, and then returns a number. If so, then it should be no different from calling a function or method and getting back a return value: b = foo(3) puts b or b = circle.getRadius() puts b In the above two cases, we can directly say puts foo(3) puts circle.getRadius() so, there is no way to make it work directly by using the following 2 ways: a = [1,2,3,4] puts a.inject do |sum, x| sum + x end but it gives ch01q2.rb:7:in `inject': no block given (LocalJumpError) from ch01q2.rb:4:in `each' from ch01q2.rb:4:in `inject' from ch01q2.rb:4 grouping the method call using ( ) doesn't work either: a = [1,2,3,4] puts (a.inject do |sum, x| sum + x end) and this gives: ch01q3.rb:4: syntax error, unexpected kDO_BLOCK, expecting ')' puts (a.inject do |sum, x| ^ ch01q3.rb:4: syntax error, unexpected '|', expecting '=' puts (a.inject do |sum, x| ^ ch01q3.rb:6: syntax error, unexpected kEND, expecting $end end) ^ finally, the following version works: a = [1,2,3,4] puts a.inject { |sum, x| sum + x } but why doesn't the grouping of the method invocation using ( ) work in the earlier example? What if a programmer insist that he uses do and end, can it be made to work?

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  • VB6 Require some help with looping

    - by k80sg
    Hi, I am trying to convert a source from C++ to vb6: C++: static double mdArray[3][3]; static double mdArray2[3][3]; for (i = 0; i < 3; i++) for (j = 0; j < 3; j++) { double sum = 0; for(k = 0; k < 3; k++) sum = sum + mdArray[k][i] * mdArray[k][k]; mdArray2[i][j] = sum } VB6: dim mdArray(0 to 2, 0 to 2) as integer dim mdArray2(0 to 2, 0 to 2) as integer for i = 0 to 2 for j = 0 to 2 dim a as double sum = 0 for k = 0 to 2 sum = sum + mdArray(k,i) * mdArray(k,j) mdArray2(i,j) = sum Next Next Next Will the vb6 version yield the same result as the C++ version? Thanks.

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  • division problems

    - by David
    This line of code: System.out.println ("aray[j], "+aray[j]+", divided by sum, "+sum+", equals: aray[j]/sum: "+ aray[j]/sum) ; is yeilding this line of text: aray[j], 21, divided by sum, 100, equals: aray[j]/sum: 0 why is it doing this? (everything is right eccept that the answer should be .21)

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  • Help with mysql sum and group query and managing jquery graph results.

    - by Scarface
    Hey guys, I have a system I am trying to design that will retrieve information from a database, so that it can be plotted in a jquery graph. I need to retrieve the information and somehow put it in the necessary format (for example two coordinates var d = [[1269417600000, 10],[1269504000000, 15]];). My table that I am selecting from is a table that stores user votes with fields: points_id (1=vote up ,2=vote down), user_id, timestamp, and topic_id. What I need to do is select all the votes and somehow group them into respective days and then sum the difference between 1 votes and 2 votes for each day. I then need to somehow display the data in the appropriate plotting format shown earlier. For example April 1, 4 votes. The data needs to be separated by commas, except the last plot entry, so I am not sure how to approach that. I showed an example below of the kind of thing I need but it is not correct, echo "var d=["; $query=mysql_query("SELECT *, SUM(IF(points_id = \"1\", 1,0))-SUM(IF([points_id = \"2\", 1,0)) AS 'total' FROM points LEFT JOIN topic on topic.topic_id=points.topic_id WHERE topic.creator='$user' GROUP by timestamp HAVING certain time interval"); while ($row=mysql_fetch_assoc($query)){ $timestamp=$row['timestamp']; $votes=$row['total']; echo "[$timestamp,$vote],"; } echo "];";

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  • Help with mysql sum and group query and managing results for jquery graph.

    - by Scarface
    I have a system I am trying to design that will retrieve information from a database, so that it can be plotted in a jquery graph. I need to retrieve the information and somehow put it in the necessary coordinates format (for example two coordinates var d = [[1269417600000, 10],[1269504000000, 15]];). My table that I am selecting from is a table that stores user votes with fields: points_id (1=vote up, 2=vote down), user_id, timestamp, topic_id What I need to do is select all the votes and somehow group them into respective days and then sum the difference between 1 votes and 2 votes for each day. I then need to somehow display the data in the appropriate plotting format shown earlier. For example April 1, 4 votes. The data needs to be separated by commas, except the last plot entry, so I am not sure how to approach that. I showed an example below of the kind of thing I need but it is not correct, echo "var d=["; $query=mysql_query( "SELECT *, SUM(IF(points_id = \"1\", 1,0))-SUM(IF([points_id = \"2\", 1,0)) AS 'total' FROM points LEFT JOIN topic ON topic.topic_id=points.topic_id WHERE topic.creator='$user' GROUP by timestamp HAVING certain time interval" ); while ($row=mysql_fetch_assoc($query)){ $timestamp=$row['timestamp']; $votes=$row['total']; echo "[$timestamp,$vote],"; } echo "];";

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  • How to find sum of node's value for given depth in binary tree?

    - by masato-san
    I've been scratching my head for several hours for this... problem: Binary Tree (0) depth 0 / \ 10 20 depth 1 / \ / \ 30 40 50 60 depth 2 I am trying to write a function that takes depth as argument and return the sum of values of nodes of the given depth. For instance, if I pass 2, it should return 180 (i.e. 30+40+50+60) I decided to use breath first search and when I find the node with desired depth, sum up the value, but I just can't figure out how to find out the way which node is in what depth. But with this approach I feel like going to totally wrong direction. function level_order($root, $targetDepth) { $q = new Queue(); $q->enqueue($root); while(!$q->isEmpty) { //how to determin the depth of the node??? $node = $q->dequeue(); if($currentDepth == $targetDepth) { $sum = $node->value; } if($node->left != null) { $q->enqueue($node->left); } if($node->right != null) { $q->enqueue($node->right); } //need to reset this somehow $currentDepth ++; } }

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  • How can I order my entries by sum from a separate table?

    - by bgadoci
    I am wondering how I can order posts in my PostController#index to display by a column total in a separate table. Here is how I have it set up. class Post < ActiveRecord::Base :has_many :votes end and Class Vote < ActiveRecord::Base :belongs_to :post end I user can either vote up or down a particular post. I know there are likely better ways to do what I am currently doing but looking for a fix given my current situation. When a user votes up a post, a value of 1 is passed to the Vote Table via a hidden field. When a user votes down a post a value of -1 is passed to the same column (names vote). I am wondering how I can display my posts in order of the sum of the vote column (in the vote table) for a particular post. Another way to say that is, if a particular post has a net vote sum of 5, I want that to appear above a post with a net vote sum of 4. I am assuming that I need to affect the PostController#index action in some fashion. But not sure how to do that.

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  • Round-twice error in .NET's Double.ToString method

    - by Jeppe Stig Nielsen
    Mathematically, consider for this question the rational number 8725724278030350 / 2**48 where ** in the denominator denotes exponentiation, i.e. the denominator is 2 to the 48th power. (The fraction is not in lowest terms, reducible by 2.) This number is exactly representable as a System.Double. Its decimal expansion is 31.0000000000000'49'73799150320701301097869873046875 (exact) where the apostrophes do not represent missing digits but merely mark the boudaries where rounding to 15 resp. 17 digits is to be performed. Note the following: If this number is rounded to 15 digits, the result will be 31 (followed by thirteen 0s) because the next digits (49...) begin with a 4 (meaning round down). But if the number is first rounded to 17 digits and then rounded to 15 digits, the result could be 31.0000000000001. This is because the first rounding rounds up by increasing the 49... digits to 50 (terminates) (next digits were 73...), and the second rounding might then round up again (when the midpoint-rounding rule says "round away from zero"). (There are many more numbers with the above characteristics, of course.) Now, it turns out that .NET's standard string representation of this number is "31.0000000000001". The question: Isn't this a bug? By standard string representation we mean the String produced by the parameterles Double.ToString() instance method which is of course identical to what is produced by ToString("G"). An interesting thing to note is that if you cast the above number to System.Decimal then you get a decimal that is 31 exactly! See this Stack Overflow question for a discussion of the surprising fact that casting a Double to Decimal involves first rounding to 15 digits. This means that casting to Decimal makes a correct round to 15 digits, whereas calling ToSting() makes an incorrect one. To sum up, we have a floating-point number that, when output to the user, is 31.0000000000001, but when converted to Decimal (where 29 digits are available), becomes 31 exactly. This is unfortunate. Here's some C# code for you to verify the problem: static void Main() { const double evil = 31.0000000000000497; string exactString = DoubleConverter.ToExactString(evil); // Jon Skeet, http://csharpindepth.com/Articles/General/FloatingPoint.aspx Console.WriteLine("Exact value (Jon Skeet): {0}", exactString); // writes 31.00000000000004973799150320701301097869873046875 Console.WriteLine("General format (G): {0}", evil); // writes 31.0000000000001 Console.WriteLine("Round-trip format (R): {0:R}", evil); // writes 31.00000000000005 Console.WriteLine(); Console.WriteLine("Binary repr.: {0}", String.Join(", ", BitConverter.GetBytes(evil).Select(b => "0x" + b.ToString("X2")))); Console.WriteLine(); decimal converted = (decimal)evil; Console.WriteLine("Decimal version: {0}", converted); // writes 31 decimal preciseDecimal = decimal.Parse(exactString, CultureInfo.InvariantCulture); Console.WriteLine("Better decimal: {0}", preciseDecimal); // writes 31.000000000000049737991503207 } The above code uses Skeet's ToExactString method. If you don't want to use his stuff (can be found through the URL), just delete the code lines above dependent on exactString. You can still see how the Double in question (evil) is rounded and cast.

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