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  • How can I sum up the lines added/removed by a user in a git repo?

    - by Mike
    I am trying to find the total number of lines added and total number of lines removed by a user in a git repository. I looked at http://stackoverflow.com/questions/1265040/how-to-count-total-lines-changed-by-a-specific-author-in-a-git-repository, which had the command git log --author="<authorname>" --pretty=tformat: --numstat, but the answer failed to give a script(however simple) to total the lines changed. What's the simplest way to sum up the lines added/removed?

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  • How can I quickly sum all numbers in a file?

    - by Mark Roberts
    I have a file which contains several thousand numbers, each on it's own line: 34 42 11 6 2 99 ... I'm looking to write a script which will print the sum of all numbers in the file. I've got a solution, but it's not very efficient. (It takes several minutes to run.) I'm looking for a more efficient solution. Any suggestions?

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  • Time complexity of a sorting algorithm

    - by Passonate Learner
    The two programs below get n integers from file and calculates the sum of ath to bth integers q(number of question) times. I think the upper program has worse time complexity than the lower, but I'm having problems calculating the time complexity of these two algorithms. [input sample] 5 3 5 4 3 2 1 2 3 3 4 2 4 [output sample] 7 5 9 Program 1: #include <stdio.h> FILE *in=fopen("input.txt","r"); FILE *out=fopen("output.txt","w"); int n,q,a,b,sum; int data[1000]; int main() int i,j; fscanf(in,"%d%d",&n,&q); for(i=1;i<=n;i++) fscanf(in,"%d",&data[i]); for i=0;i<q;i++) { fscanf(in,"%d%d",&a,&b); sum=0; for(j=a;j<=b;j++) sum+=data[j]; fprintf(out,"%d\n",sum); } return 0; } Program 2: #include <stdio.h> FILE *in=fopen("input.txt","r"); FILE *out=fopen("output.txt","w"); int n,q,a,b; int data[1000]; int sum[1000]; int main() { int i,j; fscanf(in,"%d%d",&n,&q); for(i=1;i<=n;i++) fscanf(in,"%d",&data[i]); for(i=1;i<=n;i++) sum[i]=sum[i-1]+data[i]; for(i=0;i<q;i++) { fscanf(in,"%d%d",&a,&b); fprintf(out,"%d\n",sum[b]-sum[a-1]); } return 0; } The programs below gets n integers from 1 to m and sorts them. Again, I cannot calculate the time complexity. [input sample] 5 5 2 1 3 4 5 [output sample] 1 2 3 4 5 Program: #include <stdio.h> FILE *in=fopen("input.txt","r") FILE *out=fopen("output.txt","w") int n,m; int data[1000]; int count[1000]; int main() { int i,j; fscanf(in,"%d%d",&n,&m); for(i=0;i<n;i++) { fscanf(in,"%d",&data[i]); count[data[i]]++ } for(i=1;i<=m;i++) { for(j=0;j<count[i];j++) fprintf(out,"%d ",i); } return 0; } It's ironic(or not) that I cannot calculate the time complexity of my own algorithms, but I have passions to learn, so please programming gurus, help me!

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  • Python - Blackjack

    - by user335932
    def showCards(): #SUM sum = playerCards[0] + playerCards[1] #Print cards print "Player's Hand: " + str(playerCards) + " : " + "sum print "Dealer's Hand: " + str(compCards[0]) + " : " + "sum" compCards = [Deal(),Deal()] playerCards = [Deal(),Deal()] How can i add up the interger element of a list containing to values? under #SUM error is can combine lists like ints...

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  • OBIA on Teradata - Part 2 Teradata DB Utilization for ETL

    - by Mohan Ramanuja
    Techniques to Monitor Queries and ETL Load CPU and Disk I/OSelect username, processor, sum(cputime), sum(diskio) from dbc.ampusage where processor ='1-0' order by 2,3 descgroup by 1,2;UserName    Vproc    Sum(CpuTime)    Sum(DiskIO)AC00916        10    6.71            24975 List Hardware ErrorsThere is a possibility that the system might have adequate disk space but out of free cylinders. In order to monitor hardware errors, the following query was used:Select * from dbc.Software_Event_Log where Text like '%restart%' order by thedate, thetime;For active users, usage of CPU and analysis of bad CPU to I/O ratiosSelect * from DBC.AMPUSAGE where username='CRMSTGC_DEV_ID';  AND SUBSTR(ACCOUNTNAME,6,3)='006'; Usage By I/OSelect AccountName, UserName, sum(CpuTime), sum(DiskIO)  from DBC.AMPUSAGE group by AccountName, UserName Order by Sum(DiskIO) desc; AccountName                       UserName                          Sum(CpuTime)  Sum(DiskIO)$M1$10062209                      AB89487                           374628.612    7821847$M1$10062210                      AB89487                           186692.244    2799412$M1$10062213                      COC_ETL_ID                        119531.068    331100426$M1$10062200                      AB63472                           118973.316    109881984$M1$10062204                      AB63472                           110825.356    94666986$M1$10062201                      AB63472                           110797.976    75016994$M1$10062202                      AC06936                           100924.448    407839702$M1$10062204                      AB67963                           0         4$M1$10062207                      AB91990                           0         2$M1$10062208                      AB63461                           0         24$M1$10062211                      AB84332                           0         6$M1$10062214                      AB65484                           0         8$M1$10062205                      AB77529                           0         58$M1$10062210                      AC04768                           0         36$M1$10062206                      AB54940                           0         22 Usage By CPUSelect AccountName, UserName, sum(CpuTime), sum(DiskIO)  from DBC.AMPUSAGE group by AccountName, UserName Order by Sum(CpuTime) desc;AccountName                       UserName                          Sum(CpuTime)  Sum(DiskIO)$M1$10062209                      AB89487                           374628.612    7821847$M1$10062210                      AB89487                           186692.244    2799412$M1$10062213                      COC_ETL_ID                        119531.068    331100426$M1$10062200                      AB63472                           118973.316    109881984$M1$10062204                      AB63472                           110825.356    94666986$M1$10062201                      AB63472                           110797.976    75016994$M2$100622105813004760047LOAD     T23_ETLPROC_ENT                   0 6$M1$10062215                      AA37720                           0     180$M1$10062209                      AB81670                           0     6Select count(distinct vproc) from dbc.ampusage;432select * from dbc.dbcinfo;AccountName     UserName     CpuTime DiskIO  CpuTimeNorm         Vproc VprocType    Model$M1$10062205                      CRM_STGC_DEV_ID                   0.32    1764    12.7423999023438    0     AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.28    1730    11.1495999145508    3     AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.304    1736    12.1052799072266    4    AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.248    1731    9.87535992431641    7    AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.332    1731    13.2202398986816    8    AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.284    1712    11.3088799133301    11   AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.24    1757    9.55679992675781    12    AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.292    1737    11.6274399108887    15   AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.268    1753    10.6717599182129    16   AMP      2580$M1$10062205                      CRM_STGC_DEV_ID                   0.276    1732    10.9903199157715    19   AMP      2580select * from dbc.dbcinfo;InfoKey    InfoDataLANGUAGE   SUPPORT           MODE    StandardRELEASE    12.00.03.03VERSION    12.00.03.01a

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  • How to execute a Ruby file in Java, capable of calling functions from the Java program and receiving primitive-type results?

    - by Omega
    I do not fully understand what am I asking (lol!), well, in the sense of if it is even possible, that is. If it isn't, sorry. Suppose I have a Java program. It has a Main and a JavaCalculator class. JavaCalculator has some basic functions like public int sum(int a,int b) { return a + b } Now suppose I have a ruby file. Called MyProgram.rb. MyProgram.rb may contain anything you could expect from a ruby program. Let us assume it contains the following: class RubyMain def initialize print "The sum of 5 with 3 is #{sum(5,3)}" end def sum(a,b) # <---------- Something will happen here end end rubyMain = RubyMain.new Good. Now then, you might already suspect what I want to do: I want to run my Java program I want it to execute the Ruby file MyProgram.rb When the Ruby program executes, it will create an instance of JavaCalculator, execute the sum function it has, get the value, and then print it. The ruby file has been executed successfully. The Java program closes. Note: The "create an instance of JavaCalculator" is not entirely necessary. I would be satisfied with just running a sum function from, say, the Main class. My question: is such possible? Can I run a Java program which internally executes a Ruby file which is capable of commanding the Java program to do certain things and get results? In the above example, the Ruby file asks the Java program to do a sum for it and give the result. This may sound ridiculous. I am new in this kind of thing (if it is possible, that is). WHY AM I ASKING THIS? I have a Java program, which is some kind of game engine. However, my target audience is a bunch of Ruby coders. I don't want to have them learn Java at all. So I figured that perhaps the Java program could simply offer the functionality (capacity to create windows, display sprites, play sounds...) and then, my audience can simply code with Ruby the logic, which basically justs asks my Java engine to do things like displaying sprites or playing sounds. That's when I though about asking this.

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  • Customizing the Test Status on the TFS 2010 SSRS Stories Overview Report

    - by Bob Hardister
    This post shows how to customize the SQL query used by the Team Foundation Server 2010 SQL Server Reporting Services (SSRS) Stories Overview Report. The objective is to show test status for the current version while including user story status of the current and prior versions.  Why? Because we don’t copy completed user stories into the next release. We only want one instance of a user story for the product because we believe copies can get out of sync when they are supposed to be the same. In the example below, work items for the current version are on the area path root and prior versions are not on the area path root. However, you can use area path or iteration path criteria in the query as suits your needs. In any case, here’s how you do it: 1. Download a copy of the report RDL file as a backup 2. Open the report by clicking the edit down arrow and selecting “Edit in Report Builder” 3. Right click on the dsOverview Dataset and select Dataset Properties 4. Update the following SQL per the comments in the code: Customization 1 of 3 … -- Get the list deliverable workitems that have Test Cases linked DECLARE @TestCases Table (DeliverableID int, TestCaseID int); INSERT @TestCases     SELECT h.ID, flh.TargetWorkItemID     FROM @Hierarchy h         JOIN FactWorkItemLinkHistory flh             ON flh.SourceWorkItemID = h.ID                 AND flh.WorkItemLinkTypeSK = @TestedByLinkTypeSK                 AND flh.RemovedDate = CONVERT(DATETIME, '9999', 126)                 AND flh.TeamProjectCollectionSK = @TeamProjectCollectionSK         JOIN [CurrentWorkItemView] wi ON flh.TargetWorkItemID = wi.[System_ID]                  AND wi.[System_WorkItemType] = @TestCase             AND wi.ProjectNodeGUID  = @ProjectGuid              --  Customization 1 of 3: only include test status information when test case area path = root. Added the following 2 statements              AND wi.AreaPath = '{the root area path of the team project}'  …          Customization 2 of 3 … -- Get the Bugs linked to the deliverable workitems directly DECLARE @Bugs Table (ID int, ActiveBugs int, ResolvedBugs int, ClosedBugs int, ProposedBugs int) INSERT @Bugs     SELECT h.ID,         SUM (CASE WHEN wi.[System_State] = @Active THEN 1 ELSE 0 END) Active,         SUM (CASE WHEN wi.[System_State] = @Resolved THEN 1 ELSE 0 END) Resolved,         SUM (CASE WHEN wi.[System_State] = @Closed THEN 1 ELSE 0 END) Closed,         SUM (CASE WHEN wi.[System_State] = @Proposed THEN 1 ELSE 0 END) Proposed     FROM @Hierarchy h         JOIN FactWorkItemLinkHistory flh             ON flh.SourceWorkItemID = h.ID             AND flh.TeamProjectCollectionSK = @TeamProjectCollectionSK         JOIN [CurrentWorkItemView] wi             ON wi.[System_WorkItemType] = @Bug             AND wi.[System_Id] = flh.TargetWorkItemID             AND flh.RemovedDate = CONVERT(DATETIME, '9999', 126)             AND wi.[ProjectNodeGUID] = @ProjectGuid              --  Customization 2 of 3: only include test status information when test case area path = root. Added the following statement              AND wi.AreaPath = '{the root area path of the team project}'       GROUP BY h.ID … Customization 2 of 3 … -- Add the Bugs linked to the Test Cases which are linked to the deliverable workitems -- Walks the links from the user stories to test cases (via the tested by link), and then to -- bugs that are linked to the test case. We don't need to join to the test case in the work -- item history view. -- --    [WIT:User Story/Requirement] --> [Link:Tested By]--> [Link:any type] --> [WIT:Bug] INSERT @Bugs SELECT tc.DeliverableID,     SUM (CASE WHEN wi.[System_State] = @Active THEN 1 ELSE 0 END) Active,     SUM (CASE WHEN wi.[System_State] = @Resolved THEN 1 ELSE 0 END) Resolved,     SUM (CASE WHEN wi.[System_State] = @Closed THEN 1 ELSE 0 END) Closed,     SUM (CASE WHEN wi.[System_State] = @Proposed THEN 1 ELSE 0 END) Proposed FROM @TestCases tc     JOIN FactWorkItemLinkHistory flh         ON flh.SourceWorkItemID = tc.TestCaseID         AND flh.RemovedDate = CONVERT(DATETIME, '9999', 126)         AND flh.TeamProjectCollectionSK = @TeamProjectCollectionSK     JOIN [CurrentWorkItemView] wi         ON wi.[System_Id] = flh.TargetWorkItemID         AND wi.[System_WorkItemType] = @Bug         AND wi.[ProjectNodeGUID] = @ProjectGuid         --  Customization 3 of 3: only include test status information when test case area path = root. Added the following statement         AND wi.AreaPath = '{the root area path of the team project}'     GROUP BY tc.DeliverableID … 5. Save the report and you’re all set. Note: you may need to re-apply custom parameter changes like pre-selected sprints.

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  • project hours worked to be sum of tasks hours worked.

    - by silverkid
    i have a sharepoint list called project . this list has column called hours worked. Then i also have a list called tasks. this list also has a column called hours worked. the task list also has a lookup field where we select project ID from project list. Thus for each project we can have many tasks. now tasks list items are created by individual users and i have to create such a mechanism that the hours worked in project list must always be the sum of hours worked in tasks of that project. How can I achieve this.

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  • How to measure sum of collected memory of Young Generation?

    - by Marcel
    Hi, I'd like to measure memory allocation data from my java application, i.e. the sum of the size of all objects that were allocated. Since object allocation is done in young generation this seems to be the right place. I know jconsole and I know the JMX beans but I just can't find the right variable... Right at the moment we are parsing the gc log output file but that's quite hard. Ideally we'd like to measure it via JMX... How can I get this value? Thanks, Marcel

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  • How do I sum up weighted arrays in PHP?

    - by christian studer
    Hod do I multiply the values of a multi-dimensional array with weigths and sum up the results into a new array in PHP or in general? The boring way looks like this: $weights = array(0.25, 0.4, 0.2, 0.15); $values = array ( array(5,10,15), array(20,25,30), array(35,40,45), array(50,55,60) ); $result = array(); for($i = 0; $i < count($values[0]); ++$i) { $result[$i] = 0; foreach($weights as $index => $thisWeight) $result[$i] += $thisWeight * $values[$index][$i]; } Is there a more elegant solution?

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  • How to sum up values of an array in assembly?

    - by Pablo Fallas
    I have been trying to create a program which can sum up all the values of an "array" in assembly, I have done the following: ORG 1000H TABLE DB DUP(2,4,6,8,10,12,14,16,18,20) FIN DB ? TOTAL DB ? MAX DB 13 ORG 2000H MOV AL, 0 MOV CL, OFFSET FIN-OFFSET TABLE MOV BX, OFFSET TABLE LOOP: ADD AL, [BX] INC BX DEC CL JNZ LOOP HLT END BTW I am using msx88 to compile this code. But I get an error saying that the code 0 has not been recognized. Any advise?

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  • In SQL, what does Group By mean without Count(*), or Sum(), Max(), avg(), ..., and what are some use

    - by Jian Lin
    In SQL, if we use Group By without Count(*) or Sum(), etc, then the result is as follows: mysql> select * from sentGifts; +--------+------------+--------+------+---------------------+--------+ | sentID | whenSent | fromID | toID | trytryWhen | giftID | +--------+------------+--------+------+---------------------+--------+ | 1 | 2010-04-24 | 123 | 456 | 2010-04-24 01:52:20 | 100 | | 2 | 2010-04-24 | 123 | 4568 | 2010-04-24 01:56:04 | 100 | | 3 | 2010-04-24 | 123 | NULL | NULL | 1 | | 4 | 2010-04-24 | NULL | 111 | 2010-04-24 03:10:42 | 2 | | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | 6 | | 6 | 2010-04-24 | 11 | 222 | 2010-04-24 03:54:49 | 6 | | 7 | 2010-04-24 | 1 | 2 | 2010-04-24 03:58:45 | 6 | +--------+------------+--------+------+---------------------+--------+ 7 rows in set (0.00 sec) mysql> select *, count(*) from sentGifts group by whenSent; +--------+------------+--------+------+---------------------+--------+----------+ | sentID | whenSent | fromID | toID | trytryWhen | giftID | count(*) | +--------+------------+--------+------+---------------------+--------+----------+ | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | 6 | 1 | | 1 | 2010-04-24 | 123 | 456 | 2010-04-24 01:52:20 | 100 | 6 | +--------+------------+--------+------+---------------------+--------+----------+ 2 rows in set (0.00 sec) mysql> select * from sentGifts group by whenSent; +--------+------------+--------+------+---------------------+--------+ | sentID | whenSent | fromID | toID | trytryWhen | giftID | +--------+------------+--------+------+---------------------+--------+ | 5 | 2010-03-03 | 11 | 22 | 2010-03-03 00:00:00 | 6 | | 1 | 2010-04-24 | 123 | 456 | 2010-04-24 01:52:20 | 100 | +--------+------------+--------+------+---------------------+--------+ 2 rows in set (0.00 sec) Only 1 row is returned per "group". What does it mean when there is no "Count(*)", etc when using "Group By", and what are it uses? thanks.

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  • Want to calculate the sum of the count rendered by group by option..

    - by Vijay
    i have a table with the columns such id, tid, companyid, ttype etc.. the id may be same for many companyid but unique within the companyid and tid is always unique and i want to calculate the total no of transactions entered in the table, a single transaction may be inserted in more than one row, for example, id tid companyid ttype 1 1 1 xxx 1 2 1 may be null 2 3 1 yyy 2 4 1 may be null 2 5 1 may be null the above entries should be counted as only 2 transactions .. it may be repeated for many companyids.. so how do i calculate the total no of transactions entered in the table i tried select sum(count(*)) from transaction group by id,companyId; but doesn't work select count(*) from transaction group by id; wont work because the id may be repeated for different companyids.

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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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  • Count the number of ways in which a number 'A' can be broken into a sum of 'B' numbers such that all numbers are co-prime to 'C'

    - by rajneesh2k10
    I came across the solution of a problem which involve dynamic-programming approach, solved using a three dimensional matrix. Link to actual problem is: http://community.topcoder.com/stat?c=problem_statement&pm=12189&rd=15177 Solution to this problem is here under MuddyRoad2: http://apps.topcoder.com/wiki/display/tc/SRM+555 In the last paragraph of explanation, author describes a dynamic programming approach to count the number of ways in which a number 'A' can be broken into a sum of 'B' numbers (not necessarily different), such that every number is co-prime to 3 and the order in which these numbers appear does matter. I am not able to grasp that approach. Can anyone help me understand how DP is acting here. I can't understand what is a state here and how it is derived from the previous state.

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  • Shortest command to calculate the sum of a column of output on Unix?

    - by Andrew
    I'm sure there is a quick and easy way to calculate the sum of a column of values on Unix systems (using something like awk or xargs perhaps), but writing a shell script to parse the rows line by line is the only thing that comes to mind at the moment. For example, what's the simplest way to modify the command below to compute and display the total for the SEGSZ column (70300)? ipcs -mb | head -6 IPC status from /dev/kmem as of Mon Nov 17 08:58:17 2008 T ID KEY MODE OWNER GROUP SEGSZ Shared Memory: m 0 0x411c322e --rw-rw-rw- root root 348 m 1 0x4e0c0002 --rw-rw-rw- root root 61760 m 2 0x412013f5 --rw-rw-rw- root root 8192

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  • In SQL find the combination of rows whose sum add up to a specific amount (or amt in other table)

    - by SamH
    Table_1 D_ID Integer Deposit_amt integer Table_2 Total_ID Total_amt integer Is it possible to write a select statement to find all the rows in Table_1 whose Deposit_amt sum to the Total_amt in Table_2. There are multiple rows in both tables. Say the first row in Table_2 has a Total_amt=100. I would want to know that in Table_1 the rows with D_ID 2, 6, 12 summed = 100, the rows D_ID 2, 3, 42 summed = 100, etc. Help appreciated. Let me know if I need to clarify. I am asking this question as someone as part of their job has a list of transactions and a list of totals, she needs to find the possible list of transactions that could have created the total. I agree this sounds dangerous as finding a combination of transactions that sums to a total does not guarantee that they created the total. I wasn't aware it is an np-complete problem.

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  • How to use the sum the value of 2 totals in different table (Reporting Services)?

    - by dewacorp.alliances
    Hi there In report design, I have 2 tables (Current and Proposed) the structure like this: Current Parameter | Value | Rate | Total Value ... Proposed Parameter | Value | Rate | Total Value ... Each bottom of the table (Table Footer), I have something called: "Total: " which is a sum of Total field. I called these textboxes are txtbxCurrent and txtbxProposed and the format is in currency already. This thing is running well. But now I need to get a total of these txtbxCurrent and txtbxProposed. How do I do this? Can I take the value of this or not? BTW .. I am using Ms SQL Server 2005 (ReportViewer - client) Also here my dataset looks like: RecID | Type | Parameter | Value | Rate | Total 1, CURRENT, 'Param1', 100, 0.1, 10 1, CURRENT, 'Param2', 200, 0.2, 10 1, PROPOSED, 'Param1', 100, 0.2, 20 1, PROPOSED, 'Param2', 200, 0.2, 20 Thanks

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