counting

this is my DB table: CREATE TABLE IF NOT EXISTS `inspection_report` ( `Inspection_datetime` datetime NOT NULL, `Line` char(5) NOT NULL, `S` int(11) NOT NULL, `A` int(11) NOT NULL, `B` int(11) NOT NULL, `C` int(11) NOT NULL, INSERT INTO `inspection_report` (`Inspection_datetime`,`Line`,`S`, `A`, `B`, `C`) VALUES ('2010-09-01 09:08:01','FA 05',0, 0, 0, 0),('2010-09-02 14:24:35','FA 07',0, 0, 1, 0),('2010-09-01 09:08:01','fa 05',0, 1, 1, 0),('2010-09-01 16:24:04','FA 03', 0, 1, 0, 0); I have a lot of data for this table.how do i do if i want show the result like: Line 1st week 2nd week 3rd week

I have an array of shape (128, 36, 8) and I'd like to find the number of occurrences of the unique subarrays of length 8 in the last dimension. I'm aware of np.unique and np.bincount , but those seem to be for elements rather than subarrays. I've seen this question but it's about finding the first occurrence of a particular subarray, rather than the counts of all unique subarrays. The question states that the input array is of shape (128, 36, 8) and we are interested in finding unique subarrays of length 8 in the last dimension. So, I am assuming that the uniqueness is along the first two

Problem: I have seen questions like: count the number of 0s between 0 and N? count the number of 1s between 0 and N? count the number of 2s between 0 and N? ... ... These kinds of questions are very similar of asking to find the total number that Ks (i.e. K=0,1,2,...,9) are shown in number range [0, N] . Example: Input: K=2, N=35 Output: 14 Detail: list of 2 s between [0,35] : 2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32 , note that 22 will be counted as twice (as 22 contains two 2 s) What we have: There are solutions for each of them (available if you search for it). Usually, O(log N)

What is the best method to find the number of digits of a positive integer? I have found this 3 basic methods: conversion to string String s = new Integer(t).toString(); int len = s.length(); for loop for(long long int temp = number; temp >= 1;) { temp/=10; decimalPlaces++; } logaritmic calculation digits = floor( log10( number ) ) + 1; where you can calculate log10(x) = ln(x) / ln(10) in most languages. First I thought the string method is the dirtiest one but the more I think about it the more I think it's the fastest way. Or is it? Mike Dunlavey There's always this method: n = 1; if ( i >=

I need to count the characters of an inputed string in Batch. I don't want to use temporary files. Could it be done without them? If yes, explanations of your code would be greatly appreciated. Thanks SO! A simple way is to use a function @echo off set "myVar=abcdefg" call :Stringlength result myVar echo %result% exit /b :Stringlength <resultVar> <stringVar> ( setlocal EnableDelayedExpansion set "s=!%~2!#" set "len=0" for %%P in (4096 2048 1024 512 256 128 64 32 16 8 4 2 1) do ( if "!s:~%%P,1!" NEQ "" ( set /a "len+=%%P" set "s=!s:~%%P!" ) ) ) ( endlocal set "%~1=%len%" exit /b ) This can

I've been working on this for a couple days now... and still haven't been able to achieve my desired results. Any help would on this would be greatly appreciated... thank you in advance. I have a multi-array stored in $log, which displays like this when you print_r($log): Array ( [0] => Array ( [0] => Red [1] => Steel ) [1] => Array ( [0] => Red [1] => Wood ) [2] => Array ( [0] => Blue [1] => Wood ) ) Currently I have this: $counts = $log; foreach ($log as $value) { foreach ($value as $k => $v) { if (!isset($counts[$k])) $counts[$k] = array(); if (!isset($counts[$k][$v])) $counts[$k][$v] = 0;