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Pattern for "column total" of an array of arrays


Pattern for "column total" of an array of arrays

By : user2956787
Date : November 22 2020, 03:03 PM
will help you I am looking for a pattern to get the totals for each n-th item of arrays. Conceptually this would be like getting column totals of a data table. , Just iterate over your arrays:
code :
var dataRows = [
  [10, 11, 12, 13],
  [20, 21, 22, 23],
  [30, 31, 32, 33]
]

var result = [];

for(var i = 0; i < dataRows.length; i++){
    for(var j = 0; j < dataRows[i].length; j++){
        result[j] = (result[j] || 0) + dataRows[i][j];
    } // ^ Add the current value to the proper position of `result`.
}

alert(JSON.stringify(result));


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PHP: searching in the first "column" of an array of arrays

PHP: searching in the first "column" of an array of arrays


By : user2786006
Date : March 29 2020, 07:55 AM
will be helpful for those in need PHP doesn't have a good clean way of doing this that I know of. However, you can do it yourself using a binary-search since the array is already sorted by the values in the first column of its child arrays. Here is the code to accomplish this:
code :
<?php
// Binary Search Taken By greenmr:
// http://php.net/manual/en/function.array-search.php#89413
function array_bsearch( $needle, $haystack, $comparator ) {
    $high = Count( $haystack ) -1;
    $low = 0;

    while ( $high >= $low ){
        $probe = Floor( ( $high + $low ) / 2 );
        $comparison = $comparator( $haystack[$probe], $needle );
        if ( $comparison < 0 ) {
            $low = $probe +1;
        } elseif ( $comparison > 0 ) {
            $high = $probe -1;
        } else {
            return $probe;
        }
    }

  return -1;
}

// Compare the needle the first element/column 
function KeyCompare( $obj, $needle) {
    if ( $obj[0] < $needle ) {
        return -1;
    } elseif ( $obj[0] > $needle ) {
        return 1;
    } else {
        return 0;
    }
}

$arr = array(array(0=>"homer", 1=> 1, 2=> 2, 3=> 3),
        array(0 => "marge", 1=> 2, 2 => 4, 3=> 8),
        array(0 => "bart", 1 => 6, 2 => 2, 3 => 7),
        array(0 => "lisa", 1 => 16, 2 => 20, 3 => 71));

$index = array_bsearch( 'marge', $arr, 'KeyCompare' );

// prints the array containing marge
echo print_r($arr[$index]);
?>    
Merge arrays. Second array as a "column"

Merge arrays. Second array as a "column"


By : Sonia Bailey
Date : March 29 2020, 07:55 AM
To fix the issue you can do Here is an example...
code :
$c = array_map(function ($a, $b) { return "$a','$b"; }, $a, $b);
Convert a pandas "Series of pair arrays" to a "two-column DataFrame"?

Convert a pandas "Series of pair arrays" to a "two-column DataFrame"?


By : ww1
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Assuming that each element in the series is an array of pairs, and each pair is a sequence, this should work:
code :
pair_df = pd.DataFrame(np.vstack(pair_arrays.values), columns=['x','y'])
Sql Query to get total of different "StatusName" column, based on year ("CreatedDate" column)

Sql Query to get total of different "StatusName" column, based on year ("CreatedDate" column)


By : ElDorado
Date : March 29 2020, 07:55 AM
will help you I have following need: , Try This:
code :
Select RS.[StatusName], DATEPART ( datepart , R.[CreatedDate] ) as Y, COUNT(RS.[StatusName]) As Total
FROM Table1 as R
Inner Join Table2 as RS
ON R.StatusCode=RS.StatusName
GROUP BY DATEPART ( datepart , R.[CreatedDate] )
How To multiple Euro values total arrays in ios Swift 5 Like ["£179.95", "£199.95", "£89.95&quo

How To multiple Euro values total arrays in ios Swift 5 Like ["£179.95", "£199.95", "£89.95&quo


By : Ryan Patrick
Date : March 29 2020, 07:55 AM
I hope this helps you . If you're sure that the strings contained in your array always start with a £, you could do this:
code :
let sum = array.compactMap { Double($0.replacingOccurrences(of: "£", with: "")) }
               .reduce(0.0, { $0 + $1 })
let array = ["£179.95", "£199.95", "£89.95"]
let sum = array.compactMap { Double($0.replacingOccurrences(of: "£", with: "")) }
               .reduce(0.0, { $0 + $1 })
print(sum) // 469.84999999999997
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