It can be done simply manually: $arr = array(1,1,1,2,2,3,3,1,1,2,2,3); $result = array(); $prev_value = array(‘value’ => null, ‘amount’ => null); foreach ($arr as $val) { if ($prev_value[‘value’] != $val) { unset($prev_value); $prev_value = array(‘value’ => $val, ‘amount’ => 0); $result[] =& $prev_value; } $prev_value[‘amount’]++; } var_dump($result);
Just iterate over the array and use another array for the groups. It should be fast enough and is probably faster than the overhead involved when using sqlite or similar. $groups = array(); foreach ($data as $item) { $key = $item[‘key_to_group’]; if (!isset($groups[$key])) { $groups[$key] = array( ‘items’ => array($item), ‘count’ => 1, ); } … Read more
PHP has no support for a SQL where sort of thing, especially not with an array of arrays. But you can do the counting your own while you iterate over the data: $count = array(); foreach($fruit as $one) { @$count[$one[‘color’]]++; } printf(“You have %d green fruit(s).\n”, $count[‘green’]); The alternative is to write yourself some little … Read more
Just loop trough the rows, and add up the quantities in a different array indexed by the dd values. $in = array(array()); // your input $out = array(); foreach ($in as $row) { if (!isset($out[$row[‘dd’]])) { $out[$row[‘dd’]] = array( ‘dd’ => $row[‘dd’], ‘quantity’ => 0, ); } $out[$row[‘dd’]][‘quantity’] += $row[‘quantity’]; } $out = array_values($out); // … Read more
You can use array_count_values function $array = array(‘apple’, ‘orange’, ‘pear’, ‘banana’, ‘apple’, ‘pear’, ‘kiwi’, ‘kiwi’, ‘kiwi’); print_r(array_count_values($array)); will output Array ( [apple] => 2 [orange] => 1 [pear] => 2 etc… )
Use array_key_exists & increment the count, $newArray = array(); foreach ($the_outer_array as $key=>$value) { $firstValue = $value[0]; if ($foundKey = array_key_exists($firstValue,$newArray)) { $newArray[$firstValue] += 1; } else{ $newArray[$firstValue] = 1; } } DEMO.
Python 3.x does have reduce, you just have to do a from functools import reduce. It also has “dict comprehensions”, which have exactly the syntax in your example. Python 2.7 and 3.x also have a Counter class which does exactly what you want: from collections import Counter cnt = Counter(“abracadabra”) In Python 2.6 or earlier, … Read more
True is equal to 1. >>> sum([True, True, False, False, False, True]) 3
Here’s another option: s=”some string” numbers = sum(c.isdigit() for c in s) letters = sum(c.isalpha() for c in s) spaces = sum(c.isspace() for c in s) others = len(s) – numbers – letters – spaces