如何通过关联键添加所有列值?请注意, [gozhi]
键是动态。
输入数组:
Array
(
[0] => Array
(
[gozhi] => 2
[uzorong] => 1
[ngangla] => 4
[langthel] => 5
)
[1] => Array
(
[gozhi] => 5
[uzorong] => 0
[ngangla] => 3
[langthel] => 2
)
[2] => Array
(
[gozhi] => 3
[uzorong] => 0
[ngangla] => 1
[langthel] => 3
)
)
期望的结果:
Array
(
[gozhi] => 10
[uzorong] => 1
[ngangla] => 8
[langthel] => 10
)
答案 0 :(得分:159)
您可以使用array_walk_recursive()
获取问题的一般案例解决方案(每个内部数组可能具有唯一键时的解决方案。)
$final = array();
array_walk_recursive($input, function($item, $key) use (&$final){
$final[$key] = isset($final[$key]) ? $item + $final[$key] : $item;
});
Example with array_walk_recursive()
for the general case
此外,由于 PHP 5.5 ,您可以使用array_column()
功能来获得您希望获得确切密钥的结果, {{1} } ,例如:
[gozhi]
Example with array_column()
for the specified key
如果您想使用相同的键(您已发布的所需结果)获取所有内部数组的总和,您可以执行以下操作(轴承请记住,第一个内部数组必须与其他内部数组具有相同的结构):
array_sum(array_column($input, 'gozhi'));
Example with array_column()
in case all inner arrays have the same keys
如果您想要使用array_column()
的一般案例解决方案,那么首先您可以考虑获取所有唯一密钥,然后获取每个密钥的总和:
$final = array_shift($input);
foreach ($final as $key => &$value){
$value += array_sum(array_column($input, $key));
}
unset($value);
答案 1 :(得分:84)
$sumArray = array();
foreach ($myArray as $k=>$subArray) {
foreach ($subArray as $id=>$value) {
$sumArray[$id]+=$value;
}
}
print_r($sumArray);
答案 2 :(得分:27)
这是一个类似于其他两个的解决方案:
$acc = array_shift($arr);
foreach ($arr as $val) {
foreach ($val as $key => $val) {
$acc[$key] += $val;
}
}
但是这不需要检查数组键是否已经存在并且也不会抛出通知。
答案 3 :(得分:21)
也可以使用array_map
:
$rArray = array(
0 => array(
'gozhi' => 2,
'uzorong' => 1,
'ngangla' => 4,
'langthel' => 5
),
1 => array(
'gozhi' => 5,
'uzorong' => 0,
'ngangla' => 3,
'langthel' => 2
),
2 => array(
'gozhi' => 3,
'uzorong' => 0,
'ngangla' => 1,
'langthel' => 3
),
);
$sumResult = call_user_func_array('array_map', array_merge(['sum'], $rArray));
function sum()
{
return array_sum(func_get_args());
}
答案 4 :(得分:12)
$newarr=array();
foreach($arrs as $value)
{
foreach($value as $key=>$secondValue)
{
if(!isset($newarr[$key]))
{
$newarr[$key]=0;
}
$newarr[$key]+=$secondValue;
}
}
答案 5 :(得分:11)
使用此代码段:
$key = 'gozhi';
$sum = array_sum(array_column($array,$key));
答案 6 :(得分:5)
另一个版本,下面有一些好处。
$sum = ArrayHelper::copyKeys($arr[0]);
foreach ($arr as $item) {
ArrayHelper::addArrays($sum, $item);
}
class ArrayHelper {
public function addArrays(Array &$to, Array $from) {
foreach ($from as $key=>$value) {
$to[$key] += $value;
}
}
public function copyKeys(Array $from, $init=0) {
return array_fill_keys(array_keys($from), $init);
}
}
我想将Gumbo,Graviton和Chris J的最佳答案与以下目标结合起来,以便我可以在我的应用中使用它:
a)初始化循环外的'sum'数组键(Gumbo)。应该有助于在非常大的阵列上进行性能测试(尚未测试!)。消除通知。
b)主要逻辑易于理解而无需遵守手册。 (Graviton,Chris J)。
c)解决使用相同键添加任意两个数组的值的更一般问题,并使其更少依赖于子数组结构。
与Gumbo的解决方案不同,您可以在值不在子数组中的情况下重复使用它。想象一下,在下面的示例中,$arr1
和$arr2
不是硬编码的,而是作为在循环内调用函数的结果而返回。
$arr1 = array(
'gozhi' => 2,
'uzorong' => 1,
'ngangla' => 4,
'langthel' => 5
);
$arr2 = array(
'gozhi' => 5,
'uzorong' => 0,
'ngangla' => 3,
'langthel' => 2
);
$sum = ArrayHelper::copyKeys($arr1);
ArrayHelper::addArrays($sum, $arr1);
ArrayHelper::addArrays($sum, $arr2);
答案 7 :(得分:4)
也可以使用array_walk
:
function array_sum_values(array $input, $key) {
$sum = 0;
array_walk($input, function($item, $index, $params) {
if (!empty($item[$params[1]]))
$params[0] += $item[$params[1]];
}, array(&$sum, $key)
);
return $sum;
}
var_dump(array_sum_values($arr, 'gozhi'));
不像以前的解决方案那样可读,但它有效:)
答案 8 :(得分:3)
这是一个版本,两个数组的数组键可能不一样,但是你希望它们都在最终的数组中。
function array_add_by_key( $array1, $array2 ) {
foreach ( $array2 as $k => $a ) {
if ( array_key_exists( $k, $array1 ) ) {
$array1[$k] += $a;
} else {
$array1[$k] = $a;
}
}
return $array1;
}
答案 9 :(得分:2)
我们需要先检查数组键是否存在。
<强> CODE:强>
$sum = array();
foreach ($array as $key => $sub_array) {
foreach ($sub_array as $sub_key => $value) {
//If array key doesn't exists then create and initize first before we add a value.
//Without this we will have an Undefined index error.
if( ! array_key_exists($sub_key, $sum)) $sum[$sub_key] = 0;
//Add Value
$sum[$sub_key]+=$value;
}
}
print_r($sum);
带阵列密钥验证的输出:
Array
(
[gozhi] => 10
[uzorong] => 1
[ngangla] => 8
[langthel] => 10
)
OUTPUT没有数组密钥验证:
Notice: Undefined index: gozhi in F:\web\index.php on line 37
Notice: Undefined index: uzorong in F:\web\index.php on line 37
Notice: Undefined index: ngangla in F:\web\index.php on line 37
Notice: Undefined index: langthel in F:\web\index.php on line 37
Array
(
[gozhi] => 10
[uzorong] => 1
[ngangla] => 8
[langthel] => 10
)
虽然打印输出,但这是一种不好的做法。如果密钥存在,请务必先检查。
答案 10 :(得分:1)
对于那些登陆这里并正在寻找合并 N个数组的解决方案的人来说,并且还总结了在N个数组中找到的相同键的值,我已经编写了这个递归工作的函数。 (见:https://gist.github.com/Nickology/f700e319cbafab5eaedc)
示例:
$a = array( "A" => "bob", "sum" => 10, "C" => array("x","y","z" => 50) );
$b = array( "A" => "max", "sum" => 12, "C" => array("x","y","z" => 45) );
$c = array( "A" => "tom", "sum" => 8, "C" => array("x","y","z" => 50, "w" => 1) );
print_r(array_merge_recursive_numeric($a,$b,$c));
将导致:
Array
(
[A] => tom
[sum] => 30
[C] => Array
(
[0] => x
[1] => y
[z] => 145
[w] => 1
)
)
以下是代码:
<?php
/**
* array_merge_recursive_numeric function. Merges N arrays into one array AND sums the values of identical keys.
* WARNING: If keys have values of different types, the latter values replace the previous ones.
*
* Source: https://gist.github.com/Nickology/f700e319cbafab5eaedc
* @params N arrays (all parameters must be arrays)
* @author Nick Jouannem <nick@nickology.com>
* @access public
* @return void
*/
function array_merge_recursive_numeric() {
// Gather all arrays
$arrays = func_get_args();
// If there's only one array, it's already merged
if (count($arrays)==1) {
return $arrays[0];
}
// Remove any items in $arrays that are NOT arrays
foreach($arrays as $key => $array) {
if (!is_array($array)) {
unset($arrays[$key]);
}
}
// We start by setting the first array as our final array.
// We will merge all other arrays with this one.
$final = array_shift($arrays);
foreach($arrays as $b) {
foreach($final as $key => $value) {
// If $key does not exist in $b, then it is unique and can be safely merged
if (!isset($b[$key])) {
$final[$key] = $value;
} else {
// If $key is present in $b, then we need to merge and sum numeric values in both
if ( is_numeric($value) && is_numeric($b[$key]) ) {
// If both values for these keys are numeric, we sum them
$final[$key] = $value + $b[$key];
} else if (is_array($value) && is_array($b[$key])) {
// If both values are arrays, we recursively call ourself
$final[$key] = array_merge_recursive_numeric($value, $b[$key]);
} else {
// If both keys exist but differ in type, then we cannot merge them.
// In this scenario, we will $b's value for $key is used
$final[$key] = $b[$key];
}
}
}
// Finally, we need to merge any keys that exist only in $b
foreach($b as $key => $value) {
if (!isset($final[$key])) {
$final[$key] = $value;
}
}
}
return $final;
}
?>
答案 11 :(得分:0)
代码在这里:
$temp_arr = [];
foreach ($a as $k => $v) {
if(!is_null($v)) {
$sum = isset($temp_arr[$v[0]]) ? ((int)$v[5] + $sum) : (int)$v[5];
$temp_arr[$v[0]] = $sum;
}
}
return $temp_arr;
结果:
{SEQ_OK: 1328,SEQ_ERROR: 561}
答案 12 :(得分:0)
遍历数组的每个项目,如果不存在赋值,则将它们的值加和到先前的值(如果存在)。
<?php
$array =
[
[
'a'=>1,
'b'=>1,
'c'=>1,
],
[
'a'=>2,
'b'=>2,
],
[
'a'=>3,
'd'=>3,
]
];
$result = array_reduce($array, function($carry, $item) {
foreach($item as $k => $v)
$carry[$k] = isset($carry[$k]) ? $carry[$k] + $v : $v;
return $carry;
}, []);
print_r($result);
输出:
Array
(
[a] => 6
[b] => 3
[c] => 1
[d] => 3
)
答案 13 :(得分:0)
这对我的laravel项目非常有用
print_r($Array); // your original array
$_SUM = [];
// count($Array[0]) => if the number of keys are equall in all arrays then do a count of index 0 etc.
for ($i=0; $i < count($Array[0]); $i++) {
$_SUM[] = $Array[0][$i] + $Array[1][$i]; // do a for loop on the count
}
print_r($_SUM); // get a sumed up array
答案 14 :(得分:0)
在这里,您通常会采用这种操作。
// We declare an empty array in wich we will store the results
$sumArray = array();
// We loop through all the key-value pairs in $myArray
foreach ($myArray as $k=>$subArray) {
// Each value is an array, we loop through it
foreach ($subArray as $id=>$value) {
// If $sumArray has not $id as key we initialize it to zero
if(!isset($sumArray[$id])){
$sumArray[$id] = 0;
}
// If the array already has a key named $id, we increment its value
$sumArray[$id]+=$value;
}
}
print_r($sumArray);
答案 15 :(得分:0)
你可以试试这个:
$c = array_map(function () {
return array_sum(func_get_args());
},$a, $b);
最后:
print_r($c);
答案 16 :(得分:-1)
let firstName = person.firstName ?? ""
答案 17 :(得分:-1)
$sumArray = array();
foreach ($myArray as $k => $subArray) {
foreach ($subArray as $id => $value) {
if (!isset($sumArray[$id])) {
$sumArray[$id] = 0;
}
$sumArray[$id]+=$value;
}
}
答案 18 :(得分:-1)
例如,您可以从下面的结果中选择所有字段。
我正在从数组中选择'balance'并保存到变量
$kii = $user->pluck('balance');
然后在下一行,您可以这样总结:
$sum = $kii->sum();
希望有帮助。