![](/img/trans.png)
[英]Group rows of data by column value then store nested data, first and last occurrences, and counts within each group
[英]Group rows of data, maintain a subarray of ids within the group, and only present the lowest id in each group as the first level key
我需要將一組行合並到組中,並使用每個組中的最低 id 作為第一級鍵。 在每個組中,所有遇到的 id(不包括最低的)都應收集在一個名為mergedWith
的子數組中。
樣本輸入:
[
1649 => ["firstName" => "jack", "lastName" => "straw"],
1650 => ["firstName" => "jack", "lastName" => "straw"],
1651 => ["firstName" => "jack", "lastName" => "straw"],
1652 => ["firstName" => "jack", "lastName" => "straw"],
]
我想要的結果:
[
1649 => [
"firstName" => "jack"
"lastName" => "straw"
"mergedWith" => [1650, 1651, 1652]
]
]
我有一個循環運行,可以提取重復項並找到組中最低的 ID,但不確定將它們合並為一個的正確方法。
我已經顯示了搜索的所需結果,該搜索已在這些特定字段中識別出具有重復條目的 id。 我只是想進一步完善它以不刪除,而是在每個組的末尾添加一個字段,上面寫着["mergedWith" => [1650, 1651, 1652]]
一種方法是按名字和姓氏分組,然后反轉分組以獲得單個ID。 事先krsort
輸入,以確保您獲得最低的ID。
krsort($input);
//group
foreach ($input as $id => $person) {
// overwrite the id each time, but since the input is sorted by id in descending order,
// the last one will be the lowest id
$names[$person['lastName']][$person['firstName']] = $id;
}
// ungroup to get the result
foreach ($names as $lastName => $firstNames) {
foreach ($firstNames as $firstName => $id) {
$result[$id] = ['firstName' => $firstName, 'lastName' => $lastName];
}
}
編輯:根據您更新的問題,沒有太多不同。 只保留所有ID而不是單個ID。
krsort($input);
foreach ($input as $id => $person) {
// append instead of overwrite ↓
$names[$person['lastName']][$person['firstName']][] = $id;
}
foreach ($names as $lastName => $firstNames) {
foreach ($firstNames as $firstName => $ids) {
// $ids is already in descending order based on the initial krsort
$id = array_pop($ids); // removes the last (lowest) id and returns it
$result[$id] = [
'firstName' => $firstName,
'lastName' => $lastName,
'merged_with' => implode(',', $ids)
];
}
}
ksort($resArr);
$tempArr = array_unique($resArr, SORT_REGULAR);
foreach ($tempArr as $key => $value) {
foreach ($resArr as $key1 => $value2) {
if($value['firstName'] == $value2['firstName'] && $value['lastName'] == $value2['lastName']) {
$tempArr[$key]["mergedWith"][] = $key1;
}
}
}
print_r($tempArr);
$resArr = array(1650 => array(
"firstName" => "jack",
"lastName" => "straw"
),1649 => array(
"firstName" => "jack",
"lastName" => "straw"
)
,
1651 => array(
"firstName" => "jack",
"lastName" => "straw"
),
1652 => array(
"firstName" => "jack",
"lastName" => "straw"
),
1653 => array(
"firstName" => "jack1",
"lastName" => "straw"
),
1654 => array(
"firstName" => "jack1",
"lastName" => "straw"
));
Output
Array
(
[1649] => Array
(
[firstName] => jack
[lastName] => straw
[mergedWith] => Array
(
[0] => 1649
[1] => 1650
[2] => 1651
[3] => 1652
)
)
[1653] => Array
(
[firstName] => jack1
[lastName] => straw
[mergedWith] => Array
(
[0] => 1653
[1] => 1654
)
)
)
@Don'tPanic 的答案是使用初步循環來創建查找數組,然后使用嵌套循環來形成所需的結果。
我推薦一種沒有嵌套循環的更簡單的方法。 在第一個循環中,過度填充每組中的mergedWith
元素——這將非常快,因為沒有 function 調用並且沒有條件(除了 null 合並賦值運算符??=
)。 然后使用第二個循環從mergedWith
子數組中提取第一個元素——這將應用最低的 id 作為第一級鍵,並確保第一級鍵不再存在於組的子數組中。
代碼:(演示)
ksort($array);
$temp = [];
foreach ($array as $key => $row) {
$compositeKey = $row['firstName'] . '-' . $row['firstName'];
$temp[$compositeKey] ??= $row;
$temp[$compositeKey]['mergedWith'][] = $key;
}
$result = [];
foreach ($temp as $row) {
$result[array_shift($row['mergedWith'])] = $row;
}
var_export($result);
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.