如何将第一个数组中第二个数组缺失的元素补充到第二个数组中？

你提供的数组输出格式为PHP的print_r打印结果，以下优先提供PHP实现，同时附其他常用语言的实现方案：

核心逻辑 #

• 计算第一个数组和第二个数组的差集（即只在第一个数组中存在的元素）
• 将差集元素全部追加到第二个数组中即可

PHP实现 #

第一步：定义两个测试数组 #

$firstArr = [
    '21-02',
    '21-01',
    '21-03',
    '21-04',
    '21-05',
    '21-06',
    '21-07',
    '21-08',
    '21-09'
];

$secondArr = [
    '21-01',
    '21-02'
];

第二步：计算差集并合并 #

调用PHP内置数组函数即可一行完成，无需手动循环判断：

// 取firstArr有、secondArr没有的元素
$missingItems = array_diff($firstArr, $secondArr);
// 合并到secondArr中
$secondArr = array_merge($secondArr, $missingItems);

最终$secondArr结果 #

Array
(
    [0] => 21-01
    [1] => 21-02
    [2] => 21-03
    [3] => 21-04
    [4] => 21-05
    [5] => 21-06
    [6] => 21-07
    [7] => 21-08
    [8] => 21-09
)

其他语言实现参考 #

JavaScript #

const firstArr = ['21-02','21-01','21-03','21-04','21-05','21-06','21-07','21-08','21-09']
const secondArr = ['21-01','21-02']

// 筛选缺失元素
const missingItems = firstArr.filter(item => !secondArr.includes(item))
// 追加到第二个数组
secondArr.push(...missingItems)

Python #

first_arr = ['21-02','21-01','21-03','21-04','21-05','21-06','21-07','21-08','21-09']
second_arr = ['21-01','21-02']

# 计算差集并追加
missing_items = [item for item in first_arr if item not in second_arr]
second_arr.extend(missing_items)

内容的提问来源于stack exchange，提问作者Rob