Question

我有两个长度相同的数组：

a = [12,21,21,13,13,13,13,31];
b = [4,7,4,6,2,9,4,1];

第一个数组是键，第二个数组是值，但如果重复键，则应将值分组到相应的键中而不是替换。该对象应如下所示：

o = {
    12: [4],
    21: [7,4],
    13: [6,2,9,4],
    31: [1]
}

这就是我的尝试：

var o = {};
for ( var index in a) {
    o[a[index]] = [];
    o[a[index]].push(b[index]);
}

Answer 1

试试这个：

var o = {};
for (var i = 0; i < b.length; i++) {
    var key = a[i] + '';
    if (key in o) {
        o[key].push(b[i]);
    }
    else {
        o[key] = [b[i]];
    }
}

Answer 2

循环中的第一行是破坏该插槽中的任何现有数组。如果还没有新数组，请尝试声明一个新数组：

var o = {};
for (var index = 0; index < a.length; index++) {
    if(o[a[index]] == undefined) {
        o[a[index]] = [];
    }
    o[a[index]].push(b[index]);
}

Answer 3

不要使用for..in循环遍历数组（除非它们是稀疏数组，你知道你正在做什么; details）。

除此之外，你是在正确的轨道上，但你必须在覆盖它之前检查数组是否已经存在。所以：

var o = {}, key, entry;
for (index = 0; index < a.length; ++index) {
    // Get the key
    key = a[index];

    // Get the entry's array if it already exists
    entry = o[key];
    if (!entry) {
        // It doesn't exist, create it and remember it in the object
        o[key] = entry = [];
    }

    // Put this value in it
    entry.push(b[index]);
}

或者几个小优化：

var o = {}, key, entry, len;
for (index = 0, len = a.length; index < len; ++index) {
    // Get the key
    key = a[index];

    // Get the entry's array if it already exists
    entry = o[key];
    if (!entry) {
        // It doesn't exist, create it and remember it in the object,
        // including this value as we go
        o[key] = [b[index]];
    }
    else {
        // Already existed, add this value to it
        entry.push(b[index]);
    }
}

如果您使用的是启用了ES5的环境（或者包含ES5垫片），则可以使用forEach：

var o = {};
a.forEach(function(key, index) {
    var entry;

    // Get the entry's array if it already exists
    entry = o[key];
    if (!entry) {
        // It doesn't exist, create it and remember it in the object,
        // including this value as we go
        o[key] = [b[index]];
    }
    else {
        // Already existed, add this value to it
        entry.push(b[index]);
    }
});

Answer 4

以下是您可以在phpjs.com中使用此功能的功能

function array_combine (keys, values) {
    // Creates an array by using the elements of the first parameter as keys and the elements of the second as the corresponding values  
    var new_array = {},        keycount = keys && keys.length,
        i = 0;

    // input sanitation
    if (typeof keys !== 'object' || typeof values !== 'object' || // Only accept arrays or array-like objects    typeof keycount !== 'number' || typeof values.length !== 'number' || !keycount) { // Require arrays to have a count
        return false;
    }

    // number of elements does not match    if (keycount != values.length) {
        return false;
    }

    for (i = 0; i < keycount; i++) {        new_array[keys[i]] = values[i];
    }

    return new_array;
}

使用重复键在对象中连接2个数组

4 个答案: