如何将数组转换为JavaScript对象数组。
例如我有一个数组
data = [
["fruits","frozen","fresh","rotten"],
["apples",884,494,494],
["oranges",4848,494,4949],
["kiwi",848,33,33]
]
我想将其转换为名称值对。
例如,结果集合中的第一个对象是
{"fruits": "apple", "frozen": 884, "fresh": 494, "rotten": 494}
以及其他数据。
答案 0 :(得分:8)
使用您提供的数据:
var data = [
["fruits","frozen","fresh","rotten"],
["apples",884,494,494],
["oranges",4848,494,4949],
["kiwi",848,33,33]
]
以下函数将数组的第一个元素视为对象属性的键。然后它将遍历其余元素,并使用这些键将它们转换为对象。最后,它将返回这些新对象的数组。
function convertToArrayOfObjects(data) {
var keys = data.shift(),
i = 0, k = 0,
obj = null,
output = [];
for (i = 0; i < data.length; i++) {
obj = {};
for (k = 0; k < keys.length; k++) {
obj[keys[k]] = data[i][k];
}
output.push(obj);
}
return output;
}
[
{ fruits: 'apples', fresh: 494, frozen: 884, rotten: 494 },
{ fruits: 'oranges', fresh: 494, frozen: 4848, rotten: 4949 },
{ fruits: 'kiwi', fresh: 33, frozen: 848, rotten: 33 }
]
答案 1 :(得分:7)
如果同时使用map和reduce,解决方案将变得非常优雅:
var collection = [
["fruits","frozen","fresh","rotten"],
["apples",884,494,494],
["oranges",4848,494,4949],
["kiwi",848,33,33]
];
var keys = collection.shift();
collection = collection.map(function (row) {
return keys.reduce(function (obj, key, i) {
obj[key] = row[i];
return obj;
}, {});
});
输出:
[
{ fruits: 'apples', frozen: 884, fresh: 494, rotten: 494 },
{ fruits: 'oranges', frozen: 4848, fresh: 494, rotten: 4949 },
{ fruits: 'kiwi', frozen: 848, fresh: 33, rotten: 33 }
]
答案 2 :(得分:6)
可以避免使用shift()
+ map()
+ forEach()
进行两次迭代:
var data = [
["fruits","frozen","fresh","rotten"],
["apples",884,494,494],
["oranges",4848,494,4949],
["kiwi",848,33,33]
];
var collection = data.slice(); // make a copy
var keys = collection.shift();
collection = collection.map(function (e) {
var obj = {};
keys.forEach(function (key, i) {
obj[key] = e[i];
});
return obj;
});
答案 3 :(得分:1)
破坏:
const data = [
["fruits","frozen","fresh","rotten"],
["apples",884,494,494],
["oranges",4848,494,4949],
["kiwi",848,33,33]
]
const titles = data[0]
const obj = data.slice(1).map(([fruits,frozen,fresh,rotten]) => ({ fruits,frozen,fresh,rotten }) )
console.log(obj)