我想从数组中的每个对象中删除坏属性。有没有比使用for循环并从每个对象中删除它更好的方法呢?
var array = [{"bad": "something", "good":"something"},{"bad":"something", "good":"something"},...];
for (var i = 0, len = array.length; i < len; i++) {
delete array[i].bad;
}
似乎应该有一种方法来使用原型之类的。我不知道。想法吗?
当前回答
这个问题现在有点老了，但我想提供一个替代解决方案，它不会改变源数据，并且只需要最少的手工工作:
function mapOut(sourceObject, removeKeys = []) {
const sourceKeys = Object.keys(sourceObject);
const returnKeys = sourceKeys.filter(k => !removeKeys.includes(k));
let returnObject = {};
returnKeys.forEach(k => {
returnObject[k] = sourceObject[k];
});
return returnObject;
}
const array = [
{"bad": "something", "good":"something"},
{"bad":"something", "good":"something"},
];
const newArray = array.map(obj => mapOut(obj, [ "bad", ]));
它仍然不是完美的，但是保持了某种程度的不可变性，并且可以灵活地命名您想要删除的多个属性。(建议欢迎)
其他回答
在我看来，这是最简单的变体
array.map(({good}) => ({good}))
ES6:
const newArray = array.map(({keepAttr1, keepAttr2}) => ({keepAttr1, newPropName: keepAttr2}))
通过减少:
const newArray = oldArray.reduce((acc, curr) => {
const { remove_one, remove_two, ...keep_data } = curr;
acc.push(keep_data);
return acc;
}, []);
这对我来说很有效!
export function removePropertiesFromArrayOfObjects(arr = [], properties = []) {
return arr.map(i => {
const newItem = {}
Object.keys(i).map(key => {
if (properties.includes(key)) { newItem[key] = i[key] }
})
return newItem
})
}
如果你使用的是underscore.JS库:
let asdf = [{"asd": 12, "asdf": 123}, {"asd": 121, "asdf": 1231}, {"asd": 142, "asdf": 1243}]
_.map(asdf, function (row) {
return _.omit(row, ['asd'])
})
