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
通过减少:
const newArray = oldArray.reduce((acc, curr) => {
const { remove_one, remove_two, ...keep_data } = curr;
acc.push(keep_data);
return acc;
}, []);
其他回答
在我看来，这是最简单的变体
array.map(({good}) => ({good}))
ES6中最短的方法:
array.forEach(e => {delete e.someKey});
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;
}, []);
如果你使用underscore.js:
var strippedRows = _.map(rows, function (row) {
return _.omit(row, ['bad', 'anotherbad']);
});
