从数组中的所有对象中删除键 [英] Remove key from all objects in array

问题描述

我有以下对象数组:

[{id:1, value:"100", name:"dog" ...},
{id:2, value:"200", name:"cat" ...},
{id:3, value:"300", name:"fish"....},
{id:4, value:"400", name:"mouse" ...},
{id:5, value:"500", name:"snake"...}]

我想过滤对象数组并只保留两个键，id 和 value 以获得如下内容:

I want to filter the object array and keep only two keys, id and value to get something like this:

[{id:1, value:"100"},
{id:2, value:"200"},
{id:3, value:"300"},
{id:4, value:"400"},
{id:5, value:"500"}]

目前，我正在使用 for 循环遍历对象数组，并对带有新变量的空数组执行 push() 操作.有没有更简单的方法来做到这一点?

Currently, I'm traversing through the object array with a for loop and doing a push() to an empty array with the new variables. Is there an easier way to do this?

我想像 _.pluck(PetList, 'id', 'value'); 一样使用 Lodash，但 lodash 只提取值，而不是键.

I wanted to use Lodash like _.pluck(PetList, 'id', 'value'); but lodash extracts the value only, not the key.

推荐答案

Array#filter 从数组中过滤单个项目，而不是数组中某个对象的某些键.您可以使用方法 Array#map 转换对象，只保留你想要的键.map 旨在将数组的每个元素转换为新值，将旧值映射"为新值:

Array#filter filters individual items out of the array, not certain keys from an object in the array. You could use the method Array#map to transform the objects and only keeping the keys you want. map is designed to transform each element of an array to something new, "mapping" the old value to a new value:

let newPetList = PetList.map(pet => ({ 
    id: pet.id,
    value: pet.value
}));

以上遍历数组，将当前对象存入pet.然后它从 箭头函数 这将是 newPetList 中相应的转换对象，只有来自 pet 的键 id 和 value.结果是旧数组中的所有对象都映射到一个没有name键的新对象.

The above traverses the array, and stores the current object in pet. It then returns a new object from an arrow function which will be the corresponding transformed object in newPetList with only keys id and value from pet. The result is that all objects in the old array are mapped to a new object with no name key.

您还可以通过对象解构过滤掉不需要的属性，如下所示:

You could also, with object destructuring, filter put unwanted properties like so:

let newPetList = petList.map(({ name, ...rest }) => rest);

这将 name 属性绑定到 name 并将其余部分保留在 rest 中，您可以返回以删除 name 键.

This binds the name property to name and keeps the rest in rest, which you can return to remove the name key.

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