Question

我需要实现一个给定多个对象数组的函数：

var list = [
  {id: "102", name: "Alice"},
  {id: "205", name: "Bob", title: "Dr."},
  {id: "592", name: "Clyde", age: 32}];

必须返回对象属性回调函数要求的内容：

{
  "3": [{id: "205", name: "Bob", title: "Dr."}],
  "5": [{id: "102", name: "Alice"},
        {id: "592", name: "Clyde", age: 32}]
 }

groupBy(list, function(i) { return i.name.length; });

例如：这个回调要求返回每个对象的name.length加上原始对象作为其值，如果它们相同，则将它返回到同一个数组中。

＆＃13;

var list = [
  {id: "102", name: "Alice"},
  {id: "205", name: "Bob", title: "Dr."},
  {id: "592", name: "Clyde", age: 32}];

function groupBy(arr, cb) {
  var newObj = {};
  var newArr = [];
  
  for (key of arr) {
    newObj = cb(key);
  }
  return newObj;
}

groupBy(list, function(i) { return i.id; });

＆＃13;

我没有太多自己的解决方案，而且我已经撞墙了。如果你们/ gals会帮助我，我会非常感激。

干杯

Answer 1

您需要为每个新密钥创建一个空数组，您想要分组。

将实际对象推送到此数组。

function groupBy(array, cb) {
    var result = {},
        key;
  
    for (object of array) {
        key = cb(object);
        result[key] = result[key] || [];
        result[key].push(object);
    }
    return result;
}

var list = [{ id: "102", name: "Alice" }, { id: "205", name: "Bob", title: "Dr." }, { id: "592", name: "Clyde", age: 32 }];

console.log(groupBy(list, function(o) { return o.name.length; }));
console.log(groupBy(list, function(o) { return o.id; }));

.as-console-wrapper { max-height: 100% !important; top: 0; }

通过传递回调函数返回新对象

1 个答案: