我会这样做:

function renameKeys(dict, keyMap) {
  return _.reduce(dict, function(newDict, val, oldKey) {
    var newKey = keyMap[oldKey] || oldKey
    newDict[newKey] = val 
    return newDict
  }, {})
}

在寻找了很多答案后，这是我最好的解决方案:

const renameKey = (oldKey, newKey) => {
  _.reduce(obj, (newObj, value, key) => {
    newObj[oldKey === key ? newKey : key] = value
    return newObj
  }, {})
}

很明显，它没有替换原来的键，而是构造了一个新对象。 问题中的方法有效，但会改变对象的顺序，因为它将新的键-值添加到最后一个对象上。

const data = res
const lista = []
let newElement: any

if (data && data.length > 0) {

  data.forEach(element => {
      newElement = element

      Object.entries(newElement).map(([key, value]) =>
        Object.assign(newElement, {
          [key.toLowerCase()]: value
        }, delete newElement[key], delete newElement['_id'])
      )
    lista.push(newElement)
  })
}
return lista

如果有人需要重命名属性列表:

function renameKeys(obj, newKeys) {
  const keyValues = Object.keys(obj).map(key => {
    const newKey = newKeys[key] || key;
    return { [newKey]: obj[key] };
  });
  return Object.assign({}, ...keyValues);
}

用法:

const obj = { a: "1", b: "2" };
const newKeys = { a: "A", c: "C" };
const renamedObj = renameKeys(obj, newKeys);
console.log(renamedObj);
// {A:"1", b:"2"}

下面是创建具有重命名键的新对象的示例。

let x = { id: "checkout", name: "git checkout", description: "checkout repository" };

let renamed = Object.entries(x).reduce((u, [n, v]) => {
  u[`__${n}`] = v;
  return u;
}, {});

使用对象解构和展开运算符的变体:

const old_obj = {
    k1: `111`,
    k2: `222`,
    k3: `333`
};    

// destructuring, with renaming. The variable 'rest' will hold those values not assigned to kA, kB, or kC.
const {
    k1: kA, 
    k2: kB, 
    k3: kC,
    ...rest
} = old_obj;
    

// now create a new object, with the renamed properties kA, kB, kC; 
// spread the remaining original properties in the 'rest' variable
const newObj = {kA, kB, kC, ...rest};

对于一个键，这可以很简单:

const { k1: kA, ...rest } = old_obj;
const new_obj = { kA, ...rest }

你也可能喜欢更“传统”的风格:

const { k1, ...rest } = old_obj
const new_obj = { kA: k1, ...rest}