Question

所以在我的JS中我有一个像这样的对象;

[
    {
        firstname: "John",
        lastname: "Smith"
    },
    {
        firstname: "Peter",
        lastname: "Gregory"
    },
    {
        firstname: "John",
        lastname: "Fisher"
    },
    {
        firstname: "Sam",
        lastname: "Fisher"
    }
]

我想在逗号分隔的字符串中显示名字。现在这很简单，但我也想在必要时显示姓氏的第一个首字母，以区分具有相同名字的两个人。

所以，最后，我有这个：

John S., Peter, John F., Sam

到目前为止，我已经能够做一个记住过去的首字母的循环，但对我来说问题成了例子中的第四个条目;姓氏必须与其他人区分的人，但不与任何人分享姓名。

最聪明的方法是什么？

Answer 1

Array.map + Array.join会这样做。如果你需要在底部，我可以为Array.map包含一个polyfill。

var people = [
{
    firstname: "John",
    lastname: "Smith"
},
{
    firstname: "Peter",
    lastname: "Gregory"
},
{
    firstname: "John",
    lastname: "Fisher"
},
{
    firstname: "Sam",
    lastname: "Fisher"
}
]

/* Count number of firstNames */
var firstnames = {};
for (var i = 0; i < people.length; i++) {
   if (!firstnames[people[i].firstname]) {
       firstnames[people[i].firstname] = 0;
   }
   firstnames[people[i].firstname] ++;
}

/* Create the string of names */
var peopleString = people.map(function (a) {
   /* Check if we need a last name here */ 
   var lastname = firstnames[a.firstname] > 1 ? (a.lastname ? ' ' + a.lastname.substr(0, 1) + '.' : '') : '';
   
   return a.firstname + lastname; 
}).join(', ');

document.write(peopleString);


/*Polyfill for Array.map taken from MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map#Polyfill*/
// Production steps of ECMA-262, Edition 5, 15.4.4.19
// Reference: http://es5.github.io/#x15.4.4.19
if (!Array.prototype.map) {

  Array.prototype.map = function(callback, thisArg) {

    var T, A, k;

    if (this == null) {
      throw new TypeError(' this is null or not defined');
    }

    // 1. Let O be the result of calling ToObject passing the |this| 
    //    value as the argument.
    var O = Object(this);

    // 2. Let lenValue be the result of calling the Get internal 
    //    method of O with the argument "length".
    // 3. Let len be ToUint32(lenValue).
    var len = O.length >>> 0;

    // 4. If IsCallable(callback) is false, throw a TypeError exception.
    // See: http://es5.github.com/#x9.11
    if (typeof callback !== 'function') {
      throw new TypeError(callback + ' is not a function');
    }

    // 5. If thisArg was supplied, let T be thisArg; else let T be undefined.
    if (arguments.length > 1) {
      T = thisArg;
    }

    // 6. Let A be a new array created as if by the expression new Array(len) 
    //    where Array is the standard built-in constructor with that name and 
    //    len is the value of len.
    A = new Array(len);

    // 7. Let k be 0
    k = 0;

    // 8. Repeat, while k < len
    while (k < len) {

      var kValue, mappedValue;

      // a. Let Pk be ToString(k).
      //   This is implicit for LHS operands of the in operator
      // b. Let kPresent be the result of calling the HasProperty internal 
      //    method of O with argument Pk.
      //   This step can be combined with c
      // c. If kPresent is true, then
      if (k in O) {

        // i. Let kValue be the result of calling the Get internal 
        //    method of O with argument Pk.
        kValue = O[k];

        // ii. Let mappedValue be the result of calling the Call internal 
        //     method of callback with T as the this value and argument 
        //     list containing kValue, k, and O.
        mappedValue = callback.call(T, kValue, k, O);

        // iii. Call the DefineOwnProperty internal method of A with arguments
        // Pk, Property Descriptor
        // { Value: mappedValue,
        //   Writable: true,
        //   Enumerable: true,
        //   Configurable: true },
        // and false.

        // In browsers that support Object.defineProperty, use the following:
        // Object.defineProperty(A, k, {
        //   value: mappedValue,
        //   writable: true,
        //   enumerable: true,
        //   configurable: true
        // });

        // For best browser support, use the following:
        A[k] = mappedValue;
      }
      // d. Increase k by 1.
      k++;
    }

    // 9. return A
    return A;
  };
}

Answer 2

我想你可以这样做：

var o = {};
var arr = [];
people.forEach(function(val) {
    if (!o.hasOwnProperty(val.firstname)) {
       // didn't see this name before,
       // keeping it so we know we saw it
       // and keeping its index, in case we see the same name again
       var idx = arr.push(val.firstname);
        o[val.firstname] = { idx: idx - 1, val: val };

    }
    else {
        // we saw this name, adding the current with a last name
        // also adding last name to the origninal we saved before
        arr.push(val.firstname + " " + val.lastname.substr(0, 1));
        arr[o[val.firstname].idx] = o[val.firstname].val.firstname + " " + o[val.firstname].val.lastname.substr(0, 1);
    }
})
console.log(arr);

Answer 3

如果您想要一个（几乎）内联解决方案......

var arr = [
    {
        firstname: "John",
        lastname: "Smith"
    },
    {
        firstname: "Peter",
        lastname: "Gregory"
    },
    {
        firstname: "John",
        lastname: "Fisher"
    },
    {
        firstname: "Sam",
        lastname: "Fisher"
    }
];

var out = arr.map (function(i){
    return arr.filter(function(e){
        return e.firstname == i.firstname && true;
    }, []).length > 1 ? (i.firstname + " " + i.lastname.charAt(0)) : (i.firstname);
}, []).toString();

console.log(out);

小提琴http://jsfiddle.net/owqc35p3/2/

与其他许多人相比，这并不是那么高效，因为我们每次都在当前的arr地图中循环arr（同样，.toString()可以被.join(" ,");替换）

如有必要，获取姓氏首字母的名字列表

3 个答案: