Question

大家好，我对javascript来说还很陌生，所以我寻求您的帮助。我的问题如下：

如果有类似数组

     var arr1 = ["a", "b", "a" "c"];

我想找到每个“ a”并将其替换为另一个数组的某个元素

     var arr2 = [1, 2, 3, 4, 5, 6,];

现在如何用arr2的第一个元素替换每个“ a”？每个“ b”都有第二个，依此类推。

这有可能吗？保持位置吗？

Answer 1

您可以使用一个对象将每个字符与特定的整数关联：obj：

const obj = {a:1, b:2, c:3, d:4, e:5, f:6};

然后使用.map方法与辅助对象arr1重新映射数组obj：

arr1.map(e => obj[e])

.map方法创建一个新数组，其结果是在调用数组中的每个元素上调用提供的函数。

const arr1 = ["a", "b", "a", "c"];

const obj = {a:1, b:2, c:3, d:4, e:5, f:6};

console.log(
  arr1.map(e => obj[e])
)

这样，您可以通过更改obj来更改字符和数字之间的关系。您甚至可以将变量设置为obj的值，例如：

const arr1 = ["a", "b", "a", "c"];
const otherLetter = "z";
const obj = {a:otherLetter, b:2, c:3, d:4, e:5, f:6};

console.log(
  arr1.map(e => obj[e])
)

Answer 2

您可以采用动态方法并建立成功的哈希表并采用该值。

var array1 = ["a", "b", "a", "c"],
    array2 = [1, 2, 3, 4, 5, 6],
    hash = Object.create(null),
    index = 0;
    result = array1.map(v => v in hash ? hash[v] : hash[v] = array2[index++]);

console.log(result);

没有index但带有array2变异的版本

var array1 = ["a", "b", "a", "c"],
    array2 = [1, 2, 3, 4, 5, 6],
    hash = Object.create(null),
    result = array1.map(v => v in hash ? hash[v] : hash[v] = array2.shift());

console.log(result);

Answer 3

这可能会帮助您：

var arr1 = ["a", "b", "a", "c", "f"];
var arr2 = [1, 2, 3, 4];
//print out arrays
console.log("arr1: ", arr1);
console.log("arr2: ", arr2);

//loop through the arr1
var i;
//loop until you reach the length of array1.
for (i = 0; i < arr1.length; i++) {
  //check type of item in arr1
  switch (arr1[i]) {
    case "a":
      arr1[i] = arr2[0];
      break;
    case "b":
      arr1[i] = arr2[1];
      break;
    case "c":
      arr1[i] = arr2[2];
      break;
    default:
      //default if none of the above cases match with entry in array1
      arr1[i] = -1;
  }
}

//print out arrays
console.log("arr1: ", arr1);
console.log("arr2: ", arr2);

Answer 4

您可以使用parseInt从arr1中的字符生成索引，如下所示：

let indices = arr1.map(str => parseInt(str, 36) - 10);

parseInt的第二个参数是基数，36允许使用0-9和a-z指定数字。因为a = 10我们需要减去10才能获得正确的索引（a->索引0）。

然后使用这些索引从arr2中获取相应的元素：

let result = indices.map(index => arr1[index]);

一个简短的示例：

let arr1 = ['a', 'b', 'a', 'c'],
    arr2 = [1, 2, 3, 4, 5, 6],
    result = arr1.map(s => parseInt(s, 36) - 10).map(i => arr2[i]);
    
console.log(result);

Answer 5

您可以创建arr1的字符串表示形式并替换字符：

var arr1 = ["a", "b", "a", "c"];
var arr2 = [1, 2, 3, 4, 5, 6,];
var str = arr1.join(',');
arr2.forEach((number, index)=>{
  var character = arr1[index];
  if(character && isNaN(arr1[index])){
      var re = new RegExp(character,"g");
      str = str.replace(re, number);
  }
});

arr1 = str.split(',');
console.log(arr1);

Answer 6

这是我（简单得多）的答案：

var arr1 = ["a", "b", "a", "c"];

//I'd like to find and replace each "a" 
//with a certain element of an other array
var arr2 = [1, 2, 3, 4, 5, 6,];  

for(var i in arr1) {
  arr1[i] = arr2[arr1[i].charCodeAt(0) - 97];
}
console.log('arr1 is now: [' + arr1 + ']');

Answer 7

    var arr1 = ["a", "b", "a", "c"];
    var arr2 = [1, 2, 3, 4, 5, 6];
         
    var uniqueItems = Array.from(new Set(arr1));
    for(i=0;i<uniqueItems.length;i++){
      arr1.forEach(function(item, j) { if (item == uniqueItems[i]) arr1[j] = arr2[i]; });
    }

    console.log(arr1);

在保持位置的同时替换数组的多个元素

7 个答案: