Javascript 替换嵌套数组中的所有元素_Javascript

Javascript 替换嵌套数组中的所有元素

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Javascript 替换嵌套数组中的所有元素,javascript,Javascript,我有一个挑战，我已经与之斗争了一段时间。这是关于替换网格形式的数组中的所有元素，但我的解决方案只是替换它所选择的元素，而不是我想要的在这个挑战中，我想用字符串“偶数”替换可被2整除的整数值，而用字符串“奇数”替换其余的整数值 /* *-numbers变量是一个数组数组。 *-一个嵌套的，用于循环数字。 *-将每个偶数转换为字符串“偶数” *-并将每个奇数转换为字符串“奇数” */ var myNumbers=[ [243, 12, 23, 12, 45, 45, 78, 66, 223, 3]

我有一个挑战，我已经与之斗争了一段时间。这是关于替换网格形式的数组中的所有元素，但我的解决方案只是替换它所选择的元素，而不是我想要的

在这个挑战中，我想用字符串“偶数”替换可被2整除的整数值，而用字符串“奇数”替换其余的整数值

/* *-

numbers

变量是一个数组数组。 *-一个嵌套的

，用于循环数字。
*-将每个偶数转换为字符串“偶数”
*-并将每个奇数转换为字符串“奇数”
*/
var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
[67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
[76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];
对于（var row=0；row可能更容易使用嵌套的map
：

var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
[67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
[76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];
const output=myNumbers.map（行=>row.map（num=>
数值%2==0
“甚至”
：“奇数”
));
console.log（输出）；
使用嵌套的映射可能更容易：

var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
[67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
[76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];
const output=myNumbers.map（行=>row.map（num=>
数值%2==0
“甚至”
：“奇数”
));
console.log（output）；
在内部循环中，您需要使用两个索引（即myNumbers[row][column]
）访问值。目前，您只使用列索引，因此将值拼接到包含行的数组中。
在内部循环中，您需要使用两个索引访问值（即myNumbers[row][column]
）。目前，您仅使用列索引，因此将值拼接到包含行的数组中。
您对行索引使用了错误的变量
myNumbers[column]
//needs to be
myNumbers[row]

另外，if条件使用了错误的行索引，并试图与整个数组而不是数组中的值进行比较
if(myNumbers[column]%2===0)
//needs to be
if(myNumbers[row][column]%2===0)

演示

var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
[67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
[76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];
对于（变量行=0；行console.log（myNumbers）；
您对行索引使用了错误的变量
myNumbers[column]
//needs to be
myNumbers[row]

另外，if条件使用了错误的行索引，并试图与整个数组而不是数组中的值进行比较
if(myNumbers[column]%2===0)
//needs to be
if(myNumbers[row][column]%2===0)

演示

var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
[67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
[76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];
对于（变量行=0；行console.log（myNumbers）；
为什么不简单地设置myNumbers[row][column]
的值，而不是使用splice

var myNumbers = [
    [243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
    [34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
    [67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
    [12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
    [4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
    [5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
    [74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
    [53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
    [67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
    [76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];

    for(var row=0; row<myNumbers.length; row++) {
         for(var column=0;column<myNumbers[row].length;column++) {
                if(myNumbers[row][column]%2===0) {
                  myNumbers[row][column] = "even";
                } else{ 
                    myNumbers[row][column] = "odd";
                }
            console.log(myNumbers[row][column]);
     }
    }

var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
[67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
[76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];
对于（var row=0；row为什么不简单地设置myNumbers[row][column]
的值，而不是使用splice

var myNumbers = [
    [243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
    [34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
    [67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
    [12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
    [4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
    [5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
    [74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
    [53, 6, 89, 10, 23, 52, 111, 44, 109, 80],
    [67, 6, 53, 537, 2, 168, 16, 2, 1, 8],
    [76, 7, 9, 6, 3, 73, 77, 100, 56, 100]
];

    for(var row=0; row<myNumbers.length; row++) {
         for(var column=0;column<myNumbers[row].length;column++) {
                if(myNumbers[row][column]%2===0) {
                  myNumbers[row][column] = "even";
                } else{ 
                    myNumbers[row][column] = "odd";
                }
            console.log(myNumbers[row][column]);
     }
    }

var myNumbers=[
[243, 12, 23, 12, 45, 45, 78, 66, 223, 3],
[34, 2, 1, 553, 23, 4, 66, 23, 4, 55],
[67, 56, 45, 553, 44, 55, 5, 428, 452, 3],
[12, 31, 55, 445, 79, 44, 674, 224, 4, 21],
[4, 2, 3, 52, 13, 51, 44, 1, 67, 5],
[5, 65, 4, 5, 5, 6, 5, 43, 23, 4424],
[74, 532, 6, 7, 35, 17, 89, 43, 43, 66],
[53, 6