我有下面的字符串

var a = "1,2,3,4";

当我…

var b = a.split(',');

我得到b为["1","2","3","4"]

我能做点什么让b等于[1,2,3,4]吗?


当前回答

一个更简短的解决方案:映射并传递参数给Number:

Var a = "1,2,3,4"; Var b = a.s split(','); console.log (b); var c = b.map(Number); console.log (c);

其他回答

由于所有的答案都允许包含NaN,所以我想补充一句,如果您想快速地将一个混合值的数组转换为数字,您可以这样做。

var a = "1,2,3,4,foo,bar";

var b = a.split(',');

var result = b.map(_=>_|0) // Floors the number (32-bit signed integer) so this wont work if you need all 64 bits.

// or b.map(_=>_||0) if you know your array is just numbers but may include NaN.

没有必要使用lambdas和/或给parseInt的基数参数,只需使用parseFloat或Number代替。

原因:

It's working: var src = "1,2,5,4,3"; var ids = src.split(',').map(parseFloat); // [1, 2, 5, 4, 3] var obj = {1: ..., 3: ..., 4: ..., 7: ...}; var keys= Object.keys(obj); // ["1", "3", "4", "7"] var ids = keys.map(parseFloat); // [1, 3, 4, 7] var arr = ["1", 5, "7", 11]; var ints= arr.map(parseFloat); // [1, 5, 7, 11] ints[1] === "5" // false ints[1] === 5 // true ints[2] === "7" // false ints[2] === 7 // true It's shorter. It's a tiny bit quickier and takes advantage of cache, when parseInt-approach - doesn't: // execution time measure function // keep it simple, yeah? > var f = (function (arr, c, n, m) { var i,t,m,s=n(); for(i=0;i++<c;)t=arr.map(m); return n()-s }).bind(null, "2,4,6,8,0,9,7,5,3,1".split(','), 1000000, Date.now); > f(Number) // first launch, just warming-up cache > 3971 // nice =) > f(Number) > 3964 // still the same > f(function(e){return+e}) > 5132 // yup, just little bit slower > f(function(e){return+e}) > 5112 // second run... and ok. > f(parseFloat) > 3727 // little bit quicker than .map(Number) > f(parseFloat) > 3737 // all ok > f(function(e){return parseInt(e,10)}) > 21852 // awww, how adorable... > f(function(e){return parseInt(e)}) > 22928 // maybe, without '10'?.. nope. > f(function(e){return parseInt(e)}) > 22769 // second run... and nothing changes. > f(Number) > 3873 // and again > f(parseFloat) > 3583 // and again > f(function(e){return+e}) > 4967 // and again > f(function(e){return parseInt(e,10)}) > 21649 // dammit 'parseInt'! >_<

注意:在Firefox中,parseInt的工作速度大约快4倍,但仍然比其他的慢。总共:+e < Number < parseFloat < parseInt

作为一种变体,你可以使用组合_。Map和_。来自lodash库的多种方法。整个转型将更加紧凑。以下是官方文档中的一个例子:

_.map(['6', '8', '10'], _.ary(parseInt, 1));
// → [6, 8, 10]

我对高尔夫球手的看法:

b="1,2,3,4".split`,`.map(x=>+x)

反引号是字符串字面值,所以我们可以省略圆括号(因为split函数的性质),但它相当于split(',')。字符串现在是一个数组,我们只需要用一个返回字符串整数的函数来映射每个值,因此x=>+x(它甚至比Number函数还要短(5个字符而不是6个字符))相当于:

function(x){return parseInt(x,10)}// version from techfoobar
(x)=>{return parseInt(x)}         // lambda are shorter and parseInt default is 10
(x)=>{return +x}                  // diff. with parseInt in SO but + is better in this case
x=>+x                             // no multiple args, just 1 function call

我希望它能更清楚一点。

使用arraw函数的Matt Zeunert版本(ES6)

const nums = a.split(',').map(x => parseInt(x, 10));