我想把一个非常大的字符串(比如10,000个字符)分割成n大小的块。

就性能而言,最好的方法是什么?

例如: "1234567890"除以2将变成["12","34","56","78","90"]。

使用string。prototype。match可以实现这样的事情吗如果可以,从性能来看,这是最好的方式吗?


当前回答

惊喜!你可以使用split来分割。

var parts = "1234567890 ".split(/(.{2})/).filter(O=>O)

结果显示['12','34','56','78','90',' ']

其他回答

const getChunksFromString = (str, chunkSize) => {
    var regexChunk = new RegExp(`.{1,${chunkSize}}`, 'g')   // '.' represents any character
    return str.match(regexChunk)
}

根据需要调用它

console.log(getChunksFromString("Hello world", 3))   // ["Hel", "lo ", "wor", "ld"]

这是一个快速而直接的解决方案

function chunkString (str, len) { const size = Math.ceil(str.length/len) const r = Array(size) let offset = 0 for (let i = 0; i < size; i++) { r[i] = str.substr(offset, len) offset += len } return r } console.log(chunkString("helloworld", 3)) // => [ "hel", "low", "orl", "d" ] // 10,000 char string const bigString = "helloworld".repeat(1000) console.time("perf") const result = chunkString(bigString, 3) console.timeEnd("perf") console.log(result) // => perf: 0.385 ms // => [ "hel", "low", "orl", "dhe", "llo", "wor", ... ]

你绝对可以做一些

let pieces = "1234567890 ".split(/(.{2})/).filter(x => x.length == 2);

要得到这个:

[ '12', '34', '56', '78', '90' ]

如果你想动态输入/调整数据块的大小,使数据块的大小为n,你可以这样做:

n = 2;
let pieces = "1234567890 ".split(new RegExp("(.{"+n.toString()+"})")).filter(x => x.length == n);

要在原始字符串中找到所有大小为n的块,尝试这样做:

let subs = new Set();
let n = 2;
let str = "1234567890 ";
let regex = new RegExp("(.{"+n.toString()+"})");     //set up regex expression dynamically encoded with n

for (let i = 0; i < n; i++){               //starting from all possible offsets from position 0 in the string
    let pieces = str.split(regex).filter(x => x.length == n);    //divide the string into chunks of size n...
    for (let p of pieces)                 //...and add the chunks to the set
        subs.add(p);
    str = str.substr(1);    //shift the string reading frame
}

你应该得到:

[ '12', '23', '34', '45', '56', '67', '78', '89', '90', '0 ' ]

惊喜!你可以使用split来分割。

var parts = "1234567890 ".split(/(.{2})/).filter(O=>O)

结果显示['12','34','56','78','90',' ']

这是一个解决方案,我想出了一个模板字符串后,一个小实验:

用法:

chunkString(5)`testing123`

函数chunkString(nSize) return (strToChunk) => { Let result = []; let chars = String(strToChunk).split("); 对于(设I = 0;i < (String(strToChunk)。length / nSize);我+ +){ result = result.concat(char .slice(i*nSize,(i+1)*nSize).join(")); } 返回结果 } } document . write (chunkString(5)“testing123”); //返回:testi,ng123 document . write (chunkString(3)“testing123”); //返回:tes,tin,g12,3