我对上述解决方案的问题是，不管在句子中的位置如何，它都将字符串划分为正式的大小块。

我认为下面的方法比较好;虽然它需要一些性能调整:

 static chunkString(str, length, size,delimiter='\n' ) {
        const result = [];
        for (let i = 0; i < str.length; i++) {
            const lastIndex = _.lastIndexOf(str, delimiter,size + i);
            result.push(str.substr(i, lastIndex - i));
            i = lastIndex;
        }
        return result;
    }

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"]

包括左版本和右版本的预分配。 对于小块，这和RegExp impl一样快，但是随着块大小的增加，速度会更快。它的内存效率很高。

function chunkLeft (str, size = 3) {
  if (typeof str === 'string') {
    const length = str.length
    const chunks = Array(Math.ceil(length / size))
    for (let i = 0, index = 0; index < length; i++) {
      chunks[i] = str.slice(index, index += size)
    }
    return chunks
  }
}

function chunkRight (str, size = 3) {
  if (typeof str === 'string') {
    const length = str.length
    const chunks = Array(Math.ceil(length / size))
    if (length) {
      chunks[0] = str.slice(0, length % size || size)
      for (let i = 1, index = chunks[0].length; index < length; i++) {
        chunks[i] = str.slice(index, index += size)
      }
    }
    return chunks
  }
}

console.log(chunkRight())  // undefined
console.log(chunkRight(''))  // []
console.log(chunkRight('1'))  // ["1"]
console.log(chunkRight('123'))  // ["123"]
console.log(chunkRight('1234'))  // ["1", "234"]
console.log(chunkRight('12345'))  // ["12", "345"]
console.log(chunkRight('123456'))  // ["123", "456"]
console.log(chunkRight('1234567'))  // ["1", "234", "567"]

我对上述解决方案的问题是，不管在句子中的位置如何，它都将字符串划分为正式的大小块。

我认为下面的方法比较好;虽然它需要一些性能调整:

 static chunkString(str, length, size,delimiter='\n' ) {
        const result = [];
        for (let i = 0; i < str.length; i++) {
            const lastIndex = _.lastIndexOf(str, delimiter,size + i);
            result.push(str.substr(i, lastIndex - i));
            i = lastIndex;
        }
        return result;
    }

function chunkString(str, length = 10) {
    let result = [],
        offset = 0;
    if (str.length <= length) return result.push(str) && result;
    while (offset < str.length) {
        result.push(str.substr(offset, length));
        offset += length;
    }
    return result;
}

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 ' ]