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
使用npm库"chkchars" 但是请记住，要确保给定的字符串长度完全除以“number”参数。
const phrase = "1110010111010011100101110100010000011100101110100111001011101001011101001110010111010001000001110010111010011100101110100"
const number = 7
chkchars.splitToChunks(phrase, number)
// result => ['1110010', '1110100','1110010', '1110100','0100000', '1110010','1110100', '1110010','1110100', '1011101','0011100', '1011101','0001000','0011100','1011101', '0011100','1011101']
// perf => 0.287ms
其他回答
window.format = function(b, a) {
if (!b || isNaN(+a)) return a;
var a = b.charAt(0) == "-" ? -a : +a,
j = a < 0 ? a = -a : 0,
e = b.match(/[^\d\-\+#]/g),
h = e && e[e.length - 1] || ".",
e = e && e[1] && e[0] || ",",
b = b.split(h),
a = a.toFixed(b[1] && b[1].length),
a = +a + "",
d = b[1] && b[1].lastIndexOf("0"),
c = a.split(".");
if (!c[1] || c[1] && c[1].length <= d) a = (+a).toFixed(d + 1);
d = b[0].split(e);
b[0] = d.join("");
var f = b[0] && b[0].indexOf("0");
if (f > -1)
for (; c[0].length < b[0].length - f;) c[0] = "0" + c[0];
else +c[0] == 0 && (c[0] = "");
a = a.split(".");
a[0] = c[0];
if (c = d[1] && d[d.length -
1].length) {
for (var d = a[0], f = "", k = d.length % c, g = 0, i = d.length; g < i; g++) f += d.charAt(g), !((g - k + 1) % c) && g < i - c && (f += e);
a[0] = f
}
a[1] = b[1] && a[1] ? h + a[1] : "";
return (j ? "-" : "") + a[0] + a[1]
};
var str="1234567890";
var formatstr=format( "##,###.", str);
alert(formatstr);
This will split the string in reverse order with comma separated after 3 char's. If you want you can change the position.
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", ... ]
使用npm库"chkchars" 但是请记住，要确保给定的字符串长度完全除以“number”参数。
const phrase = "1110010111010011100101110100010000011100101110100111001011101001011101001110010111010001000001110010111010011100101110100"
const number = 7
chkchars.splitToChunks(phrase, number)
// result => ['1110010', '1110100','1110010', '1110100','0100000', '1110010','1110100', '1110010','1110100', '1011101','0011100', '1011101','0001000','0011100','1011101', '0011100','1011101']
// perf => 0.287ms
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'，' ']
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 ' ]
