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?
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slice()方法:
function returnChunksArray(str, chunkSize) {
var arr = [];
while(str !== '') {
arr.push(str.slice(0, chunkSize));
str = str.slice(chunkSize);
}
return arr;
}
使用substring()方法也可以做到这一点。
function returnChunksArray(str, chunkSize) {
var arr = [];
while(str !== '') {
arr.push(str.substring(0, chunkSize));
str = str.substring(chunkSize);
}
return arr;
}
其他回答
它将大字符串拆分为给定单词的小字符串。
function chunkSubstr(str, words) {
var parts = str.split(" ") , values = [] , i = 0 , tmpVar = "";
$.each(parts, function(index, value) {
if(tmpVar.length < words){
tmpVar += " " + value;
}else{
values[i] = tmpVar.replace(/\s+/g, " ");
i++;
tmpVar = value;
}
});
if(values.length < 1 && parts.length > 0){
values[0] = tmpVar;
}
return values;
}
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.
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 ' ]
var str = "123456789";
var chunks = [];
var chunkSize = 2;
while (str) {
if (str.length < chunkSize) {
chunks.push(str);
break;
}
else {
chunks.push(str.substr(0, chunkSize));
str = str.substr(chunkSize);
}
}
alert(chunks); // chunks == 12,34,56,78,9
var l = str.length, lc = 0, chunks = [], c = 0, chunkSize = 2;
for (; lc < l; c++) {
chunks[c] = str.slice(lc, lc += chunkSize);
}