在JavaScript中将数字转换为基数64的最快方法?
在JavaScript中,您可以将数字转换为字符串表示形式,具体如下:在JavaScript中将数字转换为基数64的最快方法?,javascript,Javascript,在JavaScript中,您可以将数字转换为字符串表示形式,具体如下: (12345).toString(36) // "9ix" …您可以将其转换回常规数字,如下所示: parseInt("9ix", 36) // 12345 36是可以指定的最高基数。它显然使用字符0-9和a-z作为数字(总共36个) 我的问题:将数字转换为64进制表示的最快方法是什么(例如,对于额外的28位数字,使用a-Z,以及-和) 更新:有四个人发布了回复,说这个问题重复了,或者说我在找Base64。我不是 “”
(12345).toString(36) // "9ix"
…您可以将其转换回常规数字,如下所示:
parseInt("9ix", 36) // 12345
36是可以指定的最高基数。它显然使用字符0-9
和a-z
作为数字(总共36个)
我的问题:将数字转换为64进制表示的最快方法是什么(例如,对于额外的28位数字,使用a-Z
,以及-
和
)
更新:有四个人发布了回复,说这个问题重复了,或者说我在找Base64。我不是 “”是一种用简单的ASCII字符集对二进制数据进行编码的方法,以确保通过网络等进行传输的安全性(这样纯文本系统就不会篡改二进制数据) 这不是我要问的。我问的是如何将数字转换为基数为64的字符串表示形式。(JavaScript的
toString(radix)
会自动为任何最大为36的基数执行此操作;我需要一个自定义函数来获取基数64。)
更新2:以下是一些输入和输出示例
0 → "0"
1 → "1"
9 → "9"
10 → "a"
35 → "z"
61 → "Z"
62 → "-"
63 → "_"
64 → "10"
65 → "11"
128 → "20"
etc.
在mozilla或webkit浏览器中,您可以使用btoa()和atob()对base64进行编码和解码。好的,您可以使用任何Javascript base64库:也许可以回答它
编辑:二进制数据本质上只是一个字节序列。如果假设字节表示单个数字,则可以将字节序列表示为基64字符串。对它们进行解码,然后对字节进行一些简单的数学运算,得到一个数字。将数字转换为字节序列并编码以获得字符串。似乎很合理,除非您以某种方式投资于字符串中使用的特定字符。以下是数字(而不是字节数组)解决方案的草图:) 仅适用于正数,忽略小数部分,并且没有真正测试——只是一个草图
Base64 = {
_Rixits :
// 0 8 16 24 32 40 48 56 63
// v v v v v v v v v
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz+/",
// You have the freedom, here, to choose the glyphs you want for
// representing your base-64 numbers. The ASCII encoding guys usually
// choose a set of glyphs beginning with ABCD..., but, looking at
// your update #2, I deduce that you want glyphs beginning with
// 0123..., which is a fine choice and aligns the first ten numbers
// in base 64 with the first ten numbers in decimal.
// This cannot handle negative numbers and only works on the
// integer part, discarding the fractional part.
// Doing better means deciding on whether you're just representing
// the subset of javascript numbers of twos-complement 32-bit integers
// or going with base-64 representations for the bit pattern of the
// underlying IEEE floating-point number, or representing the mantissae
// and exponents separately, or some other possibility. For now, bail
fromNumber : function(number) {
if (isNaN(Number(number)) || number === null ||
number === Number.POSITIVE_INFINITY)
throw "The input is not valid";
if (number < 0)
throw "Can't represent negative numbers now";
var rixit; // like 'digit', only in some non-decimal radix
var residual = Math.floor(number);
var result = '';
while (true) {
rixit = residual % 64
// console.log("rixit : " + rixit);
// console.log("result before : " + result);
result = this._Rixits.charAt(rixit) + result;
// console.log("result after : " + result);
// console.log("residual before : " + residual);
residual = Math.floor(residual / 64);
// console.log("residual after : " + residual);
if (residual == 0)
break;
}
return result;
},
toNumber : function(rixits) {
var result = 0;
// console.log("rixits : " + rixits);
// console.log("rixits.split('') : " + rixits.split(''));
rixits = rixits.split('');
for (var e = 0; e < rixits.length; e++) {
// console.log("_Rixits.indexOf(" + rixits[e] + ") : " +
// this._Rixits.indexOf(rixits[e]));
// console.log("result before : " + result);
result = (result * 64) + this._Rixits.indexOf(rixits[e]);
// console.log("result after : " + result);
}
return result;
}
}
Base64={
_脱欧:
// 0 8 16 24 32 40 48 56 63
//v v v v v v v v
“0123456789ABCDEFGHIjklmnopqrstuvxyzabCDEFGHIjklmnopqrstuvxyz+/”,
//在这里,您可以自由选择所需的字形
//代表您的基64数字。ASCII编码的人通常
//选择一组以ABCD…开头的图示符,但是,请查看
//您的更新#2,我推断您需要以
//0123…,这是一个很好的选择,它与前十个数字对齐
//以64为基数,前十个数字以十进制表示。
//这不能处理负数,只能在
//整数部分,丢弃小数部分。
//做得更好意味着决定你是否只是代表
//两个补32位整数的javascript数的子集
//或者使用base-64表示法来表示
//基础IEEE浮点数,或表示尾数
//或者其他一些可能性。现在,保释
fromNumber:函数(数字){
如果(isNaN(数字(数字))| |数字===null||
数字===数字。正(无穷大)
抛出“输入无效”;
如果(数字<0)
抛出“现在不能表示负数”;
var rixit;//类似于“digit”,仅在某些非十进制基数中
var剩余=数学楼层(编号);
var结果=“”;
while(true){
rixit=剩余的%64
//log(“rixit:+rixit”);
//console.log(“前面的结果:+result”);
结果=这个。_Rixits.charAt(rixit)+结果;
//console.log(“结果在:+结果之后”);
//控制台日志(“剩余前:”+剩余);
剩余=数学楼层(剩余/64);
//控制台日志(“剩余后:”+剩余);
如果(残差==0)
打破
}
返回结果;
},
toNumber:函数(rixits){
var结果=0;
//log(“rixits:+rixits”);
//console.log(“rixits.split(“”):“+rixits.split(“”));
rixits=rixits.split(“”);
对于(var e=0;e
更新:这里有一些(非常轻量级的)上述测试,用于在有console.log的NodeJs中运行
function testBase64(x) {
console.log("My number is " + x);
var g = Base64.fromNumber(x);
console.log("My base-64 representation is " + g);
var h = Base64.toNumber(g);
console.log("Returning from base-64, I get " + h);
if (h !== Math.floor(x))
throw "TEST FAILED";
}
testBase64(0);
try {
testBase64(-1);
}
catch (err) {
console.log("caught >>>>>> " + err);
}
try {
testBase64(undefined);
}
catch (err) {
console.log("caught >>>>>> " + err);
}
try {
testBase64(null);
}
catch (err) {
console.log("caught >>>>>> " + err);
}
try {
testBase64(Number.NaN);
}
catch (err) {
console.log("caught >>>>>> " + err);
}
try {
testBase64(Number.POSITIVE_INFINITY);
}
catch (err) {
console.log("caught >>>>>> " + err);
}
try {
testBase64(Number.NEGATIVE_INFINITY);
}
catch (err) {
console.log("caught >>>>>> " + err);
}
for(i=0; i<100; i++)
testBase64(Math.random()*1e14);
函数testBase64(x){
console.log(“我的号码是”+x);
var g=Base64.从编号(x);
log(“我的base-64表示为”+g);
var h=Base64.toNumber(g);
log(“从base-64返回,我得到”+h);
如果(h!==数学楼层(x))
抛出“测试失败”;
}
testBase64(0);
试一试{
testBase64(-1);
}
捕捉(错误){
console.log(“捕获>>>>”+错误);
}
试一试{
testBase64(未定义);
}
捕捉(错误){
console.log(“捕获>>>>”+错误);
}
试一试{
testBase64(空);
}
捕捉(错误){
console.log(“捕获>>>>”+错误);
}
试一试{
testBase64(Number.NaN);
}
捕捉(错误){
console.log(“捕获>>>>”+错误);
}
试一试{
testBase64(数字正无穷大);
}
捕捉(错误){
console.log(“捕获>>>>”+错误);
}
试一试{
testBase64(数字负无穷大);
}
捕捉(错误){
console.log(“捕获>>>>”+错误);
}
对于(i=0;i,这里有一个不同的假设
function base64(value) {
if (typeof(value) === 'number') {
return base64.getChars(value, '');
}
if (typeof(value) === 'string') {
if (value === '') { return NaN; }
return value.split('').reverse().reduce(function(prev, cur, i) {
return prev + base64.chars.indexOf(cur) * Math.pow(64, i);
}, 0);
}
}
base64.chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_";
base64.getChars = function(num, res) {
var mod = num % 64,
remaining = Math.floor(num / 64),
chars = base64.chars.charAt(mod) + res;
if (remaining <= 0) { return chars; }
return base64.getChars(remaining, chars);
};
函数base64(值){
如果(类型(值)=“数字”){
返回base64.getChars(值为“”);
}
如果(类型(值)=“字符串”){
如果(值==''){返回Na
const STR64:Array = ('ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/').split( '' );
// TRANSFORM NUMBERS BETWEEN radix 10 AND radix 64
/** Methods based on answers shared in:
* @url http://stackoverflow.com/questions/6213227/fastest-way-to-convert-a-number-to-radix-64-in-javascript
*/
// METHODS
/** to64String: Method to transform a radix 10 number to radix 64 number (as string)
* @param input Number for transform to radix 64 (as String)
* @param current String data (don't needed in request)
* @return String Number in radix 64 as String;
*
* @based http://stackoverflow.com/users/383780/monocle
* @based base64( Method for number to string - NOT string part )
*/
function to64String( input:Number, current:String = '' ):String
{
if ( input < 0 && current.length == 0 ){
input = input * - 1;
}
var modify:Number = input % 64;
var remain:Number = Math.floor( input / 64 );
var result:String = STR64[ modify ] + current;
return ( remain <= 0 ) ? result : to64String( remain, result );
}
/** to64Parse: Method for transform a number in radix 64 (as string) in radix 10 number
* @param input Number in radix 64 (as String) to transform in radix 10
* @return Number in radix 10
*
* @based http://stackoverflow.com/users/520997/reb-cabin
* @based Base64.toNumber( Method for string to number )
*/
function to64Parse ( input:String ):Number
{
var result:Number = 0;
var toProc:Array = input.split( '' );
var e:String;
for ( e in toProc ){
result = ( result * 64 ) + STR64.indexOf( toProc[ e ] );
}
return result;
}
// TEST
var i:int = 0;
var max:Number = 1000000000000;
var min:Number = 0;
for ( i == 0; i < 20; i++ ){
var num:Number = ( Math.ceil( Math.random() * ( max - min + 1 ) ) + min );
var s64:String = to64String( num );
var ret:Number = to64Parse ( s64 );
trace( i + '\t# ' + num + '\t' + s64 + '\t' + ret + '\t' + ( ret == num ) )
}
// TEST RESULT
/*
0 # 808936734685 LxYYv/d 808936734685 true
1 # 931332556532 NjXvwb0 931332556532 true
2 # 336368837395 E5RJSMT 336368837395 true
3 # 862123347820 Mi6jk9s 862123347820 true
4 # 174279278611 CiT2sAT 174279278611 true
5 # 279361353722 EELO/f6 279361353722 true
6 # 435602995568 GVr9jlw 435602995568 true
7 # 547163526063 H9lfNOv 547163526063 true
8 # 188017380425 CvGtYxJ 188017380425 true
9 # 720098771622 KepO0Km 720098771622 true
10 # 408089106903 F8EAZnX 408089106903 true
11 # 293941423763 ERwRi6T 293941423763 true
12 # 383302396164 Fk+mmkE 383302396164 true
13 # 695998940618 KIMxQXK 695998940618 true
14 # 584515331314 IgX1CTy 584515331314 true
15 # 528965753970 Hso0Nxy 528965753970 true
16 # 5324317143 E9WqHX 5324317143 true
17 # 772389841267 LPWBalz 772389841267 true
18 # 954212692102 N4rgjCG 954212692102 true
19 # 867031893694 MnfIMa+ 867031893694 true
*/
Base64 = (function () {
var digitsStr =
// 0 8 16 24 32 40 48 56 63
// v v v v v v v v v
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz+-";
var digits = digitsStr.split('');
var digitsMap = {};
for (var i = 0; i < digits.length; i++) {
digitsMap[digits[i]] = i;
}
return {
fromInt: function(int32) {
var result = '';
while (true) {
result = digits[int32 & 0x3f] + result;
int32 >>>= 6;
if (int32 === 0)
break;
}
return result;
},
toInt: function(digitsStr) {
var result = 0;
var digits = digitsStr.split('');
for (var i = 0; i < digits.length; i++) {
result = (result << 6) + digitsMap[digits[i]];
}
return result;
}
};
})();
Base64.fromInt(-2147483648); // gives "200000"
Base64.toInt("200000"); // gives -2147483648
var base = ['Q', 'W', 'E', 'R', 'T', 'Y', 'I', 'O', 'U'];
new PowerRadix([1, 0], 10).toArray(base); // ['W', 'Q']
new PowerRadix('10', 10).toArray(base); // ['W', 'Q']
new PowerRadix(10, 10).toArray(base); // ['W', 'Q']
new PowerRadix([1, 0], 10).toString(base); // "WQ"
new PowerRadix('10', 10).toString(base); // "WQ"
new PowerRadix(10, 10).toString(base); // "WQ"
new PowerRadix('ba', ['a', 'b']); // base 2 source radix, uses 'a' = 0 & 'b' = 1 character set.
new PowerRadix('ba', ['a', 'b']).toString(10); // returns "2"
private String toShortString(BigInteger value, String language) {
StringBuilder stringBuilder = new StringBuilder();
BigInteger length = BigInteger.valueOf(language.length());
while (value.compareTo(BigInteger.ZERO) > 0){
int index = value.mod(length).intValue();
stringBuilder.append(language.charAt(index));
value = value.divide(length);
}
return stringBuilder.reverse().toString();
}
BigInteger value = BigInteger.valueOf(2).pow(128);
System.out.println(value);
System.out.println(value.toString(16));
System.out.println(toShortString(value, "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!@#$%^&*()-=_+"));
var base64 = Buffer.from([i>>24,i>>16,i>>8,i]).toString('base64').substr(0,6);
const alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/';
// binary to string lookup table
const b2s = alphabet.split('');
// string to binary lookup table
// 123 == 'z'.charCodeAt(0) + 1
const s2b = new Array(123);
for (let i = 0; i < alphabet.length; i++) {
s2b[alphabet.charCodeAt(i)] = i;
}
// number to base64
const ntob = (number) => {
if (number < 0) return `-${ntob(-number)}`;
let lo = number >>> 0;
let hi = (number / 4294967296) >>> 0;
let right = '';
while (hi > 0) {
right = b2s[0x3f & lo] + right;
lo >>>= 6;
lo |= (0x3f & hi) << 26;
hi >>>= 6;
}
let left = '';
do {
left = b2s[0x3f & lo] + left;
lo >>>= 6;
} while (lo > 0);
return left + right;
};
// base64 to number
const bton = (base64) => {
let number = 0;
const sign = base64.charAt(0) === '-' ? 1 : 0;
for (let i = sign; i < base64.length; i++) {
number = number * 64 + s2b[base64.charCodeAt(i)];
}
return sign ? -number : number;
};
var endex = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_";
function encode(intcode){
if(intcode < endex.length){
return endex[intcode];
}else{
return encode(Math.floor(intcode/endex.length)) + endex[intcode%endex.length];
}
}
function decode(charcode){
if(charcode.length < 2){
return endex.indexOf(charcode);
}else{
return (decode(charcode.slice(0, -1)) * endex.length) + endex.indexOf(charcode.slice(-1));
}
}