我想使用 JavaScript 解析指数值并将其转换为小数。4.65661287307739E-10应该给0.000000000465661287307739。我应该怎么做才能实现这一目标？
parseFloat(4.65661287307739E-10)返回4.65661287307739e-10。
parseInt(4.65661287307739E-10)返回4。
问问题
62943 次
8 回答
29
您可以使用toFixed()，但限制为 20 个。
前任：
(4.65661287307739E-10).toFixed(20)
"0.00000000046566128731"
但...
(4.65661287307739E-30).toFixed(20)
"0.00000000000000000000"
因此，如果您的小数位数总是少于 20 位，那就没问题了。否则，我认为您可能必须自己编写。
于 2013-09-10T13:13:46.890 回答
18
您可以显示大或小小数的字符串值：
Number.prototype.noExponents= function(){
var data= String(this).split(/[eE]/);
if(data.length== 1) return data[0];
var z= '', sign= this<0? '-':'',
str= data[0].replace('.', ''),
mag= Number(data[1])+ 1;
if(mag<0){
z= sign + '0.';
while(mag++) z += '0';
return z + str.replace(/^\-/,'');
}
mag -= str.length;
while(mag--) z += '0';
return str + z;
}
var n=4.65661287307739E-10 ;
n.noExponents()
/* returned value: (String)
0.000000000465661287307739
*/
于 2013-09-10T13:16:11.627 回答
4
这个变体非常适合我显示太小/大数字的格式化字符串值：
const exponentialToDecimal = exponential => {
let decimal = exponential.toString().toLowerCase();
if (decimal.includes('e+')) {
const exponentialSplitted = decimal.split('e+');
let postfix = '';
for (
let i = 0;
i <
+exponentialSplitted[1] -
(exponentialSplitted[0].includes('.') ? exponentialSplitted[0].split('.')[1].length : 0);
i++
) {
postfix += '0';
}
const addCommas = text => {
let j = 3;
let textLength = text.length;
while (j < textLength) {
text = `${text.slice(0, textLength - j)},${text.slice(textLength - j, textLength)}`;
textLength++;
j += 3 + 1;
}
return text;
};
decimal = addCommas(exponentialSplitted[0].replace('.', '') + postfix);
}
if (decimal.toLowerCase().includes('e-')) {
const exponentialSplitted = decimal.split('e-');
let prefix = '0.';
for (let i = 0; i < +exponentialSplitted[1] - 1; i++) {
prefix += '0';
}
decimal = prefix + exponentialSplitted[0].replace('.', '');
}
return decimal;
};
const result1 = exponentialToDecimal(5.6565e29); // "565,650,000,000,000,000,000,000,000,000"
const result2 = exponentialToDecimal(5.6565e-29); // "0.000000000000000000000000000056565"
于 2019-05-07T15:21:46.113 回答
3
一个可怕的原始转换，用 ES6 编写：
function convert(n){
var sign = +n < 0 ? "-" : "",
toStr = n.toString();
if (!/e/i.test(toStr)) {
return n;
}
var [lead,decimal,pow] = n.toString()
.replace(/^-/,"")
.replace(/^([0-9]+)(e.*)/,"$1.$2")
.split(/e|\./);
return +pow < 0
? sign + "0." + "0".repeat(Math.max(Math.abs(pow)-1 || 0, 0)) + lead + decimal
: sign + lead + (+pow >= decimal.length ? (decimal + "0".repeat(Math.max(+pow-decimal.length || 0, 0))) : (decimal.slice(0,+pow)+"."+decimal.slice(+pow)))
}
var myvar = 4.951760157141521e+27;
var myvar2 = 4.951760157141521e-2;
var myvar3 = 3.3e+3;
convert(myvar);//"4951760157141521000000000000"
convert(myvar2);//"0.04951760157141521"
convert(myvar3);//"3300"
于 2018-05-07T20:30:17.143 回答
3
这是我将指数（电子符号）数字转换为小数的简短函数代码，允许输出大量小数位。
它处理 Javascript 允许的所有电子符号语法，包括以下内容：
Valid e-notation numbers in Javascript:
123e1 ==> 1230
123E1 ==> 1230
123e+1 ==> 1230
123.e+1 ==> 1230 (with missing fractional part)
123e-1 ==> 12.3
0.1e-1 ==> 0.01
.1e-1 ==> 0.01 (with missing whole part)
-123e1 ==> -1230
0.0e1 ==> 0
-0.0e4 ==> -0
该函数不会尝试捕获 NaN 或未定义的输入，但会尝试满足正常（非电子符号）数字；这些数字按“原样”返回。
我试图在每一行上提供足够的注释，并尽量避免（尽可能！）使用短路和条件（三元）运算符，以获得更好的清晰度。
我已经使用该toLocaleString()方法自动检测小数分隔符符号，但这当然假设表示数字的输入字符串遵循机器的语言环境（尤其是手动传递给函数时）。
/********************************************************
* Converts Exponential (e-Notation) Numbers to Decimals
********************************************************
* @function numberExponentToLarge()
* @version 1.00
* @param {string} Number in exponent format.
* (other formats returned as is).
* @return {string} Returns a decimal number string.
* @author Mohsen Alyafei
* @date 12 Jan 2020
*
* Notes: No check is made for NaN or undefined inputs
*
*******************************************************/
function numberExponentToLarge(numIn) {
numIn +=""; // To cater to numric entries
var sign=""; // To remember the number sign
numIn.charAt(0)=="-" && (numIn =numIn.substring(1),sign ="-"); // remove - sign & remember it
var str = numIn.split(/[eE]/g); // Split numberic string at e or E
if (str.length<2) return sign+numIn; // Not an Exponent Number? Exit with orginal Num back
var power = str[1]; // Get Exponent (Power) (could be + or -)
var deciSp = 1.1.toLocaleString().substring(1,2); // Get Deciaml Separator
str = str[0].split(deciSp); // Split the Base Number into LH and RH at the decimal point
var baseRH = str[1] || "", // RH Base part. Make sure we have a RH fraction else ""
baseLH = str[0]; // LH base part.
if (power>=0) { // ------- Positive Exponents (Process the RH Base Part)
if (power> baseRH.length) baseRH +="0".repeat(power-baseRH.length); // Pad with "0" at RH
baseRH = baseRH.slice(0,power) + deciSp + baseRH.slice(power); // Insert decSep at the correct place into RH base
if (baseRH.charAt(baseRH.length-1) ==deciSp) baseRH =baseRH.slice(0,-1); // If decSep at RH end? => remove it
} else { // ------- Negative exponents (Process the LH Base Part)
num= Math.abs(power) - baseLH.length; // Delta necessary 0's
if (num>0) baseLH = "0".repeat(num) + baseLH; // Pad with "0" at LH
baseLH = baseLH.slice(0, power) + deciSp + baseLH.slice(power); // Insert "." at the correct place into LH base
if (baseLH.charAt(0) == deciSp) baseLH="0" + baseLH; // If decSep at LH most? => add "0"
}
// Rremove leading and trailing 0's and Return the long number (with sign)
return sign + (baseLH + baseRH).replace(/^0*(\d+|\d+\.\d+?)\.?0*$/,"$1");
}
//============ test codes ==================
function test(test,input,should){
var out=numberExponentToLarge(input);
var r = (out===should) ? true : false;
if (!r) console.log(test+" Failed: "+out+" should be: "+should);
else console.log("Passed");
}
// ------------- tests for e-notation numbers ---------------------
test(1,"123E0","123")
test(2,"123E0","123")
test(3,"-123e+0","-123")
test(4,"123e1","1230")
test(5,"123e3","123000")
test(6,"123e+3","123000")
test(7,"123E+7","1230000000")
test(8,"-123.456e+1","-1234.56")
test(9,"123.456e+4","1234560")
test(10,"123E-0","123")
test(11,"123.456e+50","12345600000000000000000000000000000000000000000000000")
test(12,"123e-0","123")
test(13,"123e-1","12.3")
test(14,"123e-3","0.123")
test(15,"-123e-7","-0.0000123")
test(16,"123.456E-1","12.3456")
test(17,"123.456e-4","0.0123456")
test(18,"123.456e-50","0.00000000000000000000000000000000000000000000000123456")
test(18-1,"-123.456e-50","-0.00000000000000000000000000000000000000000000000123456")
test(19,"1.e-5","0.00001") // handle missing base fractional part
test(20,".123e3","123") // handle missing base whole part
// The Electron's Mass:
test(21,"9.10938356e-31","0.000000000000000000000000000000910938356")
// The Earth's Mass:
test(22,"5.9724e+24","5972400000000000000000000")
// Planck constant:
test(23,"6.62607015e-34","0.000000000000000000000000000000000662607015")
test(24,"0.000e3","0")
test(25,"0.000000000000000e3","0")
test(26,"-0.0001e+9","-100000")
test(27,"-0.0e1","-0")
test(28,"-0.0000e1","-0")
test(28,"-000.0000e1","-0") // this is an invalid Javascript number
test(28,"-000.0000e-1","-0") // this is an invalid Javascript number
test(28,"-000.0000e+10","-0") // this is an invalid Javascript number
test(28,"-000.0000e+2","-0") // this is an invalid Javascript number
// ------------- testing for Non e-Notation Numbers -------------
test(29,"12345.7898","12345.7898") // no exponent
test(30,12345.7898,"12345.7898") // no exponent
test(31,0.00000000000001,"0.00000000000001") // from 1e-14
test(32,-0.0000000000000345,"-0.0000000000000345") // from -3.45e-14
test(33,-0,"0")
test(34,"1.2000e0","1.2")
test(35,"1.2000e-0","1.2")
test(35,"1.2000e+0","1.2")
test(35,"1.2000e+10","12000000000")于 2020-06-01T02:24:34.077 回答
1
使用 toFixed(number)
例如：
(+exponentialnumber).toFixed(8)
2e-7 ===> 0.00000020
于 2020-06-30T05:59:33.653 回答
0
扩展自@kennebec 的答案，但处理他失败的边缘情况更好（要点，使用 CoffeeScript）：
String.prototype.noExponents = function(explicitNum) {
var data, leader, mag, multiplier, num, sign, str, z;
if (explicitNum == null) {
explicitNum = true;
}
/*
* Remove scientific notation from a number
*
* After
* http://stackoverflow.com/a/18719988/1877527
*/
data = this.split(/[eE]/);
if (data.length === 1) {
return data[0];
}
z = "";
sign = this.slice(0, 1) === "-" ? "-" : "";
str = data[0].replace(".", "");
mag = Number(data[1]) + 1;
if (mag <= 0) {
z = sign + "0.";
while (!(mag >= 0)) {
z += "0";
++mag;
}
num = z + str.replace(/^\-/, "");
if (explicitNum) {
return parseFloat(num);
} else {
return num;
}
}
if (str.length <= mag) {
mag -= str.length;
while (!(mag <= 0)) {
z += 0;
--mag;
}
num = str + z;
if (explicitNum) {
return parseFloat(num);
} else {
return num;
}
} else {
leader = parseFloat(data[0]);
multiplier = Math.pow(10, parseInt(data[1]));
return leader * multiplier;
}
};
Number.prototype.noExponents = function() {
var strVal;
strVal = String(this);
return strVal.noExponents(true);
};于 2017-05-11T22:19:59.540 回答
-3
在 ExtJS 中，您可以使用Ext.Number.toFixed(value, precision)内部使用toFixed()方法的方法，
例如
console.log(Ext.Number.toFixed(4.65661287307739E-10, 10));
// O/p => 0.0000000005
console.log(Ext.Number.toFixed(4.65661287307739E-10, 15));
// 0.000000000465661
于 2013-09-10T14:34:00.203 回答