我正在尝试使用JavaScript获取数字的第n个根，但我没有看到使用内置Math对象的方法。我忽略了什么吗？
如果不是......
我是否可以使用具有此功能的数学库？
如果不是......
自己做这个的最佳算法是什么？
答案 0 :(得分：114)
你能用这样的东西吗？
Math.pow(n, 1/root);
例如
Math.pow(25, 1/2) == 5
答案 1 :(得分：19)
n的{{1}}根与x的{{1}}相同。x。您只需使用1/n：
Math.pow答案 2 :(得分：10)
使用Math.pow（）
请注意，它不能很好地处理否定 - 这是一个讨论和一些代码
http://cwestblog.com/2011/05/06/cube-root-an-beyond/
function nthroot(x, n) {
try {
var negate = n % 2 == 1 && x < 0;
if(negate)
x = -x;
var possible = Math.pow(x, 1 / n);
n = Math.pow(possible, n);
if(Math.abs(x - n) < 1 && (x > 0 == n > 0))
return negate ? -possible : possible;
} catch(e){}
}
答案 3 :(得分：5)
您可以使用
Math.nthroot = function(x,n) {
//if x is negative function returns NaN
return this.exp((1/n)*this.log(x));
}
//call using Math.nthroot();
答案 4 :(得分：3)
n x的根r是一个r，1/n x的力量为x。< / p>
实际上，有一些子句：
r为正数且x为偶数时，有两种解决方案（符号相反的值相同）。r为正数且x为奇数时，有一个正解。r为负数且x为奇数时，有一个否定解决方案。r为否定且Math.pow为偶数时，没有解决方案。由于function nthRoot(x, n) {
if(x < 0 && n%2 != 1) return NaN; // Not well defined
return (x < 0 ? -1 : 1) * Math.pow(Math.abs(x), 1/n);
}
不喜欢带有非整数指数的负基数，因此可以使用
nthRoot(+4, 2); // 2 (the positive is chosen, but -2 is a solution too)
nthRoot(+8, 3); // 2 (this is the only solution)
nthRoot(-8, 3); // -2 (this is the only solution)
nthRoot(-4, 2); // NaN (there is no solution)
示例：
#!/bin/bash
mkdir ~/Desktop/ror_save_unlock_all
echo "This script takes ~/Library/Application Support/com.riskofrain.riskofrain/Save.ini and backs it up to ~/Desktop/ror_save_unlock_all/originalSave. It generates a new Save.ini that is built to your specifications, unlocking all items and achievements or just the ones you choose, and replaces the existing Save.ini with the new one." >> ~/Desktop/ror_save_unlock_all/README.txt
mkdir ~/Desktop/ror_save_unlock_all/originalSave
cp ~/Library/Application\ Support/com.riskofrain.riskofrain/Save.ini ~/Desktop/ror_save_unlock_all/originalSave/
cd ~/Desktop/ror_save_unlock_all
echo -n 'How would you like the new Save.ini to be built? Press 1 for all unlocks or 2 to customize.'
read text
if [ $text = "1" ]
then
{
echo "[Achievement]" >> EOF1
count=1
while [ $count -lt 55 ]
do
echo "acheivement${count}=\"2\"" >> EOF2
count=`expr $count + 1`
done
echo "[Record]" >> EOF3
count=1
while [ $count -lt 31 ]
do
echo "mons${count}=\"1\"" >> EOF4
count=`expr $count + 1`
done
count=1
while [ $count -lt 110 ]
do
echo "item${count}=\"1\"" >> EOF5
count=`expr $count + 1`
done
count=1
while [ $count -lt 10 ]
do
echo "artifact${count}=\"1\"" >> EOF6
count=`expr $count + 1`
done
cat EOF1 EOF2 EOF3 EOF4 EOF5 EOF6 > Save.ini
rm EOF1 EOF2 EOF3 EOF4 EOF5 EOF6
cp -force ~/Desktop/ror_save_unlock_all/Save.ini ~/Library/Application\ Support/com.riskofrain.riskofrain/Save.ini
echo "Original Save.ini successfully overwritten."
exit
}
elif [ $text = "2" ]; then
{
echo "You selected customize. Now we will build a custom Save.ini"
echo -n "Press 1 if you want to unlock all achievements, press 2 to continue without unlocking all achievements."
read text
if [ $text = "1" ]; then
{
echo "[Achievement]" >> EOF1
count=1
while [ $count -lt 55 ]
do
echo "acheivement${count}=\"2\"" >> EOF2
count=`expr $count + 1`
done
echo "All achievements successfully unlocked."
echo -n "Press 1 to make the Save.ini and replace the existing one with it and then exit, or press 2 to customize it further."
read text
if [ $text = "1" ]; then
{
cat EOF1 EOF2 > Save.ini
rm EOF1 EOF2
cp -force ~/Desktop/ror_save_unlock_all/Save.ini ~/Library/Application\ Support/com.riskofrain.riskofrain/Save.ini
echo "Original Save.ini successfully overwritten."
exit
}
elif [ $text = "2" ] then;
{
echo -n "Press 1 to unlock all monster logs, or press 2 to continue without unlocking all monster logs."
read text
if [ $text = "1" ]; then
{
echo "[Record]" >> EOF3
count=1
while [ $count -lt 31 ]
do
echo "mons${count}=\"1\"" >> EOF4
count=`expr $count + 1`
done
echo "All achievements successfully unlocked."
echo -n "Press 1 to make the Save.ini and replace the existing one with it and then exit, or press 2 to customize it further."
read text
}
}
}
}
答案 5 :(得分：3)
对于square和cubic root的特殊情况，最好分别使用本机函数Math.sqrt和Math.cbrt。
从ES7开始，exponentiation operator **可用于计算 n 根作为 1 / n 非负基数的力量：
let root1 = Math.PI ** (1 / 3); // cube root of π
let root2 = 81 ** 0.25; // 4th root of 81
但这不适用于负面基础。
let root3 = (-32) ** 5; // NaN
答案 6 :(得分：0)
这是一个试图返回虚数的函数。它还首先检查一些常见的事物，例如：获得0或1的平方根，还是获得x的第0个根。
function root(x, n){
if(x == 1){
return 1;
}else if(x == 0 && n > 0){
return 0;
}else if(x == 0 && n < 0){
return Infinity;
}else if(n == 1){
return x;
}else if(n == 0 && x > 1){
return Infinity;
}else if(n == 0 && x == 1){
return 1;
}else if(n == 0 && x < 1 && x > -1){
return 0;
}else if(n == 0){
return NaN;
}
var result = false;
var num = x;
var neg = false;
if(num < 0){
//not using Math.abs because I need the function to remember if the number was positive or negative
num = num*-1;
neg = true;
}
if(n == 2){
//better to use square root if we can
result = Math.sqrt(num);
}else if(n == 3){
//better to use cube root if we can
result = Math.cbrt(num);
}else if(n > 3){
//the method Digital Plane suggested
result = Math.pow(num, 1/n);
}else if(n < 0){
//the method Digital Plane suggested
result = Math.pow(num, 1/n);
}
if(neg && n == 2){
//if square root, you can just add the imaginary number "i=√-1" to a string answer
//you should check if the functions return value contains i, before continuing any calculations
result += 'i';
}else if(neg && n % 2 !== 0 && n > 0){
//if the nth root is an odd number, you don't get an imaginary number
//neg*neg=pos, but neg*neg*neg=neg
//so you can simply make an odd nth root of a negative number, a negative number
result = result*-1;
}else if(neg){
//if the nth root is an even number that is not 2, things get more complex
//if someone wants to calculate this further, they can
//i'm just going to stop at *n√-1 (times the nth root of -1)
//you should also check if the functions return value contains * or √, before continuing any calculations
result += '*'+n+√+'-1';
}
return result;
}
答案 7 :(得分：0)
嗯，我知道这是一个老问题。但是，基于SwiftNinjaPro的答案，我简化了功能并修复了一些NaN问题。注意：此函数使用了ES6功能，箭头函数和模板字符串以及指数。因此，它可能在较旧的浏览器中不起作用：
df2.head()
col1 col2
0 NM_000000_foo bar
1 NM_000001 foo
示例：
Math.numberRoot = (x, n) => {
return (((x > 1 || x < -1) && n == 0) ? Infinity : ((x > 0 || x < 0) && n == 0) ? 1 : (x < 0 && n % 2 == 0) ? `${((x < 0 ? -x : x) ** (1 / n))}${"i"}` : (n == 3 && x < 0) ? -Math.cbrt(-x) : (x < 0) ? -((x < 0 ? -x : x) ** (1 / n)) : (n == 3 && x > 0 ? Math.cbrt(x) : (x < 0 ? -x : x) ** (1 / n)));
};
示例（虚数结果）：
Math.numberRoot(-64, 3); // Returns -4
答案 8 :(得分：0)
我已经写了一个算法，但是当您需要很多数字时，它会很慢：
https://github.com/am-trouzine/Arithmetic-algorithms-in-different-numeral-systems
NRoot(orginal, nthRoot, base, numbersAfterPoint);
该函数返回一个字符串。
例如
var original = 1000;
var fourthRoot = NRoot(original, 4, 10, 32);
console.log(fourthRoot);
//5.62341325190349080394951039776481