获取代数项系数 [英] Getting coefficients of algebraic term
问题描述
考虑到代数项的输入，我试图获取变量的系数.输入中唯一的运算符是+ -，并且只有一个变量.
Given a input of an algebraic term, I am trying to get the coefficients of the variables. The only operators in the input is + - and only one variable.
示例:
2x^2+3x+4 => [ 2, 3, 4 ]
3-x => [ -1, 3 ]
x^2+x => [ 1, 1, 0 ]
x+x^3 => [ 1, 0, 1, 0 ]
无效输入:
2x^2+2x^2
这是我第一次尝试:
var str = " 2x^4-1+9x^3-100x^2";
function getCoeff(term) {
var nterm = (term.replace(/[^0-9|-]x(?!\^)/g,"1x")).replace(/[^0-9|\+]x(?!\^)/g,"-1x"); // ==> Replace ‘-/x’ with ‘-/1x’
for ( var i = 0; i < 10; i++ ) { // ==> Loop true the regexs to replace all ‘x^n’ to ‘1x^n’
var re = new RegExp('[^0-9|\-]x\\^' + i); // ==> Regex for x^n
var re2 = new RegExp('[^0-9|]x\\^' + i); // ==> Regex for -x^n
nterm = (nterm.replace(re,"1x^" + i)).replace(re2,"-1x^" + i); }
for ( var m = 10; m > 1; m-- ) { // ==> Get the coefficients of ax^n in descending order
var re3 = new RegExp('\\W?\\d+(?=x\\^' + m + ')' );
if ( nterm.match(re3) === null ) {
var result = "";
} else {
result += ((nterm.match(re3)+', ').toString()).replace(/\+/g,""); }}
if ( nterm.match(/\W?\d+(?=x(?!\^))/g) === null ) { // Regex for coefficient x
var result2 = "";
} else {
result2 = ((nterm.match(/\W?\d+(?=x(?!\^))/g)).toString()).replace(/\+/g,"") + ','; }
if ( nterm.match(/[^\^]\d+(?!\d|x)/g) === null ) { // Regex for constant
var result3 = "";
} else {
result3 = ((nterm.match(/[^\^]\d+(?!\d|x)/g)).toString()).replace(/\+/g,""); }
console.log(('[' + ' ' + result + result2 + ' ' + result3 + ']' ).replace(/\s/g,"")); }
getCoeff(str)
问题:
- 缺少
x term时不起作用.
- Does not work when there is a missing
x term.
例如:x^4 + x + 1 ==> Expected: [1, 0, 0, 1, 1] ==> Actual: [ 1, 1 ]
-
x应该返回[ 1,0 ]，但是它返回[ 1, ]
xshould return[ 1,0 ], but it returns[ 1, ]
这是我第二次尝试.
var str = "-999x^2+x^3+x+3";
function getCoeff(string) {
if ( string.charAt(0) === 'x' ) { // If the first term is x, because of my regex it needs a space to match it
string = ' ' + string;
}
for ( var i = 0; i < 10; i++ ) { // ==> Loop true the regexs to replace all ‘x^n’ to ‘1x^n’
var re = new RegExp('[^0-9|\-]x\\^' + i);
var re2 = new RegExp('[^0-9|]x\\^' + i);
string = (string.replace(re,"+1x^" + i)).replace(re2," -1x^" + i); }
var final = string.replace(/-/g,'+-'); // ==> Spilt(‘x’) later so to retain the -ve sign
final = (final.replace(/[^0-9|-]x(?!\^)/g,"+1x")).replace(/[^0-9|+]x(?!\^)/g,"-1x"); // ==> Replace ‘-/x’ with ‘-/1x’
final = final.replace(/[^\^](\d+(?!\d|x))/g,'+$1x^0'); // ==> Replace ‘c’ with ‘cx^0’
final = final.replace(/x(?!\^)/g, "x^1"); // ==> Replace ‘x’ with ‘x^1’
final = final.split('+'); // ==> Right now array looks something like this [ ax^(n), bx^(n-1), … yx^1, zx^0]
final = final.filter(function(entry) { return entry.trim() !== ''; }); // Sorts array by the number behind in descending order
var reS = /^-?\d+/,
reE = /\d+$/;
var result = final.sort(function(a, b) {
a = reE.exec(a);
b = reE.exec(b);
return b - a;
}).reduce(function(res, str, i) {
var gap = reE.exec(final[i - 1]) - reE.exec(str);
if(gap > 0)
while(--gap) res.push(0);
res.push(+reS.exec(str));
return res;
}, []); // Return the coefficients
console.log("Result:", result);
}
getCoeff(str);
问题:
-
有没有一种方法可以在大多数情况下不使用正则表达式?
Is there a way to do it without using regular expressions mostly?
如何解决此问题?如果没有常数项
How do I fix this problem? When no constant term is
getCoeff(x^3) ==> [ 1 ] , when it should give [ 1, 0, 0 ]
如何提高代码效率?
如何使用正则表达式匹配x^n术语而不匹配-x^n术语?这是我当前的名字:[^0-9|\-]x\\^' + i，但是它前面需要一个空格.
How to match x^n terms but not -x^n terms using regex? This is my current one: [^0-9|\-]x\\^' + i, but it needs a space in front of it.
参考:
推荐答案
function getCoef(str) {
str = str.replace(/\s+/g, ""); // remove spaces (optional)
var parts = str.match(/[+\-]?[^+\-]+/g); // get the parts: see explanation bellow
// accumulate the results
return parts.reduce(function(res, part) { // for each part in parts
var coef = parseFloat(part) || +(part[0] + "1") || 1;// the coeficient is the number at the begining of each part (34x => 34), if there is no number it is assumed to be +/-1 depending on the sign (+x^2 => +1)
var x = part.indexOf('x'); // the index of "x" in this part (could be -1 if there isn't)
// calculating the power of this part
var power = x === -1 ? // if the index of "x" is -1 (there is no "x")
0: // then the power is 0 (Ex: -2)
part[x + 1] === "^" ? // otherwise (if there is an "x"), then check if the char right after "x" is "^", if so...
+part.slice(x + 2) : // then the power is the number right after it (Ex: 55x^30)
1; // otherwise it's 1 (Ex: 55x)
res[power] = (res[power] || 0) + coef; // if we have already encountered this power then add this coeficient to that, if not then just store it
return res;
}, {});
}
/** TESTS **/
[
"-999x^2 + x^3 + x + 3", "5x + 3 - 10x", "55x^3 + 1", "55.12x^4 + 20x^4 - 120x^4"
].forEach(function(test) {
console.log(test, "=>", getCoef(test));
});
输出:
函数getCoef的结果将是以下格式的对象:
The result of the function getCoef will be an object in this format:
{
"power": "coeficient",
"other power": "other coeficient",
...
}
说明:
-
str = str.replace(/\s+/g, "");:
删除空格(明显).
-
var parts = str.match(/[+\-]?[^+\-]+/g);:
将字符串分成几部分.字符串"-5x^2-3+10x"将返回["-5x^2", "-3", "+10x"].正则表达式将寻找:
Split the string into parts. The string "-5x^2-3+10x" will return ["-5x^2", "-3", "+10x"]. The regex will look for:
[+\-]? : a "+" or "-" sign (if any)
[^+\-]+ : anything that isn't a "+" nor "-" (get everything up until the new + or - or the end is reached)
g : to get all parts
-
var coef = parseFloat(part) || +(part[0] + "1") || 1;:
使用以下方法获取该部分的系数:
Get the coeficient of this part using:
parseFloat : for parts that have a number before "x" like: "+55x", "-34.22x^11", "5x", ...
+(part[0] + 1) : for parts that have only a sign like: "+x", "-x^2", ... (get the sign part[0] concatinate it with "1" and then cast the result into a number using binary +)
1 : for parts that doesn't have a number nor a sign like "x^3", "x", ...
请注意，使用上述方法，将假定像"0x^4"这样的零件的系数为1(但我不明白为什么无论如何都需要空系数)！
just note that parts like "0x^4" will be assumed as having a coeficient of 1 using the above (but I don't see why one will need a null coeficient anyway)!
-
var x = part.indexOf('x');:
获取部分中字符"x"的索引，以区分具有"3x"，"x^11"，...等字符的部分和不具有"+5"，...的部分.
Get the index of the character "x" in the part to distinguish between parts that have it like "3x", "x^11", ... and parts that doesn't like "+5", ...
-
var power = ...:
如果部件(x === -1)中没有"x"，则该部件的功效为0.
If there isn't an "x" in the part (x === -1) then the power of this part is 0.
否则(存在"x")，然后我们检查"x"(part[x + 1])之后的字符是否为"^"，如果是，则幂是其后的任意数字(将其切掉)字符串part.slice(x + 2)的一部分，并使用一元+将其转换为数字)，如果"x"之后没有"^"，则幂为1.
Otherwise (an "x" is present), then we check if the character right after "x" (part[x + 1]) is "^", if it is then the power is whaterver number that comes after it (cut that bit of the string part.slice(x + 2) and cast it to number using unary +), if there isn't a "^" after "x", then the power is 1.
-
res[power] = (res[power] || 0) + coef;:
将刚刚计算出的系数coef添加到该功率的已累积系数中(如果没有累积，则使用0).
Add the coeficient coef that has just been calculated to the already accumulated coeficient of this power (if nothing is accumulated then use 0).
此行可以这样简化:
if(res[power]) // if we already encountered this power in other parts before
res[power] += coef; // then add this coeficient to the sum of those previous coeficients
else // otherwise
res[power] = coef; // start a new sum initialized with this coeficient
这使得在相同的字符串中多次包含相同的幂成为可能，例如:"5x + 10x + 1 + x"，...
this makes it possible to include the same power more than one time in the same string like: "5x + 10x + 1 + x", ...
将生成的对象转换为所需的数组:
因此:
{
"3": 7,
"0": 19
}
将是:
[7, 0, 0, 19]
var ret = { "3": 7, "0": 19 }; // the returned object
var powers = Object.keys(ret); // get all the powers (in this case [3, 0])
var max = Math.max.apply(null, powers); // get the max power from powers array (in this case 3)
var result = [];
for(var i = max; i >= 0; i--) // from the max power to 0
result.push(ret[i] || 0); // if that power has a coeficient then push it, otherwise push 0
console.log(result);
这篇关于获取代数项系数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
