Javascript 将拼写的数字(22)转换为数字(22)
我正在尝试将拼写的数字字符串转换为数字/数字字符串 我设法做了相反的事情,如下所示,但当我尝试反转它时,我没有很好地工作,它返回undefined 00 undefined for'two'Javascript 将拼写的数字(22)转换为数字(22),javascript,math,numbers,numeric,Javascript,Math,Numbers,Numeric,我正在尝试将拼写的数字字符串转换为数字/数字字符串 我设法做了相反的事情,如下所示,但当我尝试反转它时,我没有很好地工作,它返回undefined 00 undefined for'two' var th = [ "", "thousand", "million", "billion", "trillion" ]; var dg = [ "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"
var th = [ "", "thousand", "million", "billion", "trillion" ];
var dg = [ "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" ];
var tn = [ "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" ];
var tw = [ "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" ];
function toWords(s) {
s = s.toString();
s = s.replace(/[\, ]/g, "");
if (s != parseFloat(s)) return "not a number";
var x = s.indexOf(".");
if (x == -1) x = s.length;
if (x > 15) return "too big";
var n = s.split("");
var str = "";
var sk = 0;
for (var i = 0; i < x; i++) {
if ((x - i) % 3 == 2) {
if (n[i] == "1") {
str += tn[Number(n[i + 1])] + " ";
i++;
sk = 1;
} else if (n[i] != 0) {
str += tw[n[i] - 2] + " ";
sk = 1;
}
} else if (n[i] != 0) {
str += dg[n[i]] + " ";
if ((x - i) % 3 == 0) str += "hundred ";
sk = 1;
}
if ((x - i) % 3 == 1) {
if (sk) str += th[(x - i - 1) / 3] + " ";
sk = 0;
}
}
if (x != s.length) {
var y = s.length;
str += "point ";
for (var i = x + 1; i < y; i++) str += dg[n[i]] + " ";
}
return str.replace(/\s+/g, " ");
}
有人能帮我在javascript中扭转这种局面吗?或者给我一些建议,告诉我怎样才能完成我想做的事情
我尝试过使用解析器,但我觉得我正在定义1=1,2=2,。。。十六=16,十七=17,。。。七十三等于七十三。所以本质上是用数字对应物定义许多数字字符串。必须有更好的方法来做到这一点
ML在我脑海中闪过,但为什么要学习一些东西呢?一是1。我没有得到各种输入并试图将其与单个输出关联/匹配。例如,hello,hey,hi,以及任何类似的问候语。幸运的是,除了十以外,英语的编号是一致的。您需要定义从“一”到“十九”的字符串,从“二十”到“九十”、“百”、“千”、“百万”等所有十个字符串 然后,对于解析器,将字符串解析为标记,基本上将其拆分。如果愿意,您可以将令牌转换为数字,如果愿意,也可以将其保留为字符串。我在这里用数字,因为它有点短 因此,如果您有一个四千八百八十五的字符串,您可以将其解析为令牌: 4 1000 100 80 5 您可以丢弃单词“and”,以及逗号和其他杂音 然后,您可以在下面非常简化的伪代码中从后面处理这个字符串 总数=0; 乘法器=1
repeat
Number = ReadNumber || 1;
Total += Multiplicator * Number;
Multiplicator = ReadMultiplicator;
乘法器是10,20,30。。100, 1000, 1000000
数字是1。。19这些是可选的。如果没有找到编号,则将其视为已读取编号
因此,最后,您应该将字符串视为
4100011001805
应按以下方式处理:
5+1*80+1*100+4*1000
我说伪代码非常简单。首先,你的解析需要能够“窥视”前方。如果前面的标记不是数字,则它必须是乘法器,因此不应跳过它。
此外,您还需要一些额外的支票。毕竟,100之前的数字是特殊的。您不希望解析字符串384到244
然后你需要一些额外的图层。八万五千怎么样?看起来你不应该在一个乘法器之前只读取一个数字,但实际上所有小于该乘法器的标记都应该一起解析,所以
十六万
应该被解析为
1100 160 1000
您可以稍后在该部分中添加额外的检查,尽管您可以先不添加检查。毕竟,亿实际上是十亿,但它将正确解析,甚至给出正确的结果
所以,我认为解析正确的字符串是完全可行的。检查异常会使它变得更加困难。2300是现成的,但奇怪的是,你也可以写一万两千三百,结果是12.300。但是,我想,像这样的检查一开始并不严格需要。如果您只需要整数,您可以使用键值对象从单个单词中构建数字-
function wordsToNumber(s){
var w= s.toLowerCase().replace(/(thousand|[mbr]illion),?/g, '$1,');
var n= 0, nArray= w.split(/,/),
sub= 0, tem, segment,
nwords={
billion: 1e9, eight: 8, eighteen: 18, eighty: 80,
eleven: 11, fifteen: 15, fifty: 50, five: 5, forty: 40,
four: 4, fourteen: 14, hundred: 100, million: 1e6,
nine: 9, nineteen: 19, ninety: 90, one: 1, quadrillion: 1e15,
seven: 7, seventeen: 17, seventy: 70, six: 6, sixteen: 16,
sixty: 60, ten: 10, thirteen: 13, thirty: 30, thousand: 1e3,
three: 3, trillion: 1e12,twelve: 12, twenty: 20, two: 2, zero: 0
};
while(nArray.length){
sub= 0;
segment= nArray.shift().match(/[a-z]+/g) || [];
segment.forEach(function(w2){
tem= nwords[w2.trim()];
if(isFinite(tem)){
if(tem<100) sub+= tem;
else sub*= tem;
}
});
n+= sub;
}
return n;
}
var str= 'twenty two thousand one hundred and fifty one';
wordsToNumber(str);
/* returned value: (Number)
22151
*/
为了获得更多功能,您可以编写更多代码-
<!doctype html>
<html lang="en">
<head>
<meta charset= "utf-8">
<title>Number Words</title>
<style>
</style>
<script>
Math.fraction= function(n, prec, up){
var s= String(n),
p= s.indexOf('.');
if(p== -1) return s;
var i= Math.floor(n) || '',
dec= s.substring(p),
m= prec || Math.pow(10, dec.length-1),
num= up=== 1? Math.ceil(dec*m): Math.round(dec*m),
den= m,
g= Math.gcd(num, den);
if(den/g== 1) return String(i+num);
if(i) i= i+' and ';
return i+ String(num/g)+'/'+String(den/g);
}
Number.fromFraction= function(str, prec){
var rx=/(\d+)\/(\d+)/, dec= 0, I= 0,
M= rx.exec(str);
if(M){
if(M.index>0) I= parseFloat(str);
if(M[1]) dec= M[1]/M[2];
if(!prec) prec= str.length+1;
if(typeof prec== 'number') dec= +(dec.toFixed(prec));
return I+dec;
}
return parseFloat(str);
}
Number.fromRoman= function(s){
s= String(s).toUpperCase();
if(s.length>15 || /[^MDCLXVI]/.test(s)) return NaN;
var L= s.length, sum= 0, i= 0, next, val,
R={
M: 1000, D: 500, C: 100, L: 50, X: 10, V: 5, I: 1
};
while(i<L){
val= s.charAt(i++);
if(!R[val]) return NaN;
val= R[val];
next= R[(s.charAt(i) || 'N')] || 0;
if(next>val) val*= -1;
sum+= val;
}
if(sum.toRoman()== s) return sum;
return NaN;
}
Number.prototype.toRoman= function(){
var n= Math.floor(this), val, s= '', limit= 3999, i= 0,
v= [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1],
r= ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
if(n<1 || n>limit) return '';
while(i<13){
val= v[i];
while(n>= val){
n-= val;
s+= r[i];
}
if(n== 0) return s;
++i;
}
return '';
}
var NW={
/* numbers to words */
nameNumbers: function(str){
var nw= NW.numberWords,
s= (str+'').replace(/([^\d])(?=\.\d+)/g, '$1 0').trim();
var f, w= '', x, xs, n, ns, M,
rx=/(-?\d+)(\.(\d+)([eE]([+-]?\d+))?)?/g,
rx2=/(\d+)(\.(\d+)([eE]([+-]?\d+))?)?/,
rq=/((\d+)? +(and +)?)?(\d+)\/(\d+)/g;
s= s.replace(rq, function(a){
return NW.printFraction(a);
});
s= s.replace(rx, function(q){
n= +q;
ns= n+'';
M= rx2.exec(ns) || [];
w= (n<0)? 'minus ': '';
if(M[1]) w+= nw(M[1]);
if(M[3]){
f= M[3].split('').map(function(itm){
return nw(itm);
});
w+= ' point '+ f.join(' ');
if(M[5]){
x= +M[5];
xs= x<0? 'minus ': '';
xs+= NW.ordinal(nw(x));
w+= ' times ten to the '+xs;
}
}
return w || q;
});
return s.replace(/ {2,}/g, ' ');
},
numberWords: function numberWords(x){
var n= +x;
if(isNaN(n) || n%1) return ''+x;
n= Math.abs(n);
var i= 0, p, prefix= [], num, rem,
NK= NW.numberwords_array,
w= NK[0], w1= NK[1], w2= NK[2], mag= NK[3];
while(n>99){
x= mag[i];
if(n>= x){
p= Math.floor(n/x);
n%= x;
prefix.push(numberWords(p)+w2[i]);
}
++i;
}
if(prefix.length){
prefix= prefix.join(', ');
if(n) prefix+= ' and ';
else return prefix;
}
else prefix= '';
if(n<20) num= w[n];
else{
num= w1[Math.floor(n/10)];
rem= n%10;
if(rem) num+= ' '+w[rem];
}
return prefix+num;
},
numberwords_array: [
['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven',
'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen',
'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'],
['', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty',
'seventy', 'eighty', 'ninety'],
[' quadrillion', ' trillion', ' billion', ' million',
' thousand', ' hundred'],
[1e15, 1e12, 1e9, 1e6, 1e3, 100]
],
ordinal: function(s){
var ax, rx, suffix, str= ' '+s.trim();
ax= str.lastIndexOf(' ');
suffix= str.substring(ax).toLowerCase();
if(str.charAt(str.length-1)== 'y'){
return str.slice(0, -2)+'tieth';
}
if(ax>1) str= str.substring(0, ax);
else str= ' ';
switch(suffix.trim()){
case 'one': return str+' first';
case 'two': return str+' second';
case 'three': return str+' third';
case 'five': return str+' fifth';
case 'eight': return str+' eighth';
case 'nine': return str+' ninth';
case 'twelve': return str+' twelfth';
default: return str+suffix+'th';
}
},
printFraction: function(str){
var num, RN= NW.numberWords,
M=/((\d+) +)?(\d+) *\/ *(\d+)/g.exec(str.trim());
if(!M) return str;
num= M[2]? RN(M[2])+' and ': '';
num+= RN(+M[3])+'-';
if(M[4]=== '2') num+= 'half';
else if(M[4]=== '4') num+= 'quarter';
else num+= NW.ordinal(RN(+M[4]));
if(M[3]!== '1') num+= 's';
return num;
},
/* words to numbers */
reDigit: function(n){
var sub= [], tem, d= n.match(/[a-z]+/g), L= d.length-1,
nwords= NW.wordnumber_keys;
d.map(function(w, i){
tem= nwords[w];
if(isFinite(tem)){
if(tem<20 || i== L) sub.push(tem+'');
else sub.push((tem+'').charAt(0));
}
});
return sub.join('');
},
reFraction: function(str){
var M, w= 0, n, d, f2= 0, prec, ax;
ax= str.lastIndexOf(' and ');
if(ax== -1){
ax= 0;
w= 'zero';
}
else{
w= str.substring(0, ax);
ax+= 4;
}
M= str.substring(ax).split('-');
n= M[0];
d= M[1];
if(n && d){
d= NW.wordsToNumber(d);
n= NW.wordsToNumber(n);
return [w, n/d];
}
return [w];
},
wordNumber: function(str){
var n= 0, sign= 1, whole= 0, dec= 0, frac,
suffix= 0, exp= 1, exsign= 1, prec, pt,
rx1= /(illion|thousand|hundred)/i,
rx2=/( +times ten to the( +minus)? +)/g,
rx3=/(point|times|[^a-z, ])/,
s= str.toLowerCase().trim().replace(/ieth?$/, 'y').replace(/(ths?|s)$/, '');
if(s.indexOf('minus ')== 0){
sign= -1;
s= s.substring(5);
}
if(!rx3.test(s)){
n= (!rx1.test(s))? NW.reDigit(s): NW.wordsToNumber(s);
return n*sign;
}
whole= s;
var pt= s.split('point ');
dec= pt[1];
if(dec){
whole= pt[0];
temp= rx2.exec(dec);
if(temp!= null){
if(temp[2]) exsign= -1;
suffix= dec.substring(rx2.lastIndex);
dec= dec.substring(0, temp.index);
suffix= NW.wordsToNumber(suffix);
exp= Math.pow(10, suffix*exsign);
}
dec= NW.reDigit(dec);
prec= dec.length+1;
dec= +('0.'+dec);
}
else if(whole.indexOf('-')!= -1){
pt= NW.reFraction(whole);
whole= pt[0];
dec= pt[1];
}
n= NW.wordsToNumber(whole.replace(/ +and +/g, ' '));
if(dec){
n+= dec;
if(!prec) frac= Math.min(String(n).length, 15);
n= n.toPrecision(frac);
}
if(exp!== 1) return sign*((n*exp).toExponential(prec));
return sign*n;
},
wordnumber_keys:{
billion: 1e9, eigh: 8, eight: 8, eighteen: 18, eightt: 8, eighty: 80,
eleven: 11, fif: 5, fifteen: 15, fifty: 50, first: 1, five: 5, forty: 40,
four: 4, fourteen: 14, half: 2, hundred: 100, million: 1e6, nin: 9,
nine: 9, nineteen: 19, ninety: 90, one: 1, quadrillion: 1e15, quarter: 4,
second: 2, seven: 7, seventeen: 17, seventy: 70, six: 6, sixteen: 16,
sixty: 60, ten: 10, third: 3, thirteen: 13, thirty: 30, thousand: 1e3,
three: 3, trillion: 1e12, twelf: 12, twelve: 12, twenty: 20, two: 2, zero: 0
},
wordsToNumber: function(s){
var w= s.toLowerCase().replace(/(thousand|[mbr]illion),?/g, '$1,');
var n= 0, nArray= w.split(/,/),
sub= 0, tem, segment,
nwords= NW.wordnumber_keys;
while(nArray.length){
sub= 0;
segment= nArray.shift().match(/[a-z]+/g) || [];
segment.forEach(function(w2){
tem= nwords[w2.trim()];
if(isFinite(tem)){
if(tem<100) sub+= tem;
else sub*= tem;
}
});
n+= sub;
}
return n;
}
}
onload=function(){
var nw= NW, v2= [], v;
var A= ['150', '0.12', 'fourteen hundred and ninety two', '65 1/16', '65.0045',
'thirty three and one- third', '2012', '98.6', 'MCMLIX', '897456971.25',
'1 1/2', '12 3/4', '-9', '33 1/3', Math.PI, 'nine and three- quarters',
'seventy eight point six four', 'XIX', '1.1550046210e+17', '1.222e3', 'minus two hundred',
'4.5e-10', '3.125e-3', '1.56760e+25'].map(function(sv){
if(/^[IVXLCDM]+$/.test(sv)) sv= Number.fromRoman(sv);
if(/^[a-z ,-]+$/i.test(sv)){
v2= nw.wordNumber(sv);
}
else v2= nw.nameNumbers(sv);
return '<p\>'+sv+'= '+v2+'</p\>';
});
document.getElementById('resultset').innerHTML=A.join('\n');
}
</script>
</head>
<body>
<div id="resultset">
</div>
</body>
</html>
下面是其他答案之一的实现: 我喜欢monkeypatching是的,是的,我知道…,但它可以很容易地从这个实现中删除 变量编号\u名称={ a:10,十亿:1e9,8:8,18:18,80:80, 11:11,15:15,50:50,5:5,40:40, 四:四四:一四亿:一亿:一六, 9:9,19:19,90:90,1:1,万亿元, 七点七,十七点十七,七十点七十,六点六,十六点十六, 六十:六十,十:十,十三:十三,三十:三万,一千, 三:3,万亿:1e12,十二:12,二十:20,二:2,零:0 } Array.prototype.last=函数{ 返回此[this.length-1]; } String.prototype.toNumber=函数{ var stripped=this.toLowerCase.replace/\-\124;和/g',.split'; var numbers=stripped.mapfunctionword{returnnumber\u NAMES[word]}; 如果numbers.somefunctionnumber{returnnumber==null}返回NaN; var parseNumbers=functionnumbers{ var合计=0; var readNumber=functiondef{ 返回编号。上次<20?编号。pop:def; } var readmagnity=函数{ 返回编号.last>19?编号.pop:0; } 总计+=读取编号0; 而数字。长度{ var震级=读取震级; var子_数=[]; 而numbers.length&&numbers.last<幅值{ sub_numbers.unshiftnumbers.pop; } 变量编号=子编号。长度?ParseNumbers子编号:1; 总数+=震级*数量; } 返回总数; } 返回ParseNumbersNumber; }
221.toNumber//2201“未定义”表示您没有将代码添加到问题中:在说出来的数字中,百通常跟在和后面,就像在一百零六中一样,但当数字是偶数百时,就不是了,例如七百。@RobG我想的逻辑是一个拼写数字在一个拼写数字之前,然后将它们相乘七百就变成了7*100=700。如果是在之后,那么把它们加起来,105变成100+5,这样就不需要和了。试过谷歌吗?@user3743069-907变成了九百零七,但通常被称为九百零七。一个问题。当读取数字时,我在您的逻辑中看到,当数字未读入时,使用1,但在100和80之前读入1。为什么?为什么不是141000110018015或者实际上是4100010805?每次都在读数字,为什么会有数字呢?我知道乘以1是如何使解析的数字正确的,但是这个数字是从哪里来的,为什么来的?你只需要对乘法器之前的数字这样做,但是想想看,也许这一步根本没有必要。毕竟,你写的不是“一百”,而是“一百”,不像其他语言。不会处理像2300这样的事情,但除此之外,它很好。我没想到你用这么少的代码就能做到。非常优雅。