我正在编写一些代码,将给定的数字转换为单词,这是我在谷歌搜索后得到的结果。但我认为对于这样一个简单的任务来说,这有点太长了。两个正则表达式和两个for循环,我想要更简单的东西。
我正试图在尽可能少的代码行中实现这一点。这是我到目前为止想出的:
有什么建议吗?
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) { // 0235
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,' ');
}此外,上述代码转换为英语编号系统,如百万/十亿,我需要南亚的编号系统,如拉赫斯和Crores。
发布于 2020-07-27 14:12:23
function numberToEnglish( n ) {
var string = n.toString(), units, tens, scales, start, end, chunks, chunksLen, chunk, ints, i, word, words, and = 'and';
/* Remove spaces and commas */
string = string.replace(/[, ]/g,"");
/* Is number zero? */
if( parseInt( string ) === 0 ) {
return 'zero';
}
/* Array of units as words */
units = [ '', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen' ];
/* Array of tens as words */
tens = [ '', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety' ];
/* Array of scales as words */
scales = [ '', 'thousand', 'million', 'billion', 'trillion', 'quadrillion', 'quintillion', 'sextillion', 'septillion', 'octillion', 'nonillion', 'decillion', 'undecillion', 'duodecillion', 'tredecillion', 'quatttuor-decillion', 'quindecillion', 'sexdecillion', 'septen-decillion', 'octodecillion', 'novemdecillion', 'vigintillion', 'centillion' ];
/* Split user argument into 3 digit chunks from right to left */
start = string.length;
chunks = [];
while( start > 0 ) {
end = start;
chunks.push( string.slice( ( start = Math.max( 0, start - 3 ) ), end ) );
}
/* Check if function has enough scale words to be able to stringify the user argument */
chunksLen = chunks.length;
if( chunksLen > scales.length ) {
return '';
}
/* Stringify each integer in each chunk */
words = [];
for( i = 0; i < chunksLen; i++ ) {
chunk = parseInt( chunks[i] );
if( chunk ) {
/* Split chunk into array of individual integers */
ints = chunks[i].split( '' ).reverse().map( parseFloat );
/* If tens integer is 1, i.e. 10, then add 10 to units integer */
if( ints[1] === 1 ) {
ints[0] += 10;
}
/* Add scale word if chunk is not zero and array item exists */
if( ( word = scales[i] ) ) {
words.push( word );
}
/* Add unit word if array item exists */
if( ( word = units[ ints[0] ] ) ) {
words.push( word );
}
/* Add tens word if array item exists */
if( ( word = tens[ ints[1] ] ) ) {
words.push( word );
}
/* Add 'and' string after units or tens integer if: */
if( ints[0] || ints[1] ) {
/* Chunk has a hundreds integer or chunk is the first of multiple chunks */
if( ints[2] || ! i && chunksLen ) {
words.push( and );
}
}
/* Add hundreds word if array item exists */
if( ( word = units[ ints[2] ] ) ) {
words.push( word + ' hundred' );
}
}
}
return words.reverse().join( ' ' );
}
// - - - - - Tests - - - - - -
function test(v) {
var sep = ('string'==typeof v)?'"':'';
console.log("numberToEnglish("+sep + v.toString() + sep+") = "+numberToEnglish(v));
}
test(2);
test(721);
test(13463);
test(1000001);
test("21,683,200,000,621,384");
https://stackoverflow.com/questions/14766951
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