"看似简单的任务“ - 土豆瓦特
事实上。有很多小恶魔在这个问题的细节中闲逛。解决tho非常有趣。
编辑：此更新采用更具组合性的方法。以前有一个大函数，它包装了其他几个专有函数。相反，这次我们定义了通用的可重用函数，这些函数可用于多种任务。更多关于那些之后，我们看看自己...numToWords
// numToWords :: (Number a, String a) => a -> String
let numToWords = n => {
let a = [
'', 'one', 'two', 'three', 'four',
'five', 'six', 'seven', 'eight', 'nine',
'ten', 'eleven', 'twelve', 'thirteen', 'fourteen',
'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'
];
let b = [
'', '', 'twenty', 'thirty', 'forty',
'fifty', 'sixty', 'seventy', 'eighty', 'ninety'
];
let g = [
'', 'thousand', 'million', 'billion', 'trillion', 'quadrillion',
'quintillion', 'sextillion', 'septillion', 'octillion', 'nonillion'
];
// this part is really nasty still
// it might edit this again later to show how Monoids could fix this up
let makeGroup = ([ones,tens,huns]) => {
return [
num(huns) === 0 ? '' : a[huns] + ' hundred ',
num(ones) === 0 ? b[tens] : b[tens] && b[tens] + '-' || '',
a[tens+ones] || a[ones]
].join('');
};
// "thousands" constructor; no real good names for this, i guess
let thousand = (group,i) => group === '' ? group : `${group} ${g[i]}`;
// execute !
if (typeof n === 'number') return numToWords(String(n));
if (n === '0') return 'zero';
return comp (chunk(3)) (reverse) (arr(n))
.map(makeGroup)
.map(thousand)
.filter(comp(not)(isEmpty))
.reverse()
.join(' ');
};
以下是依赖项：
您会注意到这些几乎不需要任何文档，因为它们的意图立即清晰。 可能是唯一一个需要花点时间消化的，但它真的不是太糟糕。另外，函数名称为我们提供了一个很好的指示，它可能是我们以前遇到的一个函数。chunk
const arr = x => Array.from(x);
const num = x => Number(x) || 0;
const str = x => String(x);
const isEmpty = xs => xs.length === 0;
const take = n => xs => xs.slice(0,n);
const drop = n => xs => xs.slice(n);
const reverse = xs => xs.slice(0).reverse();
const comp = f => g => x => f (g (x));
const not = x => !x;
const chunk = n => xs =>
isEmpty(xs) ? [] : [take(n)(xs), ...chunk (n) (drop (n) (xs))];
“所以这些让它变得更好？”
查看代码是如何显著清理的
// NEW CODE (truncated)
return comp (chunk(3)) (reverse) (arr(n))
.map(makeGroup)
.map(thousand)
.filter(comp(not)(isEmpty))
.reverse()
.join(' ');
// OLD CODE (truncated)
let grp = n => ('000' + n).substr(-3);
let rem = n => n.substr(0, n.length - 3);
let cons = xs => x => g => x ? [x, g && ' ' + g || '', ' ', xs].join('') : xs;
let iter = str => i => x => r => {
if (x === '000' && r.length === 0) return str;
return iter(cons(str)(fmt(x))(g[i]))
(i+1)
(grp(r))
(rem(r));
};
return iter('')(0)(grp(String(n)))(rem(String(n)));
最重要的是，我们在新代码中添加的实用程序函数可以在应用的其他位置使用。这意味着，作为以这种方式实现的副作用，我们可以免费获得其他函数。额外的苏打水！numToWords
一些测试
console.log(numToWords(11009));
//=> eleven thousand nine
console.log(numToWords(10000001));
//=> ten million one
console.log(numToWords(987));
//=> nine hundred eighty-seven
console.log(numToWords(1015));
//=> one thousand fifteen
console.log(numToWords(55111222333));
//=> fifty-five billion one hundred eleven million two hundred
// twenty-two thousand three hundred thirty-three
console.log(numToWords("999999999999999999999991"));
//=> nine hundred ninety-nine sextillion nine hundred ninety-nine
// quintillion nine hundred ninety-nine quadrillion nine hundred
// ninety-nine trillion nine hundred ninety-nine billion nine
// hundred ninety-nine million nine hundred ninety-nine thousand
// nine hundred ninety-one
console.log(numToWords(6000753512));
//=> six billion seven hundred fifty-three thousand five hundred
// twelve
可运行演示
const arr = x => Array.from(x);
const num = x => Number(x) || 0;
const str = x => String(x);
const isEmpty = xs => xs.length === 0;
const take = n => xs => xs.slice(0,n);
const drop = n => xs => xs.slice(n);
const reverse = xs => xs.slice(0).reverse();
const comp = f => g => x => f (g (x));
const not = x => !x;
const chunk = n => xs =>
isEmpty(xs) ? [] : [take(n)(xs), ...chunk (n) (drop (n) (xs))];
// numToWords :: (Number a, String a) => a -> String
let numToWords = n => {
let a = [
'', 'one', 'two', 'three', 'four',
'five', 'six', 'seven', 'eight', 'nine',
'ten', 'eleven', 'twelve', 'thirteen', 'fourteen',
'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'
];
let b = [
'', '', 'twenty', 'thirty', 'forty',
'fifty', 'sixty', 'seventy', 'eighty', 'ninety'
];
let g = [
'', 'thousand', 'million', 'billion', 'trillion', 'quadrillion',
'quintillion', 'sextillion', 'septillion', 'octillion', 'nonillion'
];
// this part is really nasty still
// it might edit this again later to show how Monoids could fix this up
let makeGroup = ([ones,tens,huns]) => {
return [
num(huns) === 0 ? '' : a[huns] + ' hundred ',
num(ones) === 0 ? b[tens] : b[tens] && b[tens] + '-' || '',
a[tens+ones] || a[ones]
].join('');
};
let thousand = (group,i) => group === '' ? group : `${group} ${g[i]}`;
if (typeof n === 'number')
return numToWords(String(n));
else if (n === '0')
return 'zero';
else
return comp (chunk(3)) (reverse) (arr(n))
.map(makeGroup)
.map(thousand)
.filter(comp(not)(isEmpty))
.reverse()
.join(' ');
};
console.log(numToWords(11009));
//=> eleven thousand nine
console.log(numToWords(10000001));
//=> ten million one
console.log(numToWords(987));
//=> nine hundred eighty-seven
console.log(numToWords(1015));
//=> one thousand fifteen
console.log(numToWords(55111222333));
//=> fifty-five billion one hundred eleven million two hundred
// twenty-two thousand three hundred thirty-three
console.log(numToWords("999999999999999999999991"));
//=> nine hundred ninety-nine sextillion nine hundred ninety-nine
// quintillion nine hundred ninety-nine quadrillion nine hundred
// ninety-nine trillion nine hundred ninety-nine billion nine
// hundred ninety-nine million nine hundred ninety-nine thousand
// nine hundred ninety-one
console.log(numToWords(6000753512));
//=> six billion seven hundred fifty-three thousand five hundred
// twelve
您可以使用 babel 转译代码.js如果您想查看 ES5 变体
