http://www.easysurf.cc/cnvert18.htm和http://www.calculatorsoup.com/calculators/conversions/numberstowords.php等网站尝试将数字字符串转换为英文字符串，但它们会提供自然的声音输出。
例如，在http://www.easysurf.cc/cnvert18.htm上：
[in]: 100456
[out]: one hundred thousand four hundred fifty-six
这个网站好一点，http://www.calculator.org/calculate-online/mathematics/text-number.aspx：
[in]: 100456
[out]: one hundred thousand, four hundred and fifty-six
[in]: 10123124001
[out]: ten billion, one hundred and twenty-three million, one hundred and twenty-four thousand, one
但它在某些时候破裂了：
[in]: 10000000001
[out]: ten billion, , , one
我已经编写了自己的版本，但它涉及很多规则，从http://pastebin.com/WwFCjYtt开始上限为10亿：
import codecs
def num2word (num):
ones = {1:"one",2:"two",3:"three",4:"four",
5:"five",6:"six",7:"seven",8:"eight",
9:"nine",0:"zero",10:"ten"}
teens = {11:"eleven",12:"twelve",13:"thirteen",
14:"fourteen",15:"fifteen"}
tens = {2:"twenty",3:"thirty",4:"forty",
5:"fifty",6:"sixty",7:"seventy",
8:"eighty",9:"ninety"}
lens = {3:"hundred",4:"thousand",6:"hundred",7:"million",
8:"million", 9:"million",10:"billion"#,13:"trillion",11:"googol",
}
if num > 999999999:
return "Number more than 1 billion"
# Ones
if num < 11:
return ones[num]
# Teens
if num < 20:
word = ones[num%10] + "teen" if num > 15 else teens[num]
return word
# Tens
if num > 19 and num < 100:
word = tens[int(str(num)[0])]
if str(num)[1] == "0":
return word
else:
word = word + " " + ones[num%10]
return word
# First digit for thousands,hundred-thousands.
if len(str(num)) in lens and len(str(num)) != 3:
word = ones[int(str(num)[0])] + " " + lens[len(str(num))]
else:
word = ""
# Hundred to Million
if num < 1000000:
# First and Second digit for ten thousands.
if len(str(num)) == 5:
word = num2word(int(str(num)[0:2])) + " thousand"
# How many hundred-thousand(s).
if len(str(num)) == 6:
word = word + " " + num2word(int(str(num)[1:3])) + \
" " + lens[len(str(num))-2]
# How many hundred(s)?
thousand_pt = len(str(num)) - 3
word = word + " " + ones[int(str(num)[thousand_pt])] + \
" " + lens[len(str(num))-thousand_pt]
# Last 2 digits.
last2 = num2word(int(str(num)[-2:]))
if last2 != "zero":
word = word + " and " + last2
word = word.replace(" zero hundred","")
return word.strip()
left, right = '',''
# Less than 1 million.
if num < 100000000:
left = num2word(int(str(num)[:-6])) + " " + lens[len(str(num))]
right = num2word(int(str(num)[-6:]))
# From 1 million to 1 billion.
if num > 100000000 and num < 1000000000:
left = num2word(int(str(num)[:3])) + " " + lens[len(str(num))]
right = num2word(int(str(num)[-6:]))
if int(str(num)[-6:]) < 100:
word = left + " and " + right
else:
word = left + " " + right
word = word.replace(" zero hundred","").replace(" zero thousand"," thousand")
return word
print num2word(int(raw_input("Give me a number:\n")))
如何制作我已写入的脚本接受> billion？
还有其他方法可以获得相同的输出吗？
我的代码可以用不那么冗长的方式编写吗？
答案 0 :(得分：3)
解决此问题的更一般方法是使用重复划分（即divmod）并仅对必要的特殊/边缘情况进行硬编码。
例如，divmod(1034393, 1000000) -> (1, 34393)，因此您可以有效地找到数百万的数字，并留下余数用于进一步计算。
可能更具说明性的例子：divmod(1034393, 1000) -> (1034, 393)允许你从右边一次取下3位十进制数字的组。
在英语中，我们倾向于将数字组成三位数，并且适用类似的规则。这应该参数化，而不是硬编码。例如，＆＃34; 303＆＃34;可能是三亿三千三百三十三，三百三十三。除了后缀之外，逻辑应该是相同的，具体取决于你所处的位置。编辑：由于递归，看起来就像这样。
这是我所说的一种方法的部分示例，使用生成器并对整数进行操作，而不是在任何地方进行大量int(str(i)[..])。
say_base = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven',
'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen',
'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen']
say_tens = ['', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy',
'eighty', 'ninety']
def hundreds_i(num):
hundreds, rest = divmod(num, 100)
if hundreds:
yield say_base[hundreds]
yield ' hundred'
if 0 < rest < len(say_base):
yield ' and '
yield say_base[rest]
elif rest != 0:
tens, ones = divmod(rest, 10)
yield ' and '
yield say_tens[tens]
if ones > 0:
yield '-'
yield say_base[ones]
assert "".join(hundreds_i(245)) == "two hundred and forty-five"
assert "".join(hundreds_i(999)) == 'nine hundred and ninety-nine'
assert "".join(hundreds_i(200)) == 'two hundred'