<p>为了您的练习,我们采用了现有的nice解决方案<a href="http://www.daniweb.com/software-development/python/code/216839/number-to-word-converter-python" rel="nofollow">ref</a>将数字转换为单词,如下所示:</p>
<pre><code>def numToWords(num,join=True):
'''words = {} convert an integer number into words'''
units = ['','one','two','three','four','five','six','seven','eight','nine']
teens = ['','eleven','twelve','thirteen','fourteen','fifteen','sixteen', \
'seventeen','eighteen','nineteen']
tens = ['','ten','twenty','thirty','forty','fifty','sixty','seventy', \
'eighty','ninety']
thousands = ['','thousand','million','billion','trillion','quadrillion', \
'quintillion','sextillion','septillion','octillion', \
'nonillion','decillion','undecillion','duodecillion', \
'tredecillion','quattuordecillion','sexdecillion', \
'septendecillion','octodecillion','novemdecillion', \
'vigintillion']
words = []
if num==0: words.append('zero')
else:
numStr = '%d'%num
numStrLen = len(numStr)
groups = (numStrLen+2)/3
numStr = numStr.zfill(groups*3)
for i in range(0,groups*3,3):
h,t,u = int(numStr[i]),int(numStr[i+1]),int(numStr[i+2])
g = groups-(i/3+1)
if h>=1:
words.append(units[h])
words.append('hundred')
if t>1:
words.append(tens[t])
if u>=1: words.append(units[u])
elif t==1:
if u>=1: words.append(teens[u])
else: words.append(tens[t])
else:
if u>=1: words.append(units[u])
if (g>=1) and ((h+t+u)>0): words.append(thousands[g]+',')
if join: return ' '.join(words)
return words
#example usages:
print numToWords(0)
print numToWords(11)
print numToWords(110)
print numToWords(1001000025)
print numToWords(123456789012)
</code></pre>
<p><strong>结果:</strong></p>
^{pr2}$
<p>注意,它适用于整数。然而,将一个浮点数分成两个整数部分并不重要。在</p>