本文實例講述了python實現(xiàn)將英文單詞表示的數(shù)字轉(zhuǎn)換成阿拉伯數(shù)字的方法。分享給大家供大家參考。具體實現(xiàn)方法如下：
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import re_known = { 'zero': 0, 'one': 1, 'two': 2, 'three': 3, 'four': 4, 'five': 5, 'six': 6, 'seven': 7, 'eight': 8, 'nine': 9, 'ten': 10, 'eleven': 11, 'twelve': 12, 'thirteen': 13, 'fourteen': 14, 'fifteen': 15, 'sixteen': 16, 'seventeen': 17, 'eighteen': 18, 'nineteen': 19, 'twenty': 20, 'thirty': 30, 'forty': 40, 'fifty': 50, 'sixty': 60, 'seventy': 70, 'eighty': 80, 'ninety': 90 }def spoken_word_to_number(n): """Assume n is a positive integer".assert _positive_integer_number('nine hundred') == 900assert spoken_word_to_number('one hundred') == 100assert spoken_word_to_number('eleven') == 11assert spoken_word_to_number('twenty two') == 22assert spoken_word_to_number('thirty-two') == 32assert spoken_word_to_number('forty two') == 42assert spoken_word_to_number('two hundred thirty two') == 232assert spoken_word_to_number('two thirty two') == 232assert spoken_word_to_number('nineteen hundred eighty nine') == 1989assert spoken_word_to_number('nineteen eighty nine') == 1989assert spoken_word_to_number('one thousand nine hundred and eighty nine') == 1989assert spoken_word_to_number('nine eighty') == 980assert spoken_word_to_number('nine two') == 92 # wont be able to convert this oneassert spoken_word_to_number('nine thousand nine hundred') == 9900assert spoken_word_to_number('one thousand nine hundred one') == 1901""" n = n.lower().strip() if n in _known: return _known[n] else: inputWordArr = re.split('[ -]', n) assert len(inputWordArr) > 1 #all single words are known #Check the pathological case where hundred is at the end or thousand is at end if inputWordArr[-1] == 'hundred': inputWordArr.append('zero') inputWordArr.append('zero') if inputWordArr[-1] == 'thousand': inputWordArr.append('zero') inputWordArr.append('zero') inputWordArr.append('zero') if inputWordArr[0] == 'hundred': inputWordArr.insert(0, 'one') if inputWordArr[0] == 'thousand': inputWordArr.insert(0, 'one') inputWordArr = [word for word in inputWordArr if word not in ['and', 'minus', 'negative']] currentPosition = 'unit' prevPosition = None output = 0 for word in reversed(inputWordArr): if currentPosition == 'unit': number = _known[word] output += number if number > 9: currentPosition = 'hundred' else: currentPosition = 'ten' elif currentPosition == 'ten': if word != 'hundred': number = _known[word] if number < 10: output += number*10 else: output += number #else: nothing special currentPosition = 'hundred' elif currentPosition == 'hundred': if word not in [ 'hundred', 'thousand']: number = _known[word] output += number*100 currentPosition = 'thousand' elif word == 'thousand': currentPosition = 'thousand' else: currentPosition = 'hundred' elif currentPosition == 'thousand': assert word != 'hundred' if word != 'thousand': number = _known[word] output += number*1000 else: assert "Can't be here" == None return(output) |
希望本文所述對大家的Python程序設(shè)計有所幫助。
