将脚本从数字转换为打印检查的文字

我想我可以避免在这里重新发明轮子。将脚本从数字转换为打印检查的文字

我需要一个Python脚本将数字转换为单词以进行打印检查。

E.g. 1,10,543应提供输出为One lac ten thousand five hundred and forty three。

来源

2012-12-17 infoadmin12345

http://www.blog。 pythonlibrary.org/2010/10/21/python-converting-number s-to-words/ – NPE

@NPE：该脚本提供的格式为“一万一五百四十三”。不太适合检查。 – infoadmin12345

看起来你需要它的印度语言环境，它很难得到一个这样的，但你可以自定义上述脚本，以适应你的目的 – Abhijit

写了一个自定义转换器与FOLL特点：

数量可用于数字从0到999999999 迎合印度次大陆即LACS和亿卢比（Word转换范围大足够以容纳大量的用例）
包括http://www.blog.pythonlibrary.org/2010/10/21/python-converting-numbers-to-words/
P派萨支持高达经启发后2位小数（四舍五入）

rofiling信息：对于正好10000次运行，此脚本在0.458秒的执行时间内与上述脚本的0.237秒相比具有较低的性能。

class Number2Words(object): 

     def __init__(self): 
      '''Initialise the class with useful data''' 

      self.wordsDict = {1: 'one', 2: 'two', 3: 'three', 4: 'four', 5: 'five', 6: 'six', 7: 'seven', 
           8: 'eight', 9: 'nine', 10: 'ten', 11: 'eleven', 12: 'twelve', 13: 'thirteen', 
           14: 'fourteen', 15: 'fifteen', 16: 'sixteen', 17: 'seventeen', 
           18: 'eighteen', 19: 'nineteen', 20: 'twenty', 30: 'thirty', 40: 'forty', 
           50: 'fifty', 60: 'sixty', 70: 'seventy', 80: 'eighty', 90: 'ninty' } 

      self.powerNameList = ['thousand', 'lac', 'crore'] 


     def convertNumberToWords(self, number): 

      # Check if there is decimal in the number. If Yes process them as paisa part. 
      formString = str(number) 
      if formString.find('.') != -1: 
       withoutDecimal, decimalPart = formString.split('.') 

       paisaPart = str(round(float(formString), 2)).split('.')[1] 
       inPaisa = self._formulateDoubleDigitWords(paisaPart) 

       formString, formNumber = str(withoutDecimal), int(withoutDecimal) 
      else: 
       # Process the number part without decimal separately 
       formNumber = int(number) 
       inPaisa = None 

      if not formNumber: 
       return 'zero' 

      self._validateNumber(formString, formNumber) 

      inRupees = self._convertNumberToWords(formString) 

      if inPaisa: 
       return 'Rs. %s and %s paisa' % (inRupees.title(), inPaisa.title()) 
      else: 
       return 'Rs. %s' % inRupees.title() 


     def _validateNumber(self, formString, formNumber): 

      assert formString.isdigit() 

      # Developed to provide words upto 999999999 
      if formNumber > 999999999 or formNumber < 0: 
       raise AssertionError('Out Of range') 


     def _convertNumberToWords(self, formString): 

      MSBs, hundredthPlace, teens = self._getGroupOfNumbers(formString) 

      wordsList = self._convertGroupsToWords(MSBs, hundredthPlace, teens) 

      return ' '.join(wordsList) 


     def _getGroupOfNumbers(self, formString): 

      hundredthPlace, teens = formString[-3:-2], formString[-2:] 

      msbUnformattedList = list(formString[:-3]) 

      #---------------------------------------------------------------------# 

      MSBs = [] 
      tempstr = '' 
      for num in msbUnformattedList[::-1]: 
       tempstr = '%s%s' % (num, tempstr) 
       if len(tempstr) == 2: 
        MSBs.insert(0, tempstr) 
        tempstr = '' 
      if tempstr: 
       MSBs.insert(0, tempstr) 

      #---------------------------------------------------------------------# 

      return MSBs, hundredthPlace, teens 


     def _convertGroupsToWords(self, MSBs, hundredthPlace, teens): 

      wordList = [] 

      #---------------------------------------------------------------------# 
      if teens: 
       teens = int(teens) 
       tensUnitsInWords = self._formulateDoubleDigitWords(teens) 
       if tensUnitsInWords: 
        wordList.insert(0, tensUnitsInWords) 

      #---------------------------------------------------------------------# 
      if hundredthPlace: 
       hundredthPlace = int(hundredthPlace) 
       if not hundredthPlace: 
        # Might be zero. Ignore. 
        pass 
       else: 
        hundredsInWords = '%s hundred' % self.wordsDict[hundredthPlace] 
        wordList.insert(0, hundredsInWords) 

      #---------------------------------------------------------------------# 
      if MSBs: 
       MSBs.reverse() 

       for idx, item in enumerate(MSBs): 
        inWords = self._formulateDoubleDigitWords(int(item)) 
        if inWords: 
         inWordsWithDenomination = '%s %s' % (inWords, self.powerNameList[idx]) 
         wordList.insert(0, inWordsWithDenomination) 

      #---------------------------------------------------------------------# 
      return wordList 


     def _formulateDoubleDigitWords(self, doubleDigit): 

      if not int(doubleDigit): 
       # Might be zero. Ignore. 
       return None 
      elif self.wordsDict.has_key(int(doubleDigit)): 
       # Global dict has the key for this number 
       tensInWords = self.wordsDict[int(doubleDigit)] 
       return tensInWords 
      else: 
       doubleDigitStr = str(doubleDigit) 
       tens, units = int(doubleDigitStr[0])*10, int(doubleDigitStr[1]) 
       tensUnitsInWords = '%s %s' % (self.wordsDict[tens], self.wordsDict[units]) 
       return tensUnitsInWords 


if __name__ == '__main__': 

    wGenerator = Number2Words() 
    print wGenerator.convertNumberToWords(100000)

来源

2012-12-18 11:58:20

以上不适用于** Python3（has_key在Python3中不可用）**，请使用Python3的以下要点：[Gist for Python 3版本的类]（https://gist.github.com/ puneetkaura/f5b7592310511db6047ded4b31c77710） – PKaura

可以使用Python库num2words（pip install num2words）：

实施例：

from num2words import num2words 
num2words(110543, to='cardinal', lang='en_IN')

输出：

'one lakh, ten thousand, five hundred and forty-three'

来源

2018-02-04 07:52:25

将脚本从数字转换为打印检查的文字

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