Mysql 1864 主從毛病處理辦法

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本文實例講述了python完成將英文單詞表現的數字轉換成阿拉伯數字的辦法。分享給年夜家供年夜家參考。詳細完成辦法以下：

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') == 900
assert spoken_word_to_number('one hundred') == 100
assert spoken_word_to_number('eleven') == 11
assert spoken_word_to_number('twenty two') == 22
assert spoken_word_to_number('thirty-two') == 32
assert spoken_word_to_number('forty two') == 42
assert spoken_word_to_number('two hundred thirty two') == 232
assert spoken_word_to_number('two thirty two') == 232
assert spoken_word_to_number('nineteen hundred eighty nine') == 1989
assert spoken_word_to_number('nineteen eighty nine') == 1989
assert spoken_word_to_number('one thousand nine hundred and eighty nine') == 1989
assert spoken_word_to_number('nine eighty') == 980
assert spoken_word_to_number('nine two') == 92 # wont be able to convert this one
assert spoken_word_to_number('nine thousand nine hundred') == 9900
assert 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)

願望本文所述對年夜家的Python法式設計有所贊助。