Question

我必须为类创建一个脚本，该脚本接受用户输入编号，脚本需要输出该编号的书面形式。这需要使用递归来完成。当我运行我在下面复制的脚本时，它不起作用，我不确定问题是什么。它还需要使用相同的列表作为上层数字，如数万及以上。我对如何完成所有这些工作毫无头绪。为了使这个脚本有效，我需要修复什么？

number = input ('Please enter an integer ')
def name(number):
  list1 = ["zero", "one", "two", "three", "four", "five", "six", "seven",
           "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen",
           "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]
  list2 = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy",
       "eighty", "ninety"]

  list3 = ['', "", 'one-hundred', 'two-hundred', 'three-hundred', 'four-hundred', 'five-hundred' , 'six-hundred', 'seven-hundred', 'eight-hundred', 'nine-hundred']
  if number < 20:
      return list1[number]

  if number < 100:
    return list3[list2[number]]

   hundreds_digit = number * 10
   tens_digit = number / 10
   ones_digit = number % 10

  if ones_digit == 0:
    return list2[tens_digit]
  else:
return list2[tens_digit] + "-" + list1[ones_digit]

  if tens-digit == 0:
     return list3[hundreds_digit]
  else:
     return list3[hundreds_digit] + '-' + list2[tens_digit] + "-" + list1[ones_digit]




################
# Main Program #
################

 name(number)

Answer 1

这是我到目前为止所得到的，但由于输入111时给予one-hundred ten one，因此可以改进。

def hundreds(num):
        obj = {1:'one-hundred', 2: 'two-hundred', 3:'three-hundred', 4:'four-hundred', 5:'five-hundred', 6:'six-hundred', 7:'seven-hundred', 8:'eight-hundred', 9:'nine-hundred'}
        return obj[int(num)]

def tens(num):
        obj = {1:"ten", 2:"twenty", 3:"thirty", 4:"forty", 5:"fifty", 6:"sixty", 7:"seventy",8:"eighty", 9:"ninety"}
        return obj[int(num)]

def ones(num):
        obj = {1:"one", 2:"two", 3:"three", 4:"four", 5:"five", 6:"six", 7:"seven", 8:"eight", 9:"nine"}
        return obj[int(num)]

def written_form(num):
    if len(str(num)) <= 0:
            return False
    else:
            if len(str(num)) == 3:
                    print hundreds(str(num)[0]),
                    try:
                            num = int(''.join(list(str(num))[1:]))
                    except ValueError:
                            return
                    written_form(num)
            elif len(str(num)) == 2:
                    print tens(str(num)[0]),
                    try:
                            num = int(''.join(list(str(num))[1:]))
                    except ValueError:
                            return
                    written_form(num)
            elif len(str(num)) == 1:
                    print ones(str(num)[0]),
                    try:
                            num = int(''.join(list(str(num))[1:]))
                    except ValueError:
                            return
                    written_form(num)

def main():
        var = int(raw_input('Enter your number: '))
        written_form(var)

我有三个函数可以通过使用字典来获取数字的数百，数十和1：

>>> def hundreds(num):
...         obj = {1:'one-hundred', 2: 'two-hundred', 3:'three-hundred', 4:'four-hundred', 5:'five-hundred', 6:'six-hundred', 7:'seven-hundred', 8:'eight-hundred', 9:'nine-hundred'}
...         return obj[int(num)]
... 
>>> hundreds(6)
'six-hundred'
>>> hundreds(4)
'four-hundred'
>>>

在＆＃39; main＆＃39;函数（ written_form() ），如果num的长度小于或等于0，我们就会破坏函数。如果不是，那么如果长度是3，我们打印第一个数字的数百，然后从整数中删除第一个数字，我们继续从那里开始。

该程序运行如下：

bash-3.2$ python written_form.py
Enter your number: 789
seven-hundred eighty nine
bash-3.2$ python written_form.py
Enter your number: 156
one-hundred fifty six
bash-3.2$ python written_form.py
Enter your number: 999
nine-hundred ninety nine
bash-3.2$ python written_form.py
Enter your number: 631
six-hundred thirty one
bash-3.2$

希望这有帮助！

python 2.7.6输入一个数字并获取一个脚本来使用递归输出书面形式，这怎么可能？

1 个答案: