递归删除所有空格(列表列表...)

时间:2013-01-17 10:09:33

标签: python recursion space

我正在尝试从列表列表列表的输入中删除所有空格...我不知道该怎么做“else:”

def removespace(lst):
   if type(lst) is str:
      return lst.replace(" ","")
   else:
      ?????

示例:

lst = [ apple, pie ,    [sth, [banana     , asd, [    sdfdsf, [fgg]]]]]

输出应为:

lst2 = [apple,pie,[sth,[banana,asd,[sdfdsf,[fgg]]]]] 

如果lst包含整数或浮点,该怎么办?我收到整数错误。

示例输入:

 L = [['apple', '2 * core+1* sth'], ['pie', '1*apple+1*sugar+1*water'], ['water', 60]]

4 个答案:

答案 0 :(得分:4)

def removespace(a):
    if type(a) is str:
        return a.replace(" ", "")
    elif type(a) is list:
        return [removespace(x) for x in a]
    elif type(a) is set:
        return {removespace(x) for x in a}
    else:
        return a

以下是一个示例:

>>> removespace([["a ",["   "]],{"b ","c d"},"e f g"])
[['a', ['']], {'b', 'cd'}, 'efg']

答案 1 :(得分:2)

我建议关注EAFP并捕获异常,而不是使用isinstance。此外,永远不要错过让函数更通用的机会:

def rreplace(it, old, new):
    try:
        return it.replace(old, new)
    except AttributeError:
        return [rreplace(x, old, new) for x in it]

示例:

a = [" foo", ["    spam", "ham"], "  bar"]
print rreplace(a, " ", "")     

或者甚至更通用,虽然这可能对你的问题有点过分:

def rapply(it, fun, *args, **kwargs):
    try:
        return fun(it, *args, **kwargs)
    except TypeError:
        return [rapply(x, fun, *args, **kwargs) for x in it]

a = [" foo", ["    spam", "ham"], "  bar"]
print rapply(a, str.replace, " ", "")     
print rapply(a, str.upper)     

答案 2 :(得分:0)

def removespace(lst):
    if type(lst) is str:
        return lst.replace(" ","")
    else:
        return [removespace(elem) for elem in lst]



lst = [' apple', 'pie ', ['sth', ['banana', 'asd', [' sdfdsf', ['fgg']]]]] 
print removespace(lst)

打印

['apple', 'pie', ['sth', ['banana', 'asd', ['sdfdsf', ['fgg']]]]]

答案 3 :(得分:0)

虽然您可以尝试使用递归解决方案,但是您可以尝试使用Python提供的精彩库,将格式良好的Python文字从字符串转换为Python文字。

  • 只需将您的列表转换为字符串
  • 即可
  • 删除所有必要的空格
  • 然后使用ast.literal_eval

    重新转换为递归列表结构
    >>> lst = [' apple', 'pie ', ['sth', ['banana', 'asd', [' sdfdsf', ['fgg']]]]]
    >>> import ast
    >>> ast.literal_eval(str(lst).translate(None,' '))
    ['apple', 'pie', ['sth', ['banana', 'asd', ['sdfdsf', ['fgg']]]]]