Removing all spaces recursively (list of list of lists of...)

Question

I'm trying to remove all spaces from the input which is list of list of lists... I don't know what to do for the "else:"

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

Example:

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

The output should be:

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

and what to do if the lst contains integers or floating points? I have received errors for integers.

example input :

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

In your example, Python will see your list as a python list, and apple, pie, etc, must be variables. There is no string here. It's certainly not what you want. — , Jan 17 '13 at 10:19

score 4 · Accepted Answer · answered Jan 17 '13 at 10:16

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

Here is a sample:

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

answered Jan 17 '13 at 10:16

@thg435. isinstance would be better, but I used the same code as the OP. For those who are interested : [link](http://stackoverflow.com/questions/1549801/differences-between-isinstance-and-type-in-python) – Jan 17 '13 at 10:36

georg · Answer 2 · 2013-01-17T10:22:17.597

I'd suggest to follow EAFP and catch an exception instead of using isinstance. Also, never miss an opportunity to make a function a bit more generic:

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

Example:

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

Or even more generic, although that might be an overkill for your problem:

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)

score 0 · Answer 3 · answered Jan 17 '13 at 10:11

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)

prints

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

score 0 · Answer 4 · answered Jan 17 '13 at 10:18

Though you may experiment with recursive solution, but you can try your hand on a wonderful library Python provides, to transform a well formed Python literal from string to Python literal.

Just convert you list to string
remove any necessary spaces

and then reconvert to the recursive list structure using ast.literal_eval

>>> lst = [' apple', 'pie ', ['sth', ['banana', 'asd', [' sdfdsf', ['fgg']]]]]
>>> import ast
>>> ast.literal_eval(str(lst).translate(None,' '))
['apple', 'pie', ['sth', ['banana', 'asd', ['sdfdsf', ['fgg']]]]]

Removing all spaces recursively (list of list of lists of...)

4 Answers4