How to Replace Characters in a List in Python?

Question

list = ["/x01", "/x30", "/x30", "/x53", "/x46", "/x46", "/x46", "/x30", "/x30", "/x30", "/x31", "/x46", "/x46", "/x0D"]

for x in list:
    x.replace("[","").replace("]","").replace('"','').replace(" ","").replace(",","").replace("[","")

This hasn't been working. Could you explain what I might be doing wrong, and tell me if there is a more effective way of getting the output: "/x01/x30/x30/x53/x46/x46/x46/x30/x30/x30/x31/x46/x46/x0D"

Answer 1

Use str.join :

>>> lis = ["/x01", "/x30", "/x30", "/x53", "/x46", "/x46", "/x46", "/x30", "/x30", "/x30", "/x31", "/x46", "/x46", "/x0D"]
>>> ''.join(lis)
'/x01/x30/x30/x53/x46/x46/x46/x30/x30/x30/x31/x46/x46/x0D'

Looking at your code, I think you were trying to apply str.replace on str version of the list. But that would be a weird way to do this, better use str.join :

>>> str(lis)
"['/x01', '/x30', '/x30', '/x53', '/x46', '/x46', '/x46', '/x30', '/x30', '/x30', '/x31', '/x46', '/x46', '/x0D']"

The above string is just a representation of the list object.

>>> str(lis).replace("[","").replace("]","").replace(" ","").replace(",","").replace("'","")
'/x01/x30/x30/x53/x46/x46/x46/x30/x30/x30/x31/x46/x46/x0D'

Answer 2

The first problem is that replace doesn't change a string in-place, it just returns a new string. And you're ignoring that new string.

What you want is:

new_list = []
for x in list:
    new_list.append(x.replace("[","").replace("]","").replace('"','').replace(" ","").replace(",","").replace("[",""))

You can simplify that with translate , or use a different way of filtering out the characters like a comprehension or a filter call. But the result will be the same.

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The bigger problem is that what you're trying to do doesn't make any sense. None of the elements in your list have a [ , ] , " , etc. character in them. You're probably confusing the string representation of the list with the list itself.

If you want to join the members of a list, or to produce any representation of the list other than the default repr , just explicitly join them. For example, this gets what you seem to want:

''.join(list)

… and this gets a different representation:

' and '.join(list)

… and this gets roughly the same thing as repr :

'[' + ', '.join(map(repr, list)) + ']'

Answer 3

You should use translate(None, nuke_list_string)

Two reasons:

1. replace performs a 1-to-1 substitution, each substitution results in the allocation of a new string which is then returned, resulting in a lot of overhead if you chain a bunch of replaces like you do in the example.

2. translate usually requires a 1-to-1 mapping, ie translate('ABC','abc') would convert the string 'a brown cow' to 'A Brown Cow'. However, if you specify None as the new mapping, it will remove the characters it finds.

Here's an example:

a = 'this is [a] string with (stuff) in {it}'
print a.translate(None, '[]{}()')

Produces:

this is a string with stuff in it