In python, how can I use regex to replace square bracket with parentheses

Question

I have a list : list = [1,2,3] . And I would like to convert that into a string with parentheses: string = (1,2,3) .

Currently I am using string replace string = str(list).replace('[','(').replace(']',')') . But I think there is a better way using regex.sub. But I have no idea how to do it. Thanks a lot

Answer 1

If you do indeed have a list, then:

>>> s  = [1,2,3]
>>> str(tuple(s))
'(1, 2, 3)'

Answer 2

You could use string.maketrans instead -- I'm betting it runs faster than a sequence of str.replace and it scales better to more single character replacements.

>>> import string
>>> table = string.maketrans('[]','()')
>>> s = "[1, 2, 3, 4]"
>>> s.translate(table)
'(1, 2, 3, 4)'

You can even use this to remove characters from the original string by passing an optional second argument to str.translate :

>>> s = str(['1','2'])
>>> s
"['1', '2']"
>>> s.translate(table,"'")
'(1, 2)'

In python3.x, the string module is gone and you gain access to maketrans via the str builtin:

table = str.maketrans('[]','()')

Answer 3

str([1,2,3]).replace('[','(').replace(']',')')

Should work for you well, and it is forward and backward compatible as far as I know.

as far as re-usability, you can use the following function for multiple different types of strings to change what they start and end with:

def change(str_obj,start,end):
    if isinstance(str_obj,str) and isinstance(start,str) and isinstance(end,str):
        pass
    else:
        raise Exception("Error, expecting a 'str' objects, got %s." % ",".join(str(type(x)) for x in [str_obj,start,end]))
    if len(str_obj)>=2:
        temp=list(str_obj)
        temp[0]=start
        temp[len(str_obj)-1]=end
        return "".join(temp)
    else:
         raise Exception("Error, string size must be greater than or equal to 2. Got a length of: %s" % len(str_obj))

Answer 4

There are a billion ways to do this, as demonstrated by all the answers. Here's another one:

my_list = [1,2,3]
my_str = "(%s)" % str(my_list).strip('[]')

or make it recyclable:

list_in_parens = lambda l: "(%s)" % str(l).strip('[]')
my_str = list_in_parens(my_list)

Answer 5

Try this:

'({0})'.format(', '.join(str(x) for x in list))

By the way, it's not a good idea to name your own variables list since it clashes with the built-in function. Also string can conflict with the module of the same name.

Answer 6

If you really want to use regex I guess this would work. But the other posted solutions are probably more efficient and/or easy to use.

import re

string = str(list)
re.sub(r"[\[]", "(", string)
re.sub(r"[\]]", ")", string)

@mgilson something like this you mean?

def replace_stuff(string):
    string_list = list(string)
    pattern = re.compile(r"[\[\]]")
    result = re.match(pattern, string)
    for i in range(len(result.groups())):
        bracket = result.group(i)
        index = bracket.start()
        print(index)
        if bracket == "[":
            string_list[index] = "("
        else:
            string_list[index] = ")"

    return str(string_list)

It doesn't quite work, for some reason len(result.groups()) is always 0 even though the regex should find matches. So couldn't test whether this is faster but because I couldn't get it to work I couldn't test it. I have to leave for bed now so if someone can fix this go ahead.

Answer 7

All you need is:

"(" + strng.strip('[]') + ")"

It works for both single element and multi-element lists.

>>> lst = [1,2,3]
>>> strng = str(lst)
>>> "(" + strng.strip('[]') + ")"
'(1, 2, 3)'


>>> lst = [1]
>>> strng = str(lst)
>>> "(" + strng.strip('[]') + ")"
'(1)'