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
If you do indeed have a list, then:
>>> s = [1,2,3]
>>> str(tuple(s))
'(1, 2, 3)'
str(tuple(s)).rstrip(',)') + ')' '(1)'
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('[]','()')
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)
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))
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.
re.sub which would then take the match object and decide whether to substitute a '(' or a ')'. It might even be slightly more efficient (you'd need to timeit to find out), but it would be slightly more code as you'd need to define the function properly, etc, etc..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)'
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.
