How to create a list of string using range() built-in function in python?
range(1,7) produces a range object equivilant to [1,2,3,4,5,6]
Desired list: ['1','2','3','4','5','6']
4 Answers
Use a cast to str() & list comprehension as per the following:
string_list = [str(x) for x in range(1, 7)]
1 Comment
CoffeeTableEspresso
you want range(1, 7) since OP wants the list to start at '1'.
With the map function:
list(map(str,range(7)))
1 Comment
user459872
Or
[*map(str,range(7))]. But your solution is much more readable.Or the new f-strings:
>>> [f'{i}' for i in range(7)]
['0', '1', '2', '3', '4', '5', '6']
1 Comment
Stephen Rauch
Note this is python 3.6+
Use string formatting & list comprehension as per the following
string_list = ["{}".format(x) for x in range(1,7)]
3 Comments
user2357112
You don't need the full functionality of
str.format here - str(x) would have worked just as well as "{}".format(x). Also, if you wanted to start from 1, your range call is wrong.
abarnert
Even if you need a format spec (which you don't, since you aren't using one), you still wouldn't want to use
str.format here; just call format(x, "03") instead of "{:03}".format(x). You only need str.format to put multiple things together. And even then, f-strings are usually better: f"{x:03}".
helcode
@user2357112, I have fixed the
range call. and you are right about the use of str(x) it is also a solution
range(7)does not output[1,2,3,4,5,6].