Prepend a string in Python

Question

I'm trying to prepend zeros to each number in a list if it isn't a the necessary number of digits.

    lst = ['1234','2345']
    for x in lst:
        while len(x) < 5:
            x = '0' + x
    print(lst)

Ideally this would print ['012345', '02345']

A better example would be if `lst = ['1234', '23456']`. How do we `print(lst)` and get `['01234', '23456']` — Coby, Jun 28 '17 at 14:30
Ultimately it was a combination of the answers that was successful. `lst = ['1234','2345'] newlst = [] for i in lst: i = i.zfill(5) newlst.append(i) print(newlst)' — Coby, Jun 28 '17 at 14:41

Psidom · Answer 1 · 2017-06-28T01:54:12.233

13

You can use zfill:

Pad a numeric string s on the left with zero digits until the given width is reached

lst = ['1234','2345']
[s.zfill(5) for s in lst]
# ['01234', '02345']

Or use format method with padding and alignment:

["{:0>5}".format(s) for s in lst]
# ['01234', '02345']

edited Jun 28 '17 at 01:54

answered Jun 28 '17 at 01:48

Psidom

1

That's a great idea indeed, perfect for this case... I like it, plus one – developer_hatch Jun 28 '17 at 02:03

developer_hatch · Answer 2 · 2017-06-28T14:55:00.197

3

You code doesn't do the job because strings in python are immutable, see this for more info Why doesn't calling a Python string method do anything unless you assign its output?

You could you enumerate in this case like this:

lst = ['1234','2345', "23456"]
for i, l in enumerate(lst):
  if len(l) < 5:
    lst[i] = '0' + l
print(lst)

['01234', '02345', '23456']

edited Jun 28 '17 at 14:55

answered Jun 28 '17 at 01:49

developer_hatch

Hi Damian, thanks for the response. Unfortunately, this doesn't seem to account for cases where a list item is long enough. I apologize for giving a poor example. – Coby Jun 28 '17 at 14:20
Sorry @ Damian, I should have given a better example. Suppose `lst = ['1234', '23456']` I was trying to get `['01234', '23456']` . The second number is not affected because it is sufficiently long. – Coby Jun 28 '17 at 14:52
@Coby aaaah now I get it, I will update the answer and let me know :) – developer_hatch Jun 28 '17 at 14:53
@Coby I think that's what you are looking at, and I did it this way to try to not modify your code so much. I hope it helps, don't forget voting and accepting if it was helpful :) – developer_hatch Jun 28 '17 at 14:56

score 2 · Answer 3 · answered Jun 28 '17 at 01:56

2

You could use a list comprehension like this:

>>> ['0' * (5-len(x)) + x for x in lst]
['01234', '02345']

Or a list + map try:

>>> list(map(lambda x: '0' * (5-len(x)) + x, lst))
['01234', '02345']

answered Jun 28 '17 at 01:56

blacksite

1

a little twisted but I like it, it does the job :) – developer_hatch Jun 28 '17 at 02:04

score 1 · Accepted Answer · answered Jun 28 '17 at 14:47

1

Ultimately, it was a combination of the answers that did the job.

lst = ['1234','2345']
newlst = []

for i in lst:
    i = i.zfill(5)
    newlst.append(i)

print(newlst)

I apologize if my example wasn't clear. Thank you to all who offered answers!

answered Jun 28 '17 at 14:47

Coby

Nice work, good for finding out the solution, is a plus one i think :) – developer_hatch Jun 28 '17 at 14:57

score 0 · Answer 5 · answered Jun 28 '17 at 01:56

0

You can do this:

>>> lst = ['1234','2345']
>>> lst = ['0' * (5 - len(i)) + i for i in lst]
>>> print(lst)
['01234', '02345']

answered Jun 28 '17 at 01:56

Dat Ha

5 Answers5