split string using regex python and "re" package

Question

I'm using Python 3 on Windows 10. Consider the following string:

import re
s = ["12345", "67891", "01112"]

I want to split these zips at the 3 character to get the zip3, but this code throws an error.

re.split("\d{3}", s)
TypeError: cannot use a string pattern on a bytes-like object

I'm not quite sure how to get around. Help appreciated. Thanks.

what should be the result? `["123","45","678","91","011","12"]` ? - Why regex? — Patrick Artner, Jan 02 '19 at 18:35
Yes the result should be the first 3 char of the string. I'm using regex because `str.split` doesn't take char counts as far as I know? If you know of better methods please advise. — Todd Shannon, Jan 02 '19 at 19:00
@WiktorStribiżew do i need an iteration? `R` does this flawlessly without iteration, so I figure `python` would be the same. — Todd Shannon, Jan 02 '19 at 19:01

Patrick Artner · Accepted Answer · 2019-01-02T19:20:04.727

To get the first three of each, simply string-slice them:

s = ["12345", "67891", "01112"]
first_tree = [p[0:3] for p in  s] 
print(first_tree)

Outtput:

['123', '678', '011'] # slicing

To split all text in threes, join it, then use chunking to get chunks of 3 letters:

s = ["12345", "67891", "01112"]
k = ''.join(s)
threesome = [k[i:i+3] for i in range(0,len(k),3)]
print(threesome)

Outtput:

['123', '456', '789', '101', '112']  # join + chunking

Slicing and chunking works on strings as well - the official doku about strings is here: about strings and slicing

To get the remainder as well:

s = ["12345", "67891", "01112"]
three_and_two = [[p[:3], p[3:]] for p in s] 
print(three_and_two) #  [['123', '45'], ['678', '91'], ['011', '12']]

1 Answers1