Python Regex Compile Split string so that words appear first

Question

Say I was given a string like so

text = "1234 I just ? shut * the door"

I want to use a regex with re.compile() such that when I split the list all of the words are in front.

Ie it should look like this.

text = ["I", "just", "shut", "the", "door", "1234", "?", "*"]

How can I use re.compile() to split the string this way?

import re
r = re.compile('regex to split string so that words are first').split(text)

Please let me know if you need any more information.

Thank you for the help.

Answer 1

IIUC, you don't need re . Just use str.split with sorted :

sorted(text.split(), key=lambda x: not x.isalpha())

Output:

['I', 'just', 'shut', 'the', 'door', '1234', '?', '*']

Answer 2

You can use sorted with re.findall :

import re
text = "1234 I just ? shut * the door"
r = sorted(text.split(), key=lambda x:(x.isalpha(), x.isdigit(), bool(re.findall('^\W+$', x))), reverse=True)

Output:

['I', 'just', 'shut', 'the', 'door', '1234', '?', '*']

Answer 3

You can't do that with a single regex. You can write one regex to get all words, then another regex to get everything else.

import re

text = "1234 I just ? shut * the door"
r = re.compile(r'[a-zA-Z]+')
words = r.findall(text)
r = re.compile(r'[^a-zA-Z\s]+')
other = r.findall(text)

print(words + other) # ['I', 'just', 'shut', 'the', 'door', '1234', '?', '*']