Get text before and after colon python

I have following example string:

'EXP DATE: 13.04.2022 PO: P101'
'LOCATION: 111 CONDITION: FN'

I need to split following strings to look like:

{"EXP DATE": "13.04.2022", "PO": "P101"}
{"LOCATION": "111", "CONDITION:" "FN"}

To achieve this i created following function:

def create_key_pair(lst):
    return {lst[i].replace(':', ''): lst[i + 1] for i in range(0, len(lst), 2)}

so if i pass

str_ = 'LOCATION: 111 CONDITION: FN'
create_key_pair(str_.split(" "))

i got

{"LOCATION": "111", "CONDITION:" "FN"}

but if i pass

str_ = 'XP DATE: 13.04.2022 PO: P101'
create_key_pair(str_.split(" "))

i got

IndexError: list index out of range

since

EXP DATE

splited by space

If the values (the part after the colon) cannot contain spaces, the following will work. The approach uses a regular expression to match any number of characters up to a colon followed by a space and then any number of non-space characters. Then split each match on the colon.

import re

def to_dict(s):
	matches = re.findall(r".+?: \S+", s)
	d = {}
	for m in matches:
		k, v = m.split(":")
		d[k.strip()] = v.strip()
	return d

If the values can contain spaces you'll have to find some other way of separating a value from the next key. JonSG gave the example in the comments of "foo: bar bat hat: 10". With the above approach, you'll get {"foo": "bar", "bat hat": "10"}, but maybe you want {"foo": "bar bat", "hat": "10"}. From the information in the question the only pattern I can see is that the values do not contain spaces.

You can use a regex:

import re

def create_key_pair(s):
    data = {}
    for kv in re.findall('([^:]+:\s*[^\s]+)', s):
        k, v = re.split(':\s*', kv)
        data[k.strip()] = v.strip()
    return data

Usage:

str_ = 'LOCATION: 111 CONDITION: FN'
create_key_pair(str_)

# Output
{'LOCATION': '111', 'CONDITION': 'FN'}


str_ = 'EXP DATE: 13.04.2022 PO: P101'
create_key_pair(str_)

# Output
{'EXP DATE': '13.04.2022', 'PO': 'P101'}