Python multiple substring index in string

You can join those patterns together using |:

import re
sub = ['ABC', 'VC', 'KI']
s = 'ABDDDABCTYYYYVCIIII'

r = '|'.join(re.escape(s) for s in sub)
for i in re.finditer(r, s):
    print(i.start(), i.end())

A substring may occur more than once in the main string (although it doesn't in the sample data). One could use a generator based around a string's built-in find() function like this:

note the source string has been modified to demonstrate repetition

sub = ['ABC', 'VC', 'KI']
s = 'ABCDDABCTYYYYVCIIII'

def find(s, sub):
    for _sub in sub:
        offset = 0
        while (idx := s[offset:].find(_sub)) >= 0:
            yield _sub, idx + offset
            offset += idx + 1

for ss, start in find(s, sub):
    print(ss, start)

Output:

ABC 0
ABC 5
VC 13

You could map over the find string method.

s = 'ABDDDABCTYYYYVCIIII'
sub = ['ABC', 'VC', 'KI']

print(*map(s.find, sub))
# Output 5 13 -1

How about using list comprehension with str.find?

s = 'ABDDDABCTYYYYVCIIII'
sub = ['ABC', 'VC', 'KI']
results = [s.find(pattern) for pattern in sub]

print(*results) # 5 13 -1

Another approach with re, if there can be multiple indices then this might be better as the list of indices is saved for each key, when there is no index found, the substring won't be in the dict.

import re
s = 'ABDDDABCTYYYYVCIIII'
sub = ['ABC', 'VC', 'KI']

# precompile regex pattern
subpat = '|'.join(sub)
pat = re.compile(rf'({subpat})')

matches = dict()
for m in pat.finditer(s):
    # append starting index of found substring to value of matched substring
    matches.setdefault(m.group(0),[]).append(m.start()) 

print(f"{matches=}")
print(f"{'KI' in matches=}")
print(f"{matches['ABC']=}")

Outputs:

matches={'ABC': [5], 'VC': [13]}
'KI' in matches=False
matches['ABC']=[5]

Just Use String index Method

list_ = ['ABC', 'VC', 'KI']

s = 'ABDDDABCTYYYYVCIIII'


for i in list_:
    if i in s:
        print(s.index(i))