How to search for one or more strings in a file using regex, and count the number of each string separately?

So I am trying to find one or more strings in each line of a file, and count the number of times each string comes up in total in the file. In some lines there is only one of the strings, however in other lines there may be multiple target strings, if that makes sense. I am trying to use a regular expression to do this.

So what I've tried is as follows (having already read the file in and separated it into lines using .readlines):

1count=0
2count=0
3count=0

Pattern=r'(?i)(\bString1\b)|(\bString2\b)|(\bString3\b)'

i=0
while i!=len(lines) 
    match=re.search(pattern, lines[i]) 

    if match:
        if match.group(1):
            1count=1count+1
        elif match.group(2):
            2count=2count+1
        elif match.group(3):
            3count=3count+1
    i=i+1

This works when there is no multiple matches in the line, however when there is it obviously only counts the first match and then moves on. Is there a way for me to scan the whole line anyway? I know re.findall finds all matches, but it then puts them into an array, and I don't know how I would reliably count the number of matches for each word, since the matches in findall would have different indexes in the array each loop through.

You can use findall and count occurrences at the end.
For example:

import re
count1=0
count2=0
count3=0
data = "String1 String2 String2 String3\nString1 String1\nString3"
Pattern=r'(?i)(\bString1\b)|(\bString2\b)|(\bString3\b)'
lines = data.split('\n')
all_matches = []
i = 0
while i!=len(lines): 
    match=re.findall(Pattern, lines[i])
    all_matches.extend(match)
    i += 1
count1 = len([el for el in all_matches if el[0] == 'String1'])
count2 = len([el for el in all_matches if el[1] == 'String2'])
count3 = len([el for el in all_matches if el[2] == 'String3'])
    
print(count1, count2, count3)

Note: findall will return a list of tuples, where first item of tuple corresponds to the first group, and so on.

all_matches will be list of tuples, each tuple is of the shape (matched item for string1, matched item for string2, matched item for string3)
if there isn't a matched, it will be ''
,something like that:

[('String1', '', ''), ('', 'String2', ''), ('', 'String2', ''), ...]

In calculating count1 for example, we are creating a list of elements which match String1(The condition here as we saw, that first element of the tuple equals 'String1') as follows:

first_group = [el for el in all_matches if el[0] == 'String1']

then we return its length as the value of count1length of those elements:

count1 = len(first_group)

In your example, the matches are all static strings, so you can just use them as dictionary keys for a Counter object.

import re
from collections import Counter

count = Counter()
for line in lines:
    for match in re.finditer(Pattern, line):
        count.update(match.group(0))

for k in count.keys():
    print(f"{c[k]} occurrences of {k}")

Part of the useful changes here is using re.finditer() instead of re.findall which returns a proper re.Match object from which you can extract the matching string with .group(0) as well as various other attributes, should you wish to.

If you need to extract matches which could contain variations, like r"c[ei]*ling" or r"\d+", you can't use the matched strings as dictionary keys (because then the Counter would count each unique string as a separate entity; so you would get "12 occurrences of 123" and "1 occurrence of 234" instead of "13 occurrences of \d+"); in that case, I would perhaps try to use named subgroups.

    for match in re.finditer(r"(?P<ceiling>c[ei]*ling)|(?P<number>\d+)", line):
        matches = match.groupdict()
        for key in matches.keys():
            if matches[key] is not None:
                count.update(key)

Just another variant is using numpy and its count_nonzero method. Since there's no need to separate the data into lines, let's assume it's all in data:

import numpy as np
# count non-empty strings along axis 0 (the matches for each word)
count = np.count_nonzero(np.array(re.findall(Pattern, data)), 0)

</details>