Python: find all occurrences in string

If we don't need to know the indexes of the occurrences we can use the method count from the String class. We can also indicate search boundaries (start, end).

sub_sting = "apple"
print(f"Count of occurrences for '{sub_sting}': {my_string.count(sub_sting)}")

start = 5
end = 25
print(f"Count of occurrences for '{sub_sting}' from the interval [{start}, {end}]: {my_string.count(sub_sting, start, end)}")

Output:

Count of occurrences for 'apple': 3
Count of occurrences for 'apple' from the interval [5, 25]: 1

Otherwise, if we also need start and end indexes of the occurrence we can use re (regular expression) package and its method finditer. Code example:

import re

sub_sting = "apple"
occurrences = list(re.finditer(sub_sting, my_string))
for o in occurrences:
    print(f"Find occurrence for '{sub_sting}': start={o.start()}, end={o.end()}")
print(f"Total occurrences count: {len(occurrences)}")

Output:

Find occurrence for 'apple': start=0, end=5
Find occurrence for 'apple': start=15, end=20
Find occurrence for 'apple': start=26, end=31
Total occurrences count: 3

This method doesn't support interval params but we can perform this method for a substring. Something like this:

import re

sub_sting = "apple"
start = 5
end = 25

occurrences = list(re.finditer(sub_sting, my_string[start:end]))
for o in occurrences:
    print(f"Find occurrence for '{sub_sting}': start={o.start()}, end={o.end()}")
print(f"Total occurrences count: {len(occurrences)}")

Output:

Find occurrence for 'apple': start=10, end=15
Total occurrences count: 1

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