Count the number of occurrences of a character in a string

Counting within a range

str.count() can be more precise, it counts within a specific portion of the string:

count = "banana".count("a", 1, -1)
print(count)  # Output: 2
# Two "a"s from 2nd to last character... still looking for the last "a"
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Let’s get comprehensive with collections.Counter

If you need a character census:

from collections import Counter
counter = Counter("banana")
print(counter['a'])  # Output: 3
# A-ha, there you are, third "a"!
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With Counter, we can even run a full-scale tally.

Ignoring case like a gentleman

Want to count regardless of case?

from collections import Counter
counter = Counter("Banana".lower())
print(counter['a'])  # Output: 3
# Capital A, lowercase a — no discrimination here!
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Use .lower() to treat 'A' and 'a' as the same character.

Regular expressions for flexing (and matching)

And if you love flexibility and pattern matching:

import re
count = len(re.findall('a', "BanAna", re.IGNORECASE))
print(count)  # Output: 4
# Regex for the win! It's raining a's and A's here.
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The re.IGNORECASE flag makes the search a case-blind investigator.

Going beyond the basics

Time is money, let’s get efficient

When processing massive strings, str.count() is usually faster — it's an efficient direct method optimized for such tasks. Regular expressions are a bit more resource-hungry due to their fancy flexibility.

Unicode? No problemo!

If you have Unicode characters in your string:

uni_count = "naïve café".count("é")
print(uni_count)  # Output: 1
# How many é's? Uno, señor!
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Unicode characters are handled seamlessly with str.count().

Additional considerations and edge cases

What’s so special about special characters?

Keep in mind, special characters need escaping in re.findall():

count = len(re.findall('\$', "Cost: $5.00"))
print(count)  # Output: 1
# One dollar sign coming right up!
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The backslash \ signals the dollar sign $ to drop its special character status.

What about overlapping occurrences?

str.count() does not consider overlapping occurrences:

overlapping_count = "aaa".count("aa")
print(overlapping_count)  # Output: 1
# Even though "aa" is there twice, let's not count our chickens before they hatch.
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Despite "aa" appearing twice with overlap, there's only one separate occurrence.