Count number of occurrences of a character in a string

Question

I'm just getting into Python and I'm building a program that analyzes a group of words and returns how many times each letter appears in the text. i.e 'A:10, B:3, C:5...etc'. So far it's working perfectly except that i am looking for a way to condense the code so i'm not writing out each part of the program 26 times. Here's what I mean..

print("Enter text to be analyzed: ")
message = input()

A = 0
b = 0
c = 0
...etc

for letter in message:

    if letter == "a":
        a += 1
    if letter == "b":
        b += 1
    if letter == "c":
        c += 1
    ...etc

print("A:", a, "B:", b, "C:", c...etc)

Use a dict instead of 26 variables.

Wooble
– Wooble

2013-09-14 03:17:50 +00:00
Commented Sep 14, 2013 at 3:17 — Wooble
– Wooble, Commented Sep 14, 2013 at 3:17

Tim Peters · Accepted Answer · 2013-09-14 03:22:19Z

13

There are many ways to do this. Most use a dictionary (dict). For example,

count = {}
for letter in message:
    if letter in count:  # saw this letter before
        count[letter] += 1
    else:   # first time we've seen this - make its first dict entry
        count[letter] = 1

There are shorter ways to write it, which I'm sure others will point out, but study this way first until you understand it. This sticks to very simple operations.

At the end, you can display it via (for example):

for letter in sorted(count):
    print(letter, count[letter])

Again, there are shorter ways to do this, but this way sticks to very basic operations.

answered Sep 14, 2013 at 3:22

Tim Peters

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2 Comments

Grijesh Chauhan Over a year ago

So we can also remove if-else as count[letter] = count.setdefault(letter, 0) + 1 . correct na?

Tim Peters Over a year ago

Yes, setdefault() instead works fine - that's what it's for. But it's harder for newcomers to understand, so I avoided it here :-)

zero323 · Accepted Answer · 2013-09-14 03:31:48Z

12

You can use Counter but @TimPeters is probably right and it is better to stick with the basics.

from collections import Counter

c = Counter([letter for letter in message if letter.isalpha()])
for k, v in sorted(c.items()):
    print('{0}: {1}'.format(k, v))

edited Sep 14, 2013 at 3:31

answered Sep 14, 2013 at 3:20

zero323

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2 Comments

Tim Peters Over a year ago

Yours is an excellent answer! It's what a Python expert would do. My hope is that studying my "sticking to the basics" answer will help the poster learn more about Python-the-language - collections.Counter pretty much solves the entire problem by magic ;-)

Kevin Arthur Over a year ago

yeah... I think i'll be sticking with the basics for now but I think I might try to experiment with this a bit! Could be useful later on. Thanks!

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