python counting letters in string without count function

Question

I am trying to write a program to count the occurrences of a specific letter in a string without the count function. I made the string into a list and set a loop to count but th

Answer 1

Instead of

count == count + 1

you need to have

count = count + 1

Answer 2

Your count is never changing because you are using == which is equality testing, where you should be using = to reassign count. Even better, you can increment with

count += 1

Also note that else: continue doesn't really do anything as you will continue with the next iteration of the loop anyways. If I were to have to come up with an alternative way to count without using the count function, I would lean towards regex:

import re
stringy = "aardvark"
print(len(re.findall("a", stringy)))

Answer 3

Be careful, you are using count == count + 1, and you must use count = count + 1

The operator to attribute a new value is =, the operator == is for compare two values

Answer 4

Although someone else has solved your problem, the simplest solution to do what you want to do is to use the Counter data type:

>>> from collections import Counter
>>> letter = 'a'
>>> myString = 'aardvark'
>>> counts = Counter(myString)
>>> print(counts)
Counter({'a': 3, 'r': 2, 'v': 1, 'k': 1, 'd': 1})
>>> count = counts[letter]
>>> print(count)
3

Or, more succinctly (if you don't want to check multiple letters):

>>> from collections import Counter
>>> letter = 'a'
>>> myString = 'aardvark'
>>> count = Counter(myString)[letter]
>>> print(count)
3

The simplest way to do your implementation would be:

count = sum(i == letter for i in myString)

or:

count = sum(1 for i in myString if i == letter)

This works because strings can be iterated just like lists, and False is counted as a 0 and True is counted as a 1 for arithmetic.

Answer 5

Use filter function like this

len(filter(lambda x: x==letter, myString))