count the number of upper case and lower case characters in word

Question

def Charater(): 
    UpperCount = 0 
    LowerCount = 0
    word = input('Enter a word: ')
    for letter in word:
        if letter == letter.upper 
            UpperCount = UpperCount + 1
            return UpperCount
        else:
            LowerCount = LowerCount + 1
            return LowerCount

print(Charater())

Please don't judge me if this looks bad. But am a beginner I am trying to make the code count how many upper and lower case characters in a word which is inputted by the user. Each time I do this it returns 1. (Its probably my if statement). can someone please point out the problem and tell me how to resolve it.

Solution 1

You may use map with str.isupper and str.islower to find the count of uppercased and lowercased characters respectively as:

>>> my_word = "HelLo WorLd"
>>> lower_count = sum(map(str.islower, my_word))
>>> lower_count
6

>>> upper_count = sum(map(str.isupper, my_word))
>>> upper_count
4

Solution 2

Count in one line using collections.Counter, generator comprehension and a nested ternary:

import collections

my_word = "HelLo WorLd"
c = collections.Counter("upper" if x.isupper() else "lower" if x.islower() else "" for x in my_word)
print(c)

result:

Counter({'lower': 6, 'upper': 4, '': 1})

Comments

[juanpa.arrivillaga]

you should just use 'some_string'.isupper()

[Elmex80s]

@juanpa.arrivillaga Thanks fixed it

[juanpa.arrivillaga]

Personally, I prefer to use sum if I just want to count when sum boolean condition is True, so sum(map(str.islower, word)) for example. It also has the advantage of being python 2-3 compatible.

[Moinuddin Quadri]

@juanpa.arrivillaga makes sense. Also as I see that OP is looking for Python3 solution, updated the answer