Finding Number of Lowercase letters in a Set

[Steven]

I have a set being passed to a function. Each element has a word in it with a mixture of upper and lower case characters. I need to find the one with the least uppercase letters and return it. I'm at a loss, but I'll show you what I've done so far. All the print statements are so I can see what I have done. 

count = str(group)
    print(count)
    count = sum(1 for character in group if character.islower())
    countString = ','.join(str(string) for string in group)
    print(countString)
    countString = countString.split()
    print(countString)
    countString = countString.strip().split(",")
    print(countString)
    print(count)
    
    #word.split(",")
    #print(word)

Any ideas?

[wavic]

Why you count the lower case characters when you need to know how many upper case characters are in each word?

[Ofnuts]

Make smaller problems:

1. create a function that counts the number of upper case characters in a single word
   
2. create a second function that
   
   1. sets some length variable some absurdly high value,
      
   2. iterates your words,
      
   3. if the count returned by the first function is smaller than the current value of length, updates length with that value and stores the corresponding word.
      
   4. Finally return the last value of length and the associated word.
      

[ankit]

hi, try this to resolve your problem and change in indentation of function fun as required:

def fun(set_value):
   dic={}
   for i in set_value:
       s=0
       for j in i:
         if j.isupper():
               s+=1
       dic[i]=s
   return [j for j in dic if dic[j]==min([dic[i] for i in dic])]

set_value =set(['qweRW','SdfA','tfgAA','llQl','llPl','llPl'])
fun(set_value)

I have tried and its working...........

[Steven]

Awesome, this is what I did.

fewestLetters = ''
    removeLetters = 26
    for i in group:
        iValue = len({x for x in i if x in ascii_uppercase})
        if iValue <= removeLetters:
            removeLetters = iValue
            fewestLetters = i
    return fewestLetters

[volcano63]

Here is a simple one-liner solution, using RE and min function

import re
shortest = min(words, key=lambda w: len(re.findall('[A-Z]', w)))

As a lambda function you may also use something similar to the way you counted lower case - with a little twist

sum(c.isupper() for c in w)

which may be or may not be more efficient   .

Here I use (abuse   ?!) the fact that Python allows implicit convert of False and True to 0 and 1  - which makes writing conditional increments very easy.

PS While I applaud using meaningful names for variables (one-letter names must be made illegal - unless in one-liner code snippets ), I would suggest to use Pythonic naming conventions - see PEP-8

[Steven]

Thanks, little late, but that's way better.