Python-从字符串列表中删除小写或大写字母？

Question

So I'm a noob programmer...dont hate my code. But I'm struggling a bit here. I'm trying to take a list of strings, and find the string that has the lowest amount of upper case letters.

I've tried a few different things. -Just counting the upper case(didnt work) -Removing lower case and then doing min(list, key=len) but that didnt work either.

I'm stuck...heres what I got as of now.

 test_set = {'MOo', 'QHue', 'ReP', 'XiIV', 'oEe'}

 def fewest_unsolved(group):
     #shortest = min(group, key = len)
     #return shortest

     for word in group:

         for i in word:
             if i == i.lower():
                 word.strip(i)
             shortest = min(group, key = len)
             return shortest   

fewest_unsolved(test_set)

Now this just returns the first string in the list that is the shortest of the list

Answer 1

Why not use only min with a lambda that counts the number of uppercases?

>>> data = {'MOo', 'QHue', 'ReP', 'XiIV', 'oEe'}
>>> min(data, key = lambda x: sum('A' <= c <= 'Z' for c in x))
'oEe'

Answer 2

You can do this using a list comprehension:

def lowercase_count(word):
  lowercase = [c for c in word if c.islower()]
  return len(lowercase)

Knowing the count of lowercase words,you can then do the following:

def fewest_unsolved(group):
  least = lowercase_count(group[0])
  current = group[0]
  for word in group[1:]:
    count = lowercase_count(word)
    if count < least:
      least = count
      current = word
  return current

This is a bit more verbose than some other solutions, but I think it might be a bit more readable :)

Oh, by the way, this crashes if you pass it an empty list, so be careful if you use it.

Answer 3

If your goal is truly well summarized by

I'm trying to take a list of strings, and find the string that has the lowest amount of upper case letters.

then probably

Removing lower or upper case letters from a list of strings

(ie your title) is not going to be the best way to do it. (This is an example of the "XY problem".)

This will do the former (but not the latter):

>>> test_set = {'MOo', 'QHue', 'ReP', 'XiIV', 'oEe'}
>>> ranked = sorted([sum('A'<=c<='Z' for c in s), s] for s in test_set) 
>>> print(ranked)
[[1, 'oEe'], [2, 'MOo'], [2, 'QHue'], [2, 'ReP'], [3, 'XiIV']]

You can see that ranked[0][1] is the string with the smallest, and ranked[-1][1] the string with the largest, number of capitals. You could directly use min() instead of sorted() to get the smallest-number string

>>> capcount, string = min( [sum('A'<=c<='Z' for c in s), s] for s in test_set )

but I figured using sorted would give you richer information that would allow you to check for ties (at the expense of time- and memory-complexity).