Find first item with alphabetical precedence in list with numbers

Question

Say I have a list object occupied with both numbers and strings. If I want to retrieve the first string item with the highest alphabetical precedence, how would I do so?

Here is an example attempt which is clearly incorrect, but corrections as to what needs to be changed in order for it to achieve the desired result would be greatly appreciated:

lst = [12, 4, 2, 15, 3, 'ALLIGATOR', 'BEAR', 'ANTEATER', 'DOG', 'CAT']

lst.sort()
for i in lst:
   if i[0] == "A":
      answer = i
print(answer)

Answer 1

First use a generator expression to filter out non-strings, and then use min() to select the string with the highest alphabetical presence:

>>> min(x for x in lst if isinstance(x, str))
'ALLIGATOR

Answer 2

IIUC you could use isinstance to get sublist of your original list with only strings, then with sorted get first element by alphabetical sorting:

sub_lst = [i for i in lst if isinstance(i, str)]
result = sorted(sub_lst)[0]


print(sub_lst)
['ALLIGATOR', 'BEAR', 'ANTEATER', 'DOG', 'CAT']

print(result)
'ALLIGATOR'

Or you could use min as @TigerhawkT3 suggested in the comment:

print(min(sub_lst))
'ALLIGATOR'

Answer 3

Another way is to filter the main list lst from intergers using filter built-in method:

>>> min(filter(lambda s:isinstance(s, str), lst))
'ALLIGATOR'