Python List Comprehension: Count the number of spaces in a string

count_spaces.py

'''
Count the number of spaces in a string
'''
some_string = 'the slow solid squid swam sumptuously through the slimy swamp'
spaces = [s for s in some_string if s == ' ']
print(len(spaces))

I try it:
list1 = [x for x in str1 if x.isspace()]

[i for i in some_string].count(' ')

len([letter for letter in some_string if letter == ' '])

print(some_string.count(" "))

Is this still list comprehension without [i for i ...] format?
But if the result we want to get is a list with a number of spaces in the string this code should be fine:

space_find = [some_string.count(' ')]
print(space_find)

l=sum([1 for i in range(len(s))if s[i]==' '])

Is this still list comprehension without [i for i ...] format? But if the result we want to get is a list with a number of spaces in the string this code should be fine:

space_find = [some_string.count(' ')] print(space_find)

here you are not using list comprehension

the_number_of_white_Spaces_in_any_string = [input(str('Waiting for a sentence...')).count(' ')]

res = sum([1 for x in s if x == ' '])

I have a doubt here, if some_string.count(' ') could already give the number of white spaces available with a simple line.
Then do we need List Comprehension ?
which one is more efficient

spaces = " "
num_of_space = [len(space) for space in spaces if space == " "]
print(num_of_space)

l = some_string.count(" ")
print(l)

l = some_string.split()
print(len(l) - 1)