I want to do a function that counts the consonants in a string, so I tried to do this:
def vowel_count(foo):
count = 0
for i in foo:
if not i == 'a' and i == 'e' and i ... and i == 'O' and i == 'U':
count += 1
return count
But this is pretty ugly and tedious to do, more so with many more conditions. Is there any way to put them together?
You are looking for the not in operator.
def vowel_count(foo):
count = 0
for i in foo.lower():
if i not in 'aeiou':
count += 1
return count
or more simply:
def vowel_count(foo):
return sum(i not in 'aeiou' for i in foo.lower()) # True == 1, False == 0
... if i not in was normally faster than cooercing i not in ... from bool to int?
set(...) -- it's much slower. It's creating a set from the string on each iteration. You can use set operations instead.
in is an O(n) operation for a string, it's such a short string that the string is still slightly faster than the set.
in works better for set or for str (I suspect set but won't state it firmly without profiling) but you can definitely build a set without it impacting performance.Use a container to store vowels in their lower form:
def vowel_count(foo):
vowel = ["a", "e", "i", "o", "u", "y"]
count = 0
for i in foo:
if i.lower() not in vowel:
count += 1
Then you could also do this a bit differently with a filter for instance.
def condition(letter):
vowel = ["a", "e", "i", "o", "u", "y"]
if letter.lower() not in vowel and letter != " ":
return True
else:
return False
foo = 'bnojouearoui okeorjaji '
count = len(list(filter(condition, foo)))
i.lower() in "aeiou"?not a and b and cis understood as(not a) and b and c, which is quite different fromnot (a and b and c). It is pretty hard for a character to be "e" and "O" and "U" all at the same time, even if it is not "a".sum(c for k, c in collections.Counter(i.lower()) if k in 'aeiou')