Multiple conditions python

Question

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?

Also note that your original code does not do what you want. `not a and b and c` is understood as `(not a) and b and c`, which is quite different from `not (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". — Amadan, Aug 13 '18 at 12:32
`sum(c for k, c in collections.Counter(i.lower()) if k in 'aeiou')` — Chris_Rands, Aug 13 '18 at 12:33

chepner · Accepted Answer · 2018-08-13T13:09:50.780

6

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

edited Aug 13 '18 at 13:09

answered Aug 13 '18 at 12:34

chepner

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is it better to build the set first once? Also I thought using `... if i not in` was normally faster than cooercing `i not in ...` from bool to int? – Chris_Rands Aug 13 '18 at 12:39
1

No need to do `set(...)` -- it's much slower. It's creating a set from the string on each iteration. You can use `set` operations instead. – warvariuc Aug 13 '18 at 12:39
@warvariuc I was working under the incorrect assumption that the optimizer would create the set at compile time. Also, though technically `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. – chepner Aug 13 '18 at 13:09
1

Or, in short, beware premature optimization ;) – chepner Aug 13 '18 at 13:10
@warvariuc: You could make a set of counted (or discounted) characters, either before the loop or even outside the function. I don't see why it would need to be constructed each iteration... The question is just whether `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. – Amadan Aug 14 '18 at 07:31
@Amadan I timed `'x' in 'aeiou'` and `'x' in set_of_vowels`, and `str.__contains__` came back slightly faster for this small set. Switching from a set of vowels to a set of consonants is sufficient for the set to win. – chepner Aug 14 '18 at 12:43

score 0 · Answer 2 · answered Aug 13 '18 at 12:37

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)))

Multiple conditions python

2 Answers2