What is a good way to do countif in Python

I want to count how many members of an iterable meet a given condition. I'd like to do it in a way that is clear and simple and preferably reasonably optimal.

My current best ideas are:

sum(meets_condition(x) for x in my_list)

and

len([x for x in my_list if meets_condition(x)])

The first one being iterator based is presumably faster for big lists. And it's the same form as you'd use for testing any and all. However it depends on the fact that int(True) == 1, which is somewhat ugly.

The second one seems easier to read to me, but it is different from the any and all forms.

Does anyone have any better suggestions? is there a library function somewhere that I am missing?

The iterator based approach is just fine. There are some slight modifications that can emphasize the fact that you are counting:

sum(1 if meets_condition(x) else 0 for x in my_list)
# or 
sum(1 for x in my_list if meets_condition(x))

And as always, if the intent isn't apparent from the code, encapsulate it in descriptively named function:

def count_matching(condition, seq):
    """Returns the amount of items in seq that return true from condition"""
    return sum(1 for item in seq if condition(item))

count_matching(meets_condition, my_list)

The first one

sum(meets_condition(x) for x in my_list)

looks perfectly readable and pythonic to me.

If you prefer the second approach I'd go for

len(filter(meets_condition, my_list))

Yet another way could be:

map(meets_condition, my_list).count(True)

countif for a list

#counting if a number or string is in a list
my_list=[1,2,3,2,3,1,1,1,1,1, "dave" , "dave"]
one=sum(1 for item in my_list if item==(1))
two=sum(1 for item in my_list if item==(2))
three=sum(1 for item in my_list if item==(3))
dave=sum(1 for item in my_list if item==("dave"))
print("number of one's in my_list > " , one)
print("number of two's in my_list > " , two)
print("number of three's in my_list > " , three)
print("number of dave's in my_list > " , dave)