Conditional counting in Python

print sum(1 for e in L if e.b == 1)

sum(x.b == 1 for x in L)

A boolean (as resulting from comparisons such as x.b == 1) is also an int, with a value of 0 for False, 1 for True, so arithmetic such as summation works just fine.

This is the simplest code, but perhaps not the speediest (only timeit can tell you for sure;-). Consider (simplified case to fit well on command lines, but equivalent):

$ py26 -mtimeit -s'L=[1,2,1,3,1]*100' 'len([x for x in L if x==1])'
10000 loops, best of 3: 56.6 usec per loop
$ py26 -mtimeit -s'L=[1,2,1,3,1]*100' 'sum(x==1 for x in L)'
10000 loops, best of 3: 87.7 usec per loop

So, for this case, the "memory wasteful" approach of generating an extra temporary list and checking its length is actually solidly faster than the simpler, shorter, memory-thrifty one I tend to prefer. Other mixes of list values, Python implementations, availability of memory to "invest" in this speedup, etc, can affect the exact performance, of course.

I would prefer the second one as it's only looping over the list once.

If you use count() you're looping over the list once to get the b values, and then looping over it again to see how many of them equal 1.

A neat way may to use reduce():

reduce(lambda x,y: x + (1 if y.b == 1 else 0),list,0)

The documentation tells us that reduce() will:

Apply function of two arguments cumulatively to the items of iterable, from left to right, so as to reduce the iterable to a single value.

So we define a lambda that adds one the accumulated value only if the list item's b attribute is 1.

To hide reduce details, you may define a count function:

def count(condition, stuff):
    return reduce(lambda s, x: \
                  s + (1 if condition(x) else 0), stuff, 0)

Then you may use it by providing the condition for counting:

n = count(lambda i: i.b, stuff)