Python: Most efficient way to get two Boolean property frequencies in a list of objects?

I think using a collections.Counter is the right idea, just do it in a more generic way with a single Counter and single loop:

from collections import Counter

user_list = [User(True, False), User(False, True), User(True, True), User(False, False)]
user_attr_count = Counter()

for user in user_list:
    user_attr_count['a_%s' % user.a] += 1
    user_attr_count['b_%s' % user.b] += 1

print user_attr_count
# Counter({'b_False': 2, 'a_True': 2, 'b_True': 2, 'a_False': 2})

Why not use two counters, and subtract from the length of user_list to find the other two values?

a_false_count = len(user_list) - a_true_count

b_false_count = len(user_list) - b_true_count

Explicitly looping like that is probably the most efficient solution time-wise, but if you're looking for something a little more succinct code-wise, you might try filter():

a_false_count = len(filter(lambda x: x.a,user_list))
b_false_count = len(filter(lambda x: x.b,user_list))

You could use bit masking:

def count(user_list,mask):
    return Counter((u.a<<1 | u.b)&mask for u in user_list)

a=0b10
b=0b01
aANDb=0b11
print count(user_list,aANDb)

from collections import Counter

c = Counter()
for u in user_list:
    c['a'] += u.a
    c['b'] += u.b

print c['a'], len(user_list) - c['a'], c['b'], len(user_list) - c['b']