Check if any value of a dictionary matches a condition

Question

How does a python programmer check if any value of a dictionary matches a condition (is greater than 0 in my case). I'm looking for the most "pythonic" way that has minimal performance-impact.

my dictionary:

pairs = { 'word1':0, 'word2':0, 'word3':2000, 'word4':64, 'word5':0, 'wordn':8 }

I used these 2 (monstrous?) methods so far.

1:

options = pairs.values() # extract values
for i in options:
    if i > 0:
        return True
return False

2:

options = sorted(pairs.items(), key=lambda e: e[1], reverse=True) # rank from max to min
if options[0][1] > 0:
    return True
else:
    return False

Answer 1

You can use any [docs]:

>>> pairs = { 'word1':0, 'word2':0, 'word3':2000, 'word4':64, 'word5':0, 'wordn':8 }
>>> any(v > 0 for v in pairs.itervalues())
True
>>> any(v > 3000 for v in pairs.itervalues())
False

See also all [docs]:

>>> all(v > 0 for v in pairs.itervalues())
False
>>> all(v < 3000 for v in pairs.itervalues())
True

Since you're using Python 2.7, .itervalues() is probably a little better than .values() because it doesn't create a new list.