One liner to determine if dictionary values are all empty lists or not

Question

I have a dict as follows:

someDict = {'a':[], 'b':[]}

I want to determine if this dictionary has any values which are not empty lists. If so, I want to return True. If not, I want to return False. Any way to make this a one liner?

Answer 1

Per my testing, the following one-liner (my original answer) has best time performance in all scenarios. See edits below for testing information. I do acknowledge that solutions using generator expressions will be much more memory efficient and should be preferred for large dicts.

EDIT: This is an aging answer and the results of my testing may not be valid for the latest version of python. Since generator expressions are the more "pythonic" way, I'd imagine their performance is improving. Please do your own testing if you're running this in a 'hot' codepath.

bool([a for a in my_dict.values() if a != []])

Edit:

Decided to have some fun. A comparison of answers, not in any particular order:

(As used below, timeit will calculate a loop order of magnitude based on what will take less than 0.2 seconds to run)

bool([a for a in my_dict.values() if a != []]) :

python -mtimeit -s"my_dict={'a':[],'b':[]}" "bool([a for a in my_dict.values() if a != []])"
1000000 loops, best of 3: 0.875 usec per loop

any([my_dict[i] != [] for i in my_dict]) :

python -mtimeit -s"my_dict={'a':[],'b':[]}" "any([my_dict[i] != [] for i in my_dict])"
1000000 loops, best of 3: 0.821 usec per loop

any(x != [] for x in my_dict.itervalues()):

python -mtimeit -s"my_dict={'a':[],'b':[]}" "any(x != [] for x in my_dict.itervalues())"
1000000 loops, best of 3: 1.03 usec per loop

all(map(lambda x: x == [], my_dict.values())):

python -mtimeit -s"my_dict={'a':[],'b':[]}" "all(map(lambda x: x == [], my_dict.values()))"
1000000 loops, best of 3: 1.47 usec per loop

filter(lambda x: x != [], my_dict.values()):

python -mtimeit -s"my_dict={'a':[],'b':[]}" "filter(lambda x: x != [], my_dict.values())"
1000000 loops, best of 3: 1.19 usec per loop

Edit again - more fun:

any() is best case O(1) (if bool(list[0]) returns True). any()'s worst case is the "positive" scenario - a long list of values for which bool(list[i]) returns False.

Check out what happens when the dict gets big:

bool([a for a in my_dict.values() if a != []]) :

#n=1000
python -mtimeit -s"my_dict=dict(zip(range(1000),[[]]*1000))" "bool([a for a in my_dict.values() if a != []])"
10000 loops, best of 3: 126 usec per loop

#n=100000
python -mtimeit -s"my_dict=dict(zip(range(100000),[[]]*100000))" "bool([a for a in my_dict.values() if a != []])"
100 loops, best of 3: 14.2 msec per loop

any([my_dict[i] != [] for i in my_dict]):

#n=1000
python -mtimeit -s"my_dict=dict(zip(range(1000),[[]]*1000))" "any([my_dict[i] != [] for i in my_dict])"
10000 loops, best of 3: 198 usec per loop

#n=100000
python -mtimeit -s"my_dict=dict(zip(range(100000),[[]]*100000))" "any([my_dict[i] != [] for i in my_dict])"
10 loops, best of 3: 21.1 msec per loop

But that's not enough - what about a worst-case 'False' scenario?

bool([a for a in my_dict.values() if a != []]) :

python -mtimeit -s"my_dict=dict(zip(range(1000),[0]*1000))" "bool([a for a in my_dict.values() if a != []])"
10000 loops, best of 3: 198 usec per loop

any([my_dict[i] != [] for i in my_dict]) :

python -mtimeit -s"my_dict=dict(zip(range(1000),[0]*1000))" "any([my_dict[i] != [] for i in my_dict])"
1000 loops, best of 3: 265 usec per loop

Answer 2

Not falsey or not empty lists:

Not falsey:

any(someDict.values())

Not empty lists:

any(a != [] for a in someDict.values())

or

any(map(lambda x: x != [], someDict.values()))

Or if you are ok with a falsey return value:

filter(lambda x: x != [], someDict.values())

Returns a list of items that are not empty lists, so if they are all empty lists it's an empty list :)

Answer 3

Quite literally:

any(x != [] for x in someDict.itervalues())

Answer 4

try this

 all([d[i] == [] for i in d])

edit: oops, i think i got you backwards. lets deMorgan that

any([d[i] != [] for i in d])

this second way has the short-circuit advantage on the first anyhow

Answer 5

>>> someDict = {'a':[], 'b':[]} 
>>> all(map(lambda x: x == [], someDict.values()))
True

Answer 6

len(filter(lambda x: x!=[], someDict.values())) != 0