python: How do I check that multiple keys are in a dict in one go?

Question

I want to do something like:

foo = {'foo':1,'zip':2,'zam':3,'bar':4}

if ("foo","bar") in foo:
    #do stuff

I'm not sure if its possible but would like to know. :-)

Answer 1

How about

if all([key in foo for key in ["foo","bar"]]):
    #do stuff
    pass

Answer 2

Well, you could do this:

>>> if all (k in foo for k in ("foo","bar")):
...     print "They're there!"
...
They're there!

Answer 3

How about using lambda?

 if reduce( (lambda x, y: x and foo.has_key(y) ), [ True, "foo", "bar"] ): # do stuff

Answer 4

if set(("foo", "bar")) <= set(myDict): ...

Answer 5

Using sets:

if set(("foo", "bar")).issubset(foo):
    #do stuff

Alternatively:

if set(("foo", "bar")) <= set(foo):
    #do stuff

Answer 6

In case you want to:

• also get the values for the keys
• check more than one dictonary

then:

from operator import itemgetter
foo = {'foo':1,'zip':2,'zam':3,'bar':4}
keys = ("foo","bar") 
getter = itemgetter(*keys) # returns all values
try:
    values = getter(foo)
except KeyError:
    # not both keys exist
    pass

Answer 7

Not to suggest that this isn't something that you haven't thought of, but I find that the simplest thing is usually the best:

if ("foo" in foo) and ("bar" in foo):
    # do stuff

Answer 8

>>> if 'foo' in foo and 'bar' in foo:
...     print 'yes'
... 
yes

Jason, () aren't necessary in Python.

Answer 9

Alex Martelli's solution set(queries) <= set(my_dict) is the shortest code but may not be the fastest. Assume Q = len(queries) and D = len(my_dict).

This takes O(Q) + O(D) to make the two sets, and then (one hopes!) only O(min(Q,D)) to do the subset test -- assuming of course that Python set look-up is O(1) -- this is worst case (when the answer is True).

The generator solution of hughdbrown (et al?) all(k in my_dict for k in queries) is worst-case O(Q).

Complicating factors:
(1) the loops in the set-based gadget are all done at C-speed whereas the any-based gadget is looping over bytecode.
(2) The caller of the any-based gadget may be able to use any knowledge of probability of failure to order the query items accordingly whereas the set-based gadget allows no such control.

As always, if speed is important, benchmarking under operational conditions is a good idea.

Answer 10

Simple benchmarking rig for 3 of the alternatives.

Put in your own values for D and Q

>>> from timeit import Timer
>>> setup='''from random import randint as R;d=dict((str(R(0,1000000)),R(0,1000000)) for i in range(D));q=dict((str(R(0,1000000)),R(0,1000000)) for i in range(Q));print("looking for %s items in %s"%(len(q),len(d)))'''

>>> Timer('set(q) <= set(d)','D=1000000;Q=100;'+setup).timeit(1)
looking for 100 items in 632499
0.28672504425048828

#This one only works for Python3
>>> Timer('set(q) <= d.keys()','D=1000000;Q=100;'+setup).timeit(1)
looking for 100 items in 632084
2.5987625122070312e-05

>>> Timer('all(k in d for k in q)','D=1000000;Q=100;'+setup).timeit(1)
looking for 100 items in 632219
1.1920928955078125e-05