如何在一次通过中检查多个键是否在字典中？

I want to do something like:

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

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

How do I check whether both foo and bar are in dict foo?

Well, you could do this:

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

if {"foo", "bar"} <= myDict.keys(): ...

If you’re still on Python 2, you can do

if {"foo", "bar"} <= myDict.viewkeys(): ...

If you’re still on a really old Python <= 2.6, you can call set on the dict, but it’ll iterate over the whole dict to build the set, and that’s slow:

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

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

You don’t have to wrap the left side in a set. You can just do this:

if {'foo', 'bar'} <= set(some_dict):
    pass

This also performs better than the all(k in d...) solution.

Using sets:

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

Alternatively:

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

How about this:

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

I think this is the smartest and pithonic.

{'key1','key2'} <= my_dict.keys()

While I like Alex Martelli’s answer, it doesn’t seem Pythonic to me. That is, I thought an important part of being Pythonic is to be easily understandable. With that goal, <= isn’t easy to understand.

While it’s more characters, using issubset() as suggested by Karl Voigtland’s answer is more understandable. Since that method can use a dictionary as an argument, a short, understandable solution is:

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

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

I’d like to use {'foo', 'bar'} in place of set(('foo', 'bar')), because it’s shorter. However, it’s not that understandable and I think the braces are too easily confused as being a dictionary.

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.

You can use .issubset() as well

>>> {"key1", "key2"}.issubset({"key1":1, "key2":2, "key3": 3})
True
>>> {"key4", "key2"}.issubset({"key1":1, "key2":2, "key3": 3})
False
>>>

How about using lambda?

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

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

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

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

Jason, () aren’t necessary in Python.

Just my take on this, there are two methods that are easy to understand of all the given options. So my main criteria is have very readable code, not exceptionally fast code. To keep code understandable, i prefer to given possibilities:

• var <= var2.keys()
• var.issubset(var2)

The fact that “var <= var2.keys()” executes faster in my testing below, i prefer this one.

import timeit

timeit.timeit('var <= var2.keys()', setup='var={"managed_ip", "hostname", "fqdn"}; var2= {"zone": "test-domain1.var23.com", "hostname": "bakje", "api_client_ip": "127.0.0.1", "request_data": "", "request_method": "GET", "request_url": "hvar2p://127.0.0.1:5000/test-domain1.var23.com/bakje", "utc_datetime": "04-Apr-2019 07:01:10", "fqdn": "bakje.test-domain1.var23.com"}; var={"managed_ip", "hostname", "fqdn"}')
0.1745898080000643

timeit.timeit('var.issubset(var2)', setup='var={"managed_ip", "hostname", "fqdn"}; var2= {"zone": "test-domain1.var23.com", "hostname": "bakje", "api_client_ip": "127.0.0.1", "request_data": "", "request_method": "GET", "request_url": "hvar2p://127.0.0.1:5000/test-domain1.var23.com/bakje", "utc_datetime": "04-Apr-2019 07:01:10", "fqdn": "bakje.test-domain1.var23.com"}; var={"managed_ip", "hostname", "fqdn"};')
0.2644960229999924

In the case of determining whether only some keys match, this works:

any_keys_i_seek = ["key1", "key2", "key3"]

if set(my_dict).intersection(any_keys_i_seek):
    # code_here
    pass

Yet another option to find if only some keys match:

any_keys_i_seek = ["key1", "key2", "key3"]

if any_keys_i_seek & my_dict.keys():
    # code_here
    pass

Another option for detecting whether all keys are in a dict:

dict_to_test = { ... }  # dict
keys_sought = { "key_sought_1", "key_sought_2", "key_sought_3" }  # set

if keys_sought & dict_to_test.keys() == keys_sought: 
    # True -- dict_to_test contains all keys in keys_sought
    # code_here
    pass

>>> ok
{'five': '5', 'two': '2', 'one': '1'}

>>> if ('two' and 'one' and 'five') in ok:
...   print "cool"
... 
cool

This seems to work