How do I check that multiple keys are in a dict in a single pass?

Question

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?

Answer 1

Well, you could do this:

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

Answer 2

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): ...

Answer 3

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

Answer 4

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.

Answer 5

Using sets:

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

Alternatively:

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

Answer 6

This should work:

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

---

Hint:

Using square brackets inside all() to make a list comprehension:

if all([key in foo for key in ["foo","bar"]]):

Is not only unnecessary, but it is positively harmful, as they impede the normal short-circuiting behavior of all().

Answer 7

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.

Answer 8

I think this is the smartest and most Pythonic.

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

Answer 9

check for existence of all keys in a dict:

{'key_1', 'key_2', 'key_3'} <= set(my_dict)

check for existence of one or more keys in a dict:

{'key_1', 'key_2', 'key_3'} & set(my_dict)

Answer 10

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 11

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
>>>

Answer 12

short and sweet

{"key1", "key2"} <= {*dict_name}

Answer 13

How about using lambda?

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

Answer 14

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 15

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 16

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

Jason, () aren't necessary in Python.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

Here's an alternative solution in case you want to get the items that didn't match...

not_existing_keys = [item for item in ["foo","bar"] if item not in foo]
if not_existing_keys:
  log.error('These items are missing', not_existing_keys)

Answer 21

my_dict = {
    'name': 'Askavy',
    'country': 'India',
    'age': 30
}

if set(('name', 'country','age')).issubset(my_dict.keys()):
     print("All keys are present in the dictionary") 
else: 
    print("All keys are not present in  the dictionary") 

Answer 22

To me, simple and easy with None key in the middle with pydash ref

import pydash as _
_.get(d, 'key1.key2.key3.whatevermaybeNone.inthemiddle', default=None) )

Answer 23

If the reason you are checking for the keys is to make sure you can access them, one option is to not check - just try to access them and then do the needful if necessary.

foo = {
    'bar': 4
}

try:
    alpha = foo['bar']
    beta = foo['leroy_jenkins']
except KeyError:
    pass # Or some other thing if you don't have the keys you expected.

Answer 24

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

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

This seems to work