Pythonic way to check if two dictionaries have the identical set of keys?

Question:

For example, let’s say I have to dictionaries:

d_1 = {'peter': 1, 'adam': 2, 'david': 3}

and

d_2 = {'peter': 14, 'adam': 44, 'david': 33, 'alan': 21}

What’s the cleverest way to check whether the two dictionaries contain the same set of keys? In the example above, it should return False because d_2 contains the 'alan' key, which d_1 doesn’t.

I am not interested in checking that the associated values match. Just want to make sure if the keys are same.

Asked By: c00kiemonster

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Answers:

>>> not set(d_1).symmetric_difference(d_2)
False
>>> not set(d_1).symmetric_difference(dict.fromkeys(d_1))
True
Answered By: SilentGhost

A quick option (not sure if its the most optimal)

len(set(d_1.keys()).difference(d_2.keys())) == 0
Answered By: Alex

In Python2,

set(d_1) == set(d_2)

In Python3, you can do this which may be a tiny bit more efficient than creating sets

d1.keys() == d2.keys()

although the Python2 way would work too

Answered By: John La Rooy

One way is to check for symmetric difference (new set with elements in either s or t but not both):

set(d_1.keys()).symmetric_difference(set(d_2.keys()))

But a shorter way it to just compare the sets:

set(d_1) == set(d_2)
Answered By: miku

You can get the keys for a dictionary with dict.keys().

You can turn this into a set with set(dict.keys())

You can compare sets with ==

To sum up:

set(d_1.keys()) == set(d_2.keys())

will give you what you want.

Answered By: xorsyst
  • In Python 3, dict.keys() returns a “view object” that can be used like a set. This is much more efficient than constructing a separate set.

    d_1.keys() == d_2.keys()
    
  • In Python 2.7, dict.viewkeys() does the same thing.

    d_1.viewkeys() == d_2.viewkeys()
    
  • In Python 2.6 and below, you have to construct a set of the keys of each dict.

    set(d_1) == set(d_2)
    

    Or you can iterate over the keys yourself for greater memory efficiency.

    len(d_1) == len(d_2) and all(k in d_2 for k in d_1)
    
Answered By: augurar
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