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, \         


        

Answer 1

• 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)
  

Answer 2

>>> not set(d_1).symmetric_difference(d_2)
False
>>> not set(d_1).symmetric_difference(dict.fromkeys(d_1))
True

Answer 3

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.

Answer 4

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)

Answer 5

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

len(set(d_1.keys()).difference(d_2.keys())) == 0

Answer 6

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