Should I use 'has_key()' or 'in' on Python dicts?



>>> d = {'a': 1, 'b': 2}

Which of the following is the best way to check if 'a' is in d?

>>> 'a' in d
>>> d.has_key('a')
Asked By: igorgue



in is definitely more pythonic.

In fact has_key() was removed in Python 3.x.

Answered By: tonfa

According to python docs:

has_key() is deprecated in favor of
key in d.

Answered By: Nadia Alramli

in wins hands-down, not just in elegance (and not being deprecated;-) but also in performance, e.g.:

$ python -mtimeit -s'd=dict.fromkeys(range(99))' '12 in d'
10000000 loops, best of 3: 0.0983 usec per loop
$ python -mtimeit -s'd=dict.fromkeys(range(99))' 'd.has_key(12)'
1000000 loops, best of 3: 0.21 usec per loop

While the following observation is not always true, you’ll notice that usually, in Python, the faster solution is more elegant and Pythonic; that’s why -mtimeit is SO helpful — it’s not just about saving a hundred nanoseconds here and there!-)

Answered By: Alex Martelli

has_key is a dictionary method, but in will work on any collection, and even when __contains__ is missing, in will use any other method to iterate the collection to find out.

Answered By: u0b34a0f6ae

Use dict.has_key() if (and only if) your code is required to be runnable by Python versions earlier than 2.3 (when key in dict was introduced).

Answered By: John Machin

There is one example where in actually kills your performance.

If you use in on a O(1) container that only implements __getitem__ and has_key() but not __contains__ you will turn an O(1) search into an O(N) search (as in falls back to a linear search via __getitem__).

Fix is obviously trivial:

def __contains__(self, x):
    return self.has_key(x)
Answered By: schlenk

Solution to dict.has_key() is deprecated, use ‘in’ — sublime text editor 3

Here I have taken an example of dictionary named ‘ages’ –

ages = {}

# Add a couple of names to the dictionary
ages['Sue'] = 23

ages['Peter'] = 19

ages['Andrew'] = 78

ages['Karren'] = 45

# use of 'in' in if condition instead of function_name.has_key(key-name).
if 'Sue' in ages:

    print "Sue is in the dictionary. She is", ages['Sue'], "years old"


    print "Sue is not in the dictionary"
Answered By: Greena modi

Expanding on Alex Martelli’s performance tests with Adam Parkin’s comments…

$ python3.5 -mtimeit -s'd=dict.fromkeys(range( 99))' 'd.has_key(12)'
Traceback (most recent call last):
  File "/usr/local/Cellar/python3/3.5.2_3/Frameworks/Python.framework/Versions/3.5/lib/python3.5/", line 301, in main
    x = t.timeit(number)
  File "/usr/local/Cellar/python3/3.5.2_3/Frameworks/Python.framework/Versions/3.5/lib/python3.5/", line 178, in timeit
    timing = self.inner(it, self.timer)
  File "<timeit-src>", line 6, in inner
AttributeError: 'dict' object has no attribute 'has_key'

$ python2.7 -mtimeit -s'd=dict.fromkeys(range(  99))' 'd.has_key(12)'
10000000 loops, best of 3: 0.0872 usec per loop

$ python2.7 -mtimeit -s'd=dict.fromkeys(range(1999))' 'd.has_key(12)'
10000000 loops, best of 3: 0.0858 usec per loop

$ python3.5 -mtimeit -s'd=dict.fromkeys(range(  99))' '12 in d'
10000000 loops, best of 3: 0.031 usec per loop

$ python3.5 -mtimeit -s'd=dict.fromkeys(range(1999))' '12 in d'
10000000 loops, best of 3: 0.033 usec per loop

$ python3.5 -mtimeit -s'd=dict.fromkeys(range(  99))' '12 in d.keys()'
10000000 loops, best of 3: 0.115 usec per loop

$ python3.5 -mtimeit -s'd=dict.fromkeys(range(1999))' '12 in d.keys()'
10000000 loops, best of 3: 0.117 usec per loop
Answered By: Bruno Bronosky

If you have something like this:


change it to below for running on Python 3.X and above:

key = ew
if key not in t
Answered By: Harshita Jhavar
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