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How to remove a key from a Python dictionary?


Question

When deleting a key from a dictionary, I use:

if 'key' in my_dict:
    del my_dict['key']

Is there a one line way of doing this?

2020/04/03
1
1888
4/3/2020 8:21:50 AM

Accepted Answer

To delete a key regardless of whether it is in the dictionary, use the two-argument form of dict.pop():

my_dict.pop('key', None)

This will return my_dict[key] if key exists in the dictionary, and None otherwise. If the second parameter is not specified (ie. my_dict.pop('key')) and key does not exist, a KeyError is raised.

To delete a key that is guaranteed to exist, you can also use

del my_dict['key']

This will raise a KeyError if the key is not in the dictionary.

2020/03/15
3037
3/15/2020 8:20:40 PM

Specifically to answer "is there a one line way of doing this?"

if 'key' in my_dict: del my_dict['key']

...well, you asked ;-)

You should consider, though, that this way of deleting an object from a dict is not atomic—it is possible that 'key' may be in my_dict during the if statement, but may be deleted before del is executed, in which case del will fail with a KeyError. Given this, it would be safest to either use dict.pop or something along the lines of

try:
    del my_dict['key']
except KeyError:
    pass

which, of course, is definitely not a one-liner.

2020/04/03

It took me some time to figure out what exactly my_dict.pop("key", None) is doing. So I'll add this as an answer to save others Googling time:

pop(key[, default])

If key is in the dictionary, remove it and return its value, else return default. If default is not given and key is not in the dictionary, a KeyError is raised.

Documentation

2019/11/07

del my_dict[key] is slightly faster than my_dict.pop(key) for removing a key from a dictionary when the key exists

>>> import timeit
>>> setup = "d = {i: i for i in range(100000)}"

>>> timeit.timeit("del d[3]", setup=setup, number=1)
1.79e-06
>>> timeit.timeit("d.pop(3)", setup=setup, number=1)
2.09e-06
>>> timeit.timeit("d2 = {key: val for key, val in d.items() if key != 3}", setup=setup, number=1)
0.00786

But when the key doesn't exist if key in my_dict: del my_dict[key] is slightly faster than my_dict.pop(key, None). Both are at least three times faster than del in a try/except statement:

>>> timeit.timeit("if 'missing key' in d: del d['missing key']", setup=setup)
0.0229
>>> timeit.timeit("d.pop('missing key', None)", setup=setup)
0.0426
>>> try_except = """
... try:
...     del d['missing key']
... except KeyError:
...     pass
... """
>>> timeit.timeit(try_except, setup=setup)
0.133
2020/04/03

If you need to remove a lot of keys from a dictionary in one line of code, I think using map() is quite succinct and Pythonic readable:

myDict = {'a':1,'b':2,'c':3,'d':4}
map(myDict.pop, ['a','c']) # The list of keys to remove
>>> myDict
{'b': 2, 'd': 4}

And if you need to catch errors where you pop a value that isn't in the dictionary, use lambda inside map() like this:

map(lambda x: myDict.pop(x,None), ['a', 'c', 'e'])
[1, 3, None] # pop returns
>>> myDict
{'b': 2, 'd': 4}

or in python3, you must use a list comprehension instead:

[myDict.pop(x, None) for x in ['a', 'c', 'e']]

It works. And 'e' did not cause an error, even though myDict did not have an 'e' key.

2019/05/07

You can use a dictionary comprehension to create a new dictionary with that key removed:

>>> my_dict = {k: v for k, v in my_dict.items() if k != 'key'}

You can delete by conditions. No error if key doesn't exist.

2020/04/03