Return copy of dictionary excluding specified keys

Question

I want to make a function that returns a copy of a dictionary excluding keys specified in a list.

Considering this dictionary:

my_dict = {
    "keyA": 1,
    "keyB": 2,
    "keyC": 3
}

A call to without_keys(my_dict, ['keyB', 'keyC']) should return:

{
    "keyA": 1
}

I would like to do this in a one-line with a neat dictionary comprehension but I'm having trouble. My attempt is this:

def without_keys(d, keys):
    return {k: d[f] if k not in keys for f in d}

which is invalid syntax. How can I do this?

Anshul Goyal · Accepted Answer · 2015-07-15T15:08:05.753

77

You were close, try the snippet below:

>>> my_dict = {
...     "keyA": 1,
...     "keyB": 2,
...     "keyC": 3
... }
>>> invalid = {"keyA", "keyB"}
>>> def without_keys(d, keys):
...     return {x: d[x] for x in d if x not in keys}
>>> without_keys(my_dict, invalid)
{'keyC': 3}

Basically, the if k not in keys will go at the end of the dict comprehension in the above case.

edited Jul 15 '15 at 15:08

answered Jul 15 '15 at 15:01

Anshul Goyal

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You declared invalid as a set. Is there a reason for that? Would a list suffice as well? – Prithvi Boinpally Oct 23 '20 at 03:10
2

Yes, list will suffice as well (you can try it), but generally when order is not important for you, `set` is preferred (even from performance side which is not important in this short example with just a few data). – Nerxis Dec 10 '20 at 09:58
Is this faster than getting all keys and then manually skipping not needed keys with a for loop? – sebko_iic Dec 24 '20 at 15:16

score 43 · Answer 2 · answered Jul 15 '15 at 15:02

43

In your dictionary comprehension you should be iterating over your dictionary (not k , not sure what that is either). Example -

return {k:v for k,v in d.items() if k not in keys}

answered Jul 15 '15 at 15:02

Anand S Kumar

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I prefer this solution a little bit more because it makes the (tiny) save of not calling the `d[x]` index operation. – Alejandro Piad Jan 17 '19 at 18:33
1

this is a shallow copy though – Eissa N. Feb 19 '21 at 17:22

score 10 · Answer 3 · answered Jul 15 '15 at 15:01

10

This should work for you.

def without_keys(d, keys):
    return {k: v for k, v in d.items() if k not in keys}

answered Jul 15 '15 at 15:01

Morgan Thrapp

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score 9 · Answer 4 · answered Sep 17 '20 at 18:50

9

Even shorter. Apparently python 3 lets you 'subtract' a list from a dict_keys.

def without_keys(d, keys):
    return {k: d[k] for k in d.keys() - keys}

answered Sep 17 '20 at 18:50

Mike Fogel

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5

Note that using `keys()` instead of `items()` is ~33% slower: https://discuss.python.org/t/copy-a-dictionary-except-some-keys/2559/5 – Tuukka Mustonen Dec 16 '20 at 10:39

score 1 · Answer 5 · edited Oct 15 '18 at 02:27

1

For those who don't like list comprehensions, this is my version:

def without_keys(d, *keys):
     return dict(filter(lambda key_value: key_value[0] not in keys, d.items()))

Usage:

>>> d={1:3, 5:7, 9:11, 13:15}
>>> without_keys(d, 1, 5, 9)
{13: 15}
>>> without_keys(d, 13)
{1: 3, 5: 7, 9: 11}
>>> without_keys(d, *[5, 7])
{1: 3, 13: 15, 9: 11}

edited Oct 15 '18 at 02:27

Xiv

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answered Jan 02 '17 at 06:50

Himel Das

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For those who don't like list comprehensions, you should learn list comprehensions. They are very 'pythonic', and we they eventually click, you will be gald you spent the time to understand them, or else you are fighting the language. – run_the_race Sep 19 '22 at 18:21

Friedrich -- Слава Україні · Answer 6 · 2021-11-29T17:13:38.613

1

Your oneliner

my_dict = {"keyA": 1, "keyB": 2, "keyC": 3}
(lambda keyB, keyC, **kw: kw)(**my_dict)

which returns {'keyA': 1}. Not very pythonic and dynamic, but hacky and short. It uses the dict unpacking (destructuring assignment) of function arguments.

See also https://stackoverflow.com/a/53851069/11769765.

edited Nov 29 '21 at 17:13

answered Nov 29 '21 at 14:33

Friedrich -- Слава Україні

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score 0 · Answer 7 · answered Nov 17 '21 at 19:54

You could this generalized for nested dictionaries solution

def copy_dict(data, strip_values=False, remove_keys=[]):
    if type(data) is dict:
        out = {}
        for key, value in data.items():
            if key not in remove_keys:
                out[key] = copy_dict(value, strip_values=strip_values, remove_keys=remove_keys)
        return out
    else:
        return [] if strip_values else data

This recursive solution works for nested dictionaries and removes keys not required from the entire nested structure. It also gives you the ability to return the nest with only keys and no values.

Return copy of dictionary excluding specified keys

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