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?

Solution 1

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.

Solution 2

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}

Solution 3

This should work for you.

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

Solution 4

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}

Solution 5

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}

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