477

Is there any other way to delete an item in a dictionary only if the given key exists, other than:

if key in mydict:
    del mydict[key]

The scenario is that I'm given a collection of keys to be removed from a given dictionary, but I am not certain if all of them exist in the dictionary. Just in case I miss a more efficient solution.

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  • 5
    @user1929959 Nope, that question wants a copy of the dictionary, and doesn't care about hte case that the key is not in the input. Commented Mar 14, 2013 at 13:52

3 Answers 3

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1004

You can use dict.pop:

 mydict.pop("key", None)

Note that if the second argument, i.e. None is not given, KeyError is raised if the key is not in the dictionary. Providing the second argument prevents the conditional exception.

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7 Comments

For reference, the second argument to .pop() is what it returns if the key is not found.
Does this only do 1 lookup?
So, just for curiosity, what is the default value of the second argument in the source code?
@CarlosPinzón There isn't one. See the docs. D.pop(key[, default]) "If key is not found, default is returned if given, otherwise KeyError is raised".
@GirishGupta: I doubt it. Python dict is written in C. I would be very surprised if not optimised for single lookup. Ref: github.com/python/cpython/blob/main/Objects/dictobject.c#L2206
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31

There is also:

try:
    del mydict[key]
except KeyError:
    pass

This only does 1 lookup instead of 2. However, except clauses are expensive, so if you end up hitting the except clause frequently, this will probably be less efficient than what you already have.

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5 Comments

that is my fear. I wasn't sure about the cost of two look-ups per deletion vs. that of using try-except.
@SimonHughes -- See this answer on programmers ...
This is called the eaiser to ask for forgiveness than permission (EAFP) Python style.
And EAFP is the recommended approach in Python. I would like to see the evidence and rationale for the statement "except clauses are expensive."
@Bobort -- This has been a pretty well explained in a number of places. See this answer on programmers or this response on the python mailing list. In short, setting up the try/except block is cheap, but handling the exception is less cheap.
7

Approach: calculate keys to remove, mutate dict

Let's call keys the list/iterator of keys that you are given to remove. I'd do this:

keys_to_remove = set(keys).intersection(set(mydict.keys()))
for key in keys_to_remove:
    del mydict[key]

You calculate up front all the affected items and operate on them.

Approach: calculate keys to keep, make new dict with those keys

I prefer to create a new dictionary over mutating an existing one, so I would probably also consider this:

keys_to_keep = set(mydict.keys()) - set(keys)
new_dict = {k: v for k, v in mydict.iteritems() if k in keys_to_keep}

or:

keys_to_keep = set(mydict.keys()) - set(keys)
new_dict = {k: mydict[k] for k in keys_to_keep}
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1 Comment

first approach doesn't look thread-safe to me

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