Delete a dictionary item if the key exists
Question:
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.
Answers:
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.
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.
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}
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.
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.
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.
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}