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?
Answer #1:
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.
Answer #2:
Specifically to answer “is there a one line way of doing this?”
if 'key' in my_dict: del my_dict['key']
…well, you asked π
You should consider, though, that this way of deleting an object from a dict
is not atomicβit 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.
Answer #3:
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, aKeyError
is raised.
Answer #4:
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
Answer #5:
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.
Answer #6:
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.
Answer #7:
Using the “del” keyword:
del dict[key]
Answer #8:
We can delete a key from a Python dictionary by the some of the following approaches.
Using the del
keyword; it’s almost the same approach like you did though –
myDict = {'one': 100, 'two': 200, 'three': 300 }
print(myDict) # {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : del myDict['one']
print(myDict) # {'two': 200, 'three': 300}
Or
We can do like the following:
But one should keep in mind that, in this process actually it won’t delete any key from the dictionary rather than making a specific key excluded from that dictionary. In addition, I observed that it returned a dictionary which was not ordered the same as myDict
.
myDict = {'one': 100, 'two': 200, 'three': 300, 'four': 400, 'five': 500}
{key:value for key, value in myDict.items() if key != 'one'}
If we run it in the shell, it’ll execute something like {'five': 500, 'four': 400, 'three': 300, 'two': 200}
– notice that it’s not the same ordered as myDict
. Again if we try to print myDict
, then we can see all keys including which we excluded from the dictionary by this approach. However, we can make a new dictionary by assigning the following statement into a variable:
var = {key:value for key, value in myDict.items() if key != 'one'}
Now if we try to print it, then it’ll follow the parent order:
print(var) # {'two': 200, 'three': 300, 'four': 400, 'five': 500}
Or
Using the pop()
method.
myDict = {'one': 100, 'two': 200, 'three': 300}
print(myDict)
if myDict.get('one') : myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
The difference between del
and pop
is that, using pop()
method, we can actually store the key’s value if needed, like the following:
myDict = {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : var = myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
print(var) # 100
Fork this gist for future reference, if you find this useful.