How can I remove a key from a Python dictionary?

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 ;-)

You should consider, though, that this way of deleting an object from a dict is not atomic—it 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, a KeyError is raised.

Documentation

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.

Answer 9

You can use exception handling if you want to be very verbose:

try: 
    del dict[key]

except KeyError: pass

This is slower, however, than the pop() method, if the key doesn't exist.

my_dict.pop('key', None)

It won't matter for a few keys, but if you're doing this repeatedly, then the latter method is a better bet.

The fastest approach is this:

if 'key' in dict: 
    del myDict['key']

But this method is dangerous because if 'key' is removed in between the two lines, a KeyError will be raised.

Answer 10

Dictionary data type has a method called dict_name.pop(item) and this can be used to delete a key:value pair from a dictionary.

a={9:4,2:3,4:2,1:3}
a.pop(9)
print(a)

This will give the output as:

{2: 3, 4: 2, 1: 3}

This way you can delete an item from a dictionary in one line.

Answer 11

Another way is by using items() + dict comprehension.

items() coupled with dict comprehension can also help us achieve the task of key-value pair deletion, but it has the drawback of not being an in place dict technique. Actually a new dict if created except for the key we don’t wish to include.

test_dict = {"sai" : 22, "kiran" : 21, "vinod" : 21, "sangam" : 21}

# Printing dictionary before removal
print ("dictionary before performing remove is : " + str(test_dict))

# Using items() + dict comprehension to remove a dict. pair
# removes  vinod
new_dict = {key:val for key, val in test_dict.items() if key != 'vinod'}

# Printing dictionary after removal
print ("dictionary after remove is : " + str(new_dict))

Output:

dictionary before performing remove is : {'sai': 22, 'kiran': 21, 'vinod': 21, 'sangam': 21}
dictionary after remove is : {'sai': 22, 'kiran': 21, 'sangam': 21}

Answer 12

I prefer the immutable version

foo = {
    1:1,
    2:2,
    3:3
}
removeKeys = [1,2]
def woKeys(dct, keyIter):
    return {
        k:v
        for k,v in dct.items() if k not in keyIter
    }

>>> print(woKeys(foo, removeKeys))
{3: 3}
>>> print(foo)
{1: 1, 2: 2, 3: 3}

Answer 13

Single filter on key

• return "key" and remove it from my_dict if "key" exists in my_dict
• return None if "key" doesn't exist in my_dict

this will change my_dict in place (mutable)

my_dict.pop('key', None)

Multiple filters on keys

generate a new dict (immutable)

dic1 = {
    "x":1,
    "y": 2,
    "z": 3
}

def func1(item):
    return  item[0]!= "x" and item[0] != "y"

print(
    dict(
        filter(
            lambda item: item[0] != "x" and item[0] != "y", 
            dic1.items()
            )
    )
)