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Delete an element from a dictionary


Question

Is there a way to delete an item from a dictionary in Python?

Additionally, how can I delete an item from a dictionary to return a copy (i.e., not modifying the original)?

2019/08/22
1
1479
8/22/2019 4:20:41 PM

Accepted Answer

The del statement removes an element:

del d[key]

However, this mutates the existing dictionary so the contents of the dictionary changes for anybody else who has a reference to the same instance. To return a new dictionary, make a copy of the dictionary:

def removekey(d, key):
    r = dict(d)
    del r[key]
    return r

The dict() constructor makes a shallow copy. To make a deep copy, see the copy module.


Note that making a copy for every dict del/assignment/etc. means you're going from constant time to linear time, and also using linear space. For small dicts, this is not a problem. But if you're planning to make lots of copies of large dicts, you probably want a different data structure, like a HAMT (as described in this answer).

2019/03/20
1815
3/20/2019 3:19:45 PM

pop mutates the dictionary.

 >>> lol = {"hello": "gdbye"}
 >>> lol.pop("hello")
     'gdbye'
 >>> lol
     {}

If you want to keep the original you could just copy it.

2020/01/06

I think your solution is best way to do it. But if you want another solution, you can create a new dictionary with using the keys from old dictionary without including your specified key, like this:

>>> a
{0: 'zero', 1: 'one', 2: 'two', 3: 'three'}
>>> {i:a[i] for i in a if i!=0}
{1: 'one', 2: 'two', 3: 'three'}
2011/04/30

The del statement is what you're looking for. If you have a dictionary named foo with a key called 'bar', you can delete 'bar' from foo like this:

del foo['bar']

Note that this permanently modifies the dictionary being operated on. If you want to keep the original dictionary, you'll have to create a copy beforehand:

>>> foo = {'bar': 'baz'}
>>> fu = dict(foo)
>>> del foo['bar']
>>> print foo
{}
>>> print fu
{'bar': 'baz'}

The dict call makes a shallow copy. If you want a deep copy, use copy.deepcopy.

Here's a method you can copy & paste, for your convenience:

def minus_key(key, dictionary):
    shallow_copy = dict(dictionary)
    del shallow_copy[key]
    return shallow_copy
2017/05/30

There're a lot of nice answers, but I want to emphasize one thing.

You can use both dict.pop() method and a more generic del statement to remove items from a dictionary. They both mutate the original dictionary, so you need to make a copy (see details below).

And both of them will raise a KeyError if the key you're providing to them is not present in the dictionary:

key_to_remove = "c"
d = {"a": 1, "b": 2}
del d[key_to_remove]  # Raises `KeyError: 'c'`

and

key_to_remove = "c"
d = {"a": 1, "b": 2}
d.pop(key_to_remove)  # Raises `KeyError: 'c'`

You have to take care of this:

by capturing the exception:

key_to_remove = "c"
d = {"a": 1, "b": 2}
try:
    del d[key_to_remove]
except KeyError as ex:
    print("No such key: '%s'" % ex.message)

and

key_to_remove = "c"
d = {"a": 1, "b": 2}
try:
    d.pop(key_to_remove)
except KeyError as ex:
    print("No such key: '%s'" % ex.message)

by performing a check:

key_to_remove = "c"
d = {"a": 1, "b": 2}
if key_to_remove in d:
    del d[key_to_remove]

and

key_to_remove = "c"
d = {"a": 1, "b": 2}
if key_to_remove in d:
    d.pop(key_to_remove)

but with pop() there's also a much more concise way - provide the default return value:

key_to_remove = "c"
d = {"a": 1, "b": 2}
d.pop(key_to_remove, None)  # No `KeyError` here

Unless you use pop() to get the value of a key being removed you may provide anything, not necessary None. Though it might be that using del with in check is slightly faster due to pop() being a function with its own complications causing overhead. Usually it's not the case, so pop() with default value is good enough.


As for the main question, you'll have to make a copy of your dictionary, to save the original dictionary and have a new one without the key being removed.

Some other people here suggest making a full (deep) copy with copy.deepcopy(), which might be an overkill, a "normal" (shallow) copy, using copy.copy() or dict.copy(), might be enough. The dictionary keeps a reference to the object as a value for a key. So when you remove a key from a dictionary this reference is removed, not the object being referenced. The object itself may be removed later automatically by the garbage collector, if there're no other references for it in the memory. Making a deep copy requires more calculations compared to shallow copy, so it decreases code performance by making the copy, wasting memory and providing more work to the GC, sometimes shallow copy is enough.

However, if you have mutable objects as dictionary values and plan to modify them later in the returned dictionary without the key, you have to make a deep copy.

With shallow copy:

def get_dict_wo_key(dictionary, key):
    """Returns a **shallow** copy of the dictionary without a key."""
    _dict = dictionary.copy()
    _dict.pop(key, None)
    return _dict


d = {"a": [1, 2, 3], "b": 2, "c": 3}
key_to_remove = "c"

new_d = get_dict_wo_key(d, key_to_remove)
print(d)  # {"a": [1, 2, 3], "b": 2, "c": 3}
print(new_d)  # {"a": [1, 2, 3], "b": 2}
new_d["a"].append(100)
print(d)  # {"a": [1, 2, 3, 100], "b": 2, "c": 3}
print(new_d)  # {"a": [1, 2, 3, 100], "b": 2}
new_d["b"] = 2222
print(d)  # {"a": [1, 2, 3, 100], "b": 2, "c": 3}
print(new_d)  # {"a": [1, 2, 3, 100], "b": 2222}

With deep copy:

from copy import deepcopy


def get_dict_wo_key(dictionary, key):
    """Returns a **deep** copy of the dictionary without a key."""
    _dict = deepcopy(dictionary)
    _dict.pop(key, None)
    return _dict


d = {"a": [1, 2, 3], "b": 2, "c": 3}
key_to_remove = "c"

new_d = get_dict_wo_key(d, key_to_remove)
print(d)  # {"a": [1, 2, 3], "b": 2, "c": 3}
print(new_d)  # {"a": [1, 2, 3], "b": 2}
new_d["a"].append(100)
print(d)  # {"a": [1, 2, 3], "b": 2, "c": 3}
print(new_d)  # {"a": [1, 2, 3, 100], "b": 2}
new_d["b"] = 2222
print(d)  # {"a": [1, 2, 3], "b": 2, "c": 3}
print(new_d)  # {"a": [1, 2, 3, 100], "b": 2222}
2018/01/23

… how can I delete an item from a dictionary to return a copy (i.e., not modifying the original)?

A dict is the wrong data structure to use for this.

Sure, copying the dict and popping from the copy works, and so does building a new dict with a comprehension, but all that copying takes time—you've replaced a constant-time operation with a linear-time one. And all those copies alive at once take space—linear space per copy.

Other data structures, like hash array mapped tries, are designed for exactly this kind of use case: adding or removing an element returns a copy in logarithmic time, sharing most of its storage with the original.1

Of course there are some downsides. Performance is logarithmic rather than constant (although with a large base, usually 32-128). And, while you can make the non-mutating API identical to dict, the "mutating" API is obviously different. And, most of all, there's no HAMT batteries included with Python.2

The pyrsistent library is a pretty solid implementation of HAMT-based dict-replacements (and various other types) for Python. It even has a nifty evolver API for porting existing mutating code to persistent code as smoothly as possible. But if you want to be explicit about returning copies rather than mutating, you just use it like this:

>>> from pyrsistent import m
>>> d1 = m(a=1, b=2)
>>> d2 = d1.set('c', 3)
>>> d3 = d1.remove('a')
>>> d1
pmap({'a': 1, 'b': 2})
>>> d2
pmap({'c': 3, 'a': 1, 'b': 2})
>>> d3
pmap({'b': 2})

That d3 = d1.remove('a') is exactly what the question is asking for.

If you've got mutable data structures like dict and list embedded in the pmap, you'll still have aliasing issues—you can only fix that by going immutable all the way down, embedding pmaps and pvectors.


1. HAMTs have also become popular in languages like Scala, Clojure, Haskell because they play very nicely with lock-free programming and software transactional memory, but neither of those is very relevant in Python.

2. In fact, there is an HAMT in the stdlib, used in the implementation of contextvars. The earlier withdrawn PEP explains why. But this is a hidden implementation detail of the library, not a public collection type.

2018/05/15

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