How to remove a key from a dictionary in Python

1del d[key]Recommended

del is the direct statement for "this key should not be here anymore". It removes the entry in place and evaluates to nothing — and it raises KeyError if the key was never there, which is exactly what you want when a missing key means a bug upstream.

user = {"name": "Ada", "role": "admin", "temp_token": "abc123"}

del user["temp_token"]

print(user)
print("temp_token" in user)

# del raises KeyError when the key isn't there
try:
    del user["nickname"]
except KeyError as exc:
    print("KeyError:", exc)

Output: {'name': 'Ada', 'role': 'admin'}, then False, then KeyError: 'nickname'. If a missing key is fine, guard with if key in d: del d[key] — or skip the guard entirely and use d.pop(key, None) from section 3.

2d.pop(key) — keep the removed value

pop() does everything del does and hands you the value it just removed. Use it whenever the value still matters — moving a field out of a payload, consuming a one-shot token, popping an option out of a **kwargs dict before passing the rest on.

user = {"name": "Ada", "role": "admin", "temp_token": "abc123"}

token = user.pop("temp_token")

print(token)
print(user)

Prints abc123 and then {'name': 'Ada', 'role': 'admin'}. With no second argument, pop() raises KeyError on a missing key, just like del.

3d.pop(key, None) — never raises

Give pop() a second argument and a missing key stops being an error: you get the default back instead. d.pop(key, None) is the idiomatic "remove it if it is there" one-liner — no in check, no try/except.

user = {"name": "Ada", "role": "admin"}

token = user.pop("temp_token", None)
role = user.pop("role", None)

print(token)
print(role)
print(user)

removed = user.pop("nickname", "(nothing to remove)")
print(removed)

Output: None, admin, {'name': 'Ada'}, (nothing to remove). The catch is that the return value alone can't tell you whether the key was absent or present-but-None — check key in d first if you need to distinguish those.

4A dict comprehension for several keys

To drop a whole group of keys, filter with a comprehension instead of looping over del. This is the one approach here that does not mutate: it returns a new dict and the original is untouched — usually what you want when you're producing a redacted or public view of a record.

record = {"id": 7, "name": "Ada", "password": "hunter2", "session": "x91", "role": "admin"}
drop = {"password", "session"}

public = {k: v for k, v in record.items() if k not in drop}

print(public)
print(record)

Prints {'id': 7, 'name': 'Ada', 'role': 'admin'} and then the original record, still complete. Keep the keys to drop in a set — membership is O(1), and the surviving keys stay in their original insertion order. If you do want the original mutated, loop instead: for k in drop: record.pop(k, None).

5Which should you use?

6Common variation: removing keys while iterating

Deleting inside for k in d: raises RuntimeError: dictionary changed size during iteration — the loop iterates over the live dict, so resizing it mid-flight invalidates the iterator. The fix is one call: list(d) snapshots the keys first, so the loop iterates a list while you mutate the dict.

scores = {"ada": 91, "bob": 45, "cleo": 78, "dan": 30}

try:
    for name in scores:
        if scores[name] < 50:
            del scores[name]
except RuntimeError as exc:
    print("RuntimeError:", exc)

print(scores)

for name in list(scores):
    if scores[name] < 50:
        del scores[name]

print(scores)

last = scores.popitem()
print(last)
print(scores)

Note the second line of output — {'ada': 91, 'cleo': 78, 'dan': 30}. The failed loop still deleted bob before it blew up, so a RuntimeError here leaves you half-mutated, not unchanged. After the list(d) pass you get {'ada': 91, 'cleo': 78}, and popitem() then removes and returns the last inserted pair, ('cleo', 78).