How to delete items from a dictionary while iterating over it?

Question

Can I delete items from a dictionary in Python while iterating over it?

I want to remove elements that don't meet a certain condition from the dictionary, instead of creating an entirely new dictionary. Is the following a good solution, or are there better ways?

for k, v in mydict.items():
    if k == val:
        del mydict[k]

Answer 1

list(mydict) can be used:

>>> mydict = {'one': 1, 'two': 2, 'three': 3, 'four': 4}
>>> for k in list(mydict):
...     if k == 'three':
...         del mydict[k]
... 
>>> mydict
{'four': 4, 'two': 2, 'one': 1}

This corresponds to a parallel structure for lists:

>>> mylist = ['one', 'two', 'three', 'four']
>>> for k in list(mylist):                            # or mylist[:]
...     if k == 'three':
...         mylist.remove(k)
... 
>>> mylist
['one', 'two', 'four']

Both work in python2 and python3.

Answer 2

Another nice and efficient way using partial function

next((partial(d.pop, k) for k in d if k == v), partial(print, f"{v} key not found"))()

Answer 3

One-liner:

my_dict = {k: v for k, v in my_dict.copy().items() if not k == value}

The my_dict.copy() object is used for the iteration only and will not be available outside the scope of the dictionary comprehension. This avoids editing the object over which you are currently iterating, as advised against by @user395760 in their answer.

You can split this over multiple lines for clarity:

my_dict = {
    k: v
    for k, v in my_dict.copy().items()
    if not k == value
}

Answer 4

For Python 3+:

>>> mydict
{'four': 4, 'three': 3, 'one': 1}

>>> for k in list(mydict.keys()):
...     if mydict[k] == 3:
...         del mydict[k]

>>> mydict
{'four': 4, 'one': 1}

The other answers work fine with Python 2 but raise a RuntimeError for Python 3:

RuntimeError: dictionary changed size during iteration.

This happens because mydict.keys() returns an iterator not a list. As pointed out in comments simply convert mydict.keys() to a list by list(mydict.keys()) and it should work.

---

For Python 2:

A simple test in the console shows you cannot modify a dictionary while iterating over it:

>>> mydict = {'one': 1, 'two': 2, 'three': 3, 'four': 4}

>>> for k, v in mydict.iteritems():
...    if k == 'two':
...        del mydict[k]

------------------------------------------------------------
Traceback (most recent call last):
  File "<ipython console>", line 1, in <module>
RuntimeError: dictionary changed size during iteration

As stated in delnan's answer, deleting entries causes problems when the iterator tries to move onto the next entry. Instead, use the keys() method to get a list of the keys and work with that:

>>> for k in mydict.keys():
...    if k == 'two':
...        del mydict[k]

>>> mydict
{'four': 4, 'three': 3, 'one': 1}

If you need to delete based on the items value, use the items() method instead:

>>> for k, v in mydict.items():
...     if v == 3:
...         del mydict[k]

>>> mydict
{'four': 4, 'one': 1}

Answer 5

With python3, iterate on dic.keys() will raise the dictionary size error. You can use this alternative way:

Tested with python3, it works fine and the Error "dictionary changed size during iteration" is not raised:

my_dic = { 1:10, 2:20, 3:30 }
# Is important here to cast because ".keys()" method returns a dict_keys object.
key_list = list( my_dic.keys() )

# Iterate on the list:
for k in key_list:
    print(key_list)
    print(my_dic)
    del( my_dic[k] )


print( my_dic )
# {}

Answer 6

There is a way that may be suitable if the items you want to delete are always at the "beginning" of the dict iteration

while mydict:
    key, value = next(iter(mydict.items()))
    if should_delete(key, value):
       del mydict[key]
    else:
       break

The "beginning" is only guaranteed to be consistent for certain Python versions/implementations. For example from What’s New In Python 3.7

the insertion-order preservation nature of dict objects has been declared to be an official part of the Python language spec.

This way avoids a copy of the dict that a lot of the other answers suggest, at least in Python 3.

Answer 7

I tried the above solutions in Python3 but this one seems to be the only one working for me when storing objects in a dict. Basically you make a copy of your dict() and iterate over that while deleting the entries in your original dictionary.

        tmpDict = realDict.copy()
        for key, value in tmpDict.items():
            if value:
                del(realDict[key])

Answer 8

Iterate over a copy instead, such as the one returned by items():

for k, v in list(mydict.items()):

Answer 9

You can use a dictionary comprehension.

d = {k:d[k] for k in d if d[k] != val}

Answer 10

You could first build a list of keys to delete, and then iterate over that list deleting them.

dict = {'one' : 1, 'two' : 2, 'three' : 3, 'four' : 4}
delete = []
for k,v in dict.items():
    if v%2 == 1:
        delete.append(k)
for i in delete:
    del dict[i]

Answer 11

You could also do it in two steps:

remove = [k for k in mydict if k == val]
for k in remove: del mydict[k]

My favorite approach is usually to just make a new dict:

# Python 2.7 and 3.x
mydict = { k:v for k,v in mydict.items() if k!=val }
# before Python 2.7
mydict = dict((k,v) for k,v in mydict.iteritems() if k!=val)

Answer 12

You can't modify a collection while iterating it. That way lies madness - most notably, if you were allowed to delete and deleted the current item, the iterator would have to move on (+1) and the next call to next would take you beyond that (+2), so you'd end up skipping one element (the one right behind the one you deleted). You have two options:

• Copy all keys (or values, or both, depending on what you need), then iterate over those. You can use .keys() et al for this (in Python 3, pass the resulting iterator to list). Could be highly wasteful space-wise though.
• Iterate over mydict as usual, saving the keys to delete in a seperate collection to_delete. When you're done iterating mydict, delete all items in to_delete from mydict. Saves some (depending on how many keys are deleted and how many stay) space over the first approach, but also requires a few more lines.