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

ScriptingDictionaryPython

Scripting Problem Overview


Is it legitimate to delete items from a dictionary in Python while iterating over it?

For example:

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

The idea is to remove elements that don't meet a certain condition from the dictionary, instead of creating a new dictionary that's a subset of the one being iterated over.

Is this a good solution? Are there more elegant/efficient ways?

Scripting Solutions


Solution 1 - Scripting

EDIT:

For Python3 (or greater):

>>> 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 rest of the answers works fine with Python2 but do not work for Python3 and raises RuntimeError. >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 python2:

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}

Solution 2 - Scripting

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)

Solution 3 - Scripting

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

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

Solution 4 - Scripting

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.

Solution 5 - Scripting

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 )
# {}

Solution 6 - Scripting

It's cleanest to use list(mydict):

>>> 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.

Solution 7 - Scripting

You can use a dictionary comprehension.

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

Solution 8 - Scripting

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]

Solution 9 - Scripting

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.

Solution 10 - Scripting

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])
              

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Questionuser248237View Question on Stackoverflow
Solution 1 - ScriptingBlairView Answer on Stackoverflow
Solution 2 - ScriptingJochen RitzelView Answer on Stackoverflow
Solution 3 - ScriptingIgnacio Vazquez-AbramsView Answer on Stackoverflow
Solution 4 - Scriptinguser395760View Answer on Stackoverflow
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Solution 9 - ScriptingMichal CharemzaView Answer on Stackoverflow
Solution 10 - ScriptingJason LandbridgeView Answer on Stackoverflow