How can I avoid "RuntimeError: dictionary changed size during iteration" error?


I have a dictionary of lists in which some of the values are empty:

d = {'a': [1], 'b': [1, 2], 'c': [], 'd':[]}

At the end of creating these lists, I want to remove these empty lists before returning my dictionary. I tried doing it like this:

for i in d:
    if not d[i]:

but I got a RuntimeError. I am aware that you cannot add/remove elements in a dictionary while iterating through it…what would be a way around this then?

See Modifying a Python dict while iterating over it for citations that this can cause problems, and why.

Asked By: user1530318



In Python 3.x and 2.x you can use use list to force a copy of the keys to be made:

for i in list(d):

In Python 2.x calling keys made a copy of the keys that you could iterate over while modifying the dict:

for i in d.keys():

But note that in Python 3.x this second method doesn’t help with your error because keys returns an a view object instead of copying the keys into a list.

Answered By: Mark Byers

Just use dictionary comprehension to copy the relevant items into a new dict:

>>> d
{'a': [1], 'c': [], 'b': [1, 2], 'd': []}
>>> d = {k: v for k, v in d.items() if v}
>>> d
{'a': [1], 'b': [1, 2]}

For this in Python 2:

>>> d
{'a': [1], 'c': [], 'b': [1, 2], 'd': []}
>>> d = {k: v for k, v in d.iteritems() if v}
>>> d
{'a': [1], 'b': [1, 2]}
Answered By: Maria Zverina

I would try to avoid inserting empty lists in the first place, but, would generally use:

d = {k: v for k,v in d.iteritems() if v} # re-bind to non-empty

If prior to 2.7:

d = dict( (k, v) for k,v in d.iteritems() if v )

or just:

empty_key_vals = list(k for k in k,v in d.iteritems() if v)
for k in empty_key_vals:
Answered By: Jon Clements

For Python 3:

{k:v for k,v in d.items() if v}
Answered By: ucyo

You only need to use copy:

This way you iterate over the original dictionary fields and on the fly can change the desired dict d.
It works on each Python version, so it’s more clear.

In [1]: d = {'a': [1], 'b': [1, 2], 'c': [], 'd':[]}

In [2]: for i in d.copy():
   ...:     if not d[i]:
   ...:         d.pop(i)

In [3]: d
Out[3]: {'a': [1], 'b': [1, 2]}

(BTW – Generally to iterate over copy of your data structure, instead of using .copy for dictionaries or slicing [:] for lists, you can use import copy -> copy.copy (for shallow copy which is equivalent to copy that is supported by dictionaries or slicing [:] that is supported by lists) or copy.deepcopy on your data structure.)

Answered By: Alon Elharar

The reason for the runtime error is that you cannot iterate through a data structure while its structure is changing during iteration.

One way to achieve what you are looking for is to use a list to append the keys you want to remove and then use the pop function on dictionary to remove the identified key while iterating through the list.

d = {'a': [1], 'b': [1, 2], 'c': [], 'd':[]}
pop_list = []

for i in d:
        if not d[i]:

for x in pop_list:
print (d)
Answered By: Rohit

For situations like this, I like to make a deep copy and loop through that copy while modifying the original dict.

If the lookup field is within a list, you can enumerate in the for loop of the list and then specify the position as the index to access the field in the original dict.

You cannot iterate through a dictionary while it’s changing during a for loop. Make a casting to list and iterate over that list. It works for me.

    for key in list(d):
        if not d[key]:
Answered By: Alvaro Romero Diaz

Python 3 does not allow deletion while iterating (using the for loop above) a dictionary. There are various alternatives to do it; one simple way is to change the line

for i in x.keys():


for i in list(x)
Answered By: Hasham Beyg

This worked for me:

d = {1: 'a', 2: '', 3: 'b', 4: '', 5: '', 6: 'c'}
for key, value in list(d.items()):
    if value == '':
        del d[key]
# {1: 'a', 3: 'b', 6: 'c'}

Casting the dictionary items to list creates a list of its items, so you can iterate over it and avoid the RuntimeError.

Answered By: singrium
for key in list(keys):
    dictc[key.upper()] ='New value'
Answered By: vaibhav.patil

To avoid "dictionary changed size during iteration error".

For example: "when you try to delete some key",

Just use ‘list’ with ‘.items()’. Here is a simple example:

my_dict = {

for key, val in list(my_dict.items()):
    if val == 2 or val == 4:



{'k1': 1, 'k2': 2, 'k3': 3, 'k4': 4}

{'k1': 1, 'k3': 3}

This is just an example. Change it based on your case/requirements.

Answered By: K.A

If the values in the dictionary were unique too, then I used this solution:

keyToBeDeleted = None
for k, v in mydict.items():
    if(v == match):
        keyToBeDeleted = k
mydict.pop(keyToBeDeleted, None)
Answered By: Ganesh S
  • The Python "RuntimeError: dictionary changed size during iteration" occurs when we change the size of a dictionary when iterating over it.

  • To solve the error, use the copy() method to create a shallow copy of the dictionary that you can iterate over, e.g., my_dict.copy().

    my_dict = {'a': 1, 'b': 2, 'c': 3}
    for key in my_dict.copy():
        if key == 'b':
            del my_dict[key]
    print(my_dict) #  ️ {'a': 1, 'c': 3}
  • You can also convert the keys of the dictionary to a list and iterate over the list of keys.

    my_dict = {'a': 1, 'b': 2, 'c': 3}
    for key in list(my_dict.keys()):
        if key == 'b':
            del my_dict[key]
    print(my_dict)  #  ️ {'a': 1, 'c': 3}
Answered By: Md Shayon

Nested null values

Let’s say we have a dictionary with nested keys, some of which are null values:

dicti = {
    "k0_l1": {
        "k0_l2": {
        "k2_l1":"not none",

Then we can remove the null values using this function:

def pop_nested_nulls(dicti):
    for k in list(dicti):
        if isinstance(dicti[k], dict):
            dicti[k] = pop_nested_nulls(dicti[k])
        elif not dicti[k]:
    return dicti

Output for pop_nested_nulls(dicti)

{'k0_l0': {'k0_l1': {'k0_l2': {'k1_1': 1,
                               'k2_2': 2.2}},
           'k2_l1': 'not '
 'k1_l0': 'l0'}
Answered By: Echo9k
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