How do I delete an item from a dictionary in Python?
Without modifying the original dictionary, how do I obtain another dict with the item removed?
See also How can I remove a key from a Python dictionary? for the specific issue of removing an item (by key) that may not already be present.
The del statement removes an element:
del d[key]
Note that 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).
pop mutates the dictionary.
>>> lol = {"hello": "gdbye"}
>>> lol.pop("hello")
'gdbye'
>>> lol
{}
If you want to keep the original you could just copy it.
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'}
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}
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
… 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.
d = {1: 2, '2': 3, 5: 7}
del d[5]
print 'd = ', d
Result: d = {1: 2, '2': 3}
Using del you can remove a dict value passing the key of that value
Link:
del method
del dictionary['key_to_del']
Simply call del d['key'].
However, in production, it is always a good practice to check if 'key' exists in d.
if 'key' in d:
del d['key']
No, there is no other way than
def dictMinus(dct, val):
copy = dct.copy()
del copy[val]
return copy
However, often creating copies of only slightly altered dictionaries is probably not a good idea because it will result in comparatively large memory demands. It is usually better to log the old dictionary(if even necessary) and then modify it.
# mutate/remove with a default
ret_val = body.pop('key', 5)
# no mutation with a default
ret_val = body.get('key', 5)
Here a top level design approach:
def eraseElement(d,k):
if isinstance(d, dict):
if k in d:
d.pop(k)
print(d)
else:
print("Cannot find matching key")
else:
print("Not able to delete")
exp = {'A':34, 'B':55, 'C':87}
eraseElement(exp, 'C')
I'm passing the dictionary and the key I want into my function, validates if it's a dictionary and if the key is okay, and if both exist, removes the value from the dictionary and prints out the left-overs.
Output: {'B': 55, 'A': 34}
Hope that helps!
>>> def delete_key(dict, key):
... del dict[key]
... return dict
...
>>> test_dict = {'one': 1, 'two' : 2}
>>> print delete_key(test_dict, 'two')
{'one': 1}
>>>
this doesn't do any error handling, it assumes the key is in the dict, you might want to check that first and raise if its not
Below code snippet will help you definitely, I have added comments in each line which will help you in understanding the code.
def execute():
dic = {'a':1,'b':2}
dic2 = remove_key_from_dict(dic, 'b')
print(dict2) # {'a': 1}
print(dict) # {'a':1,'b':2}
def remove_key_from_dict(dictionary_to_use, key_to_delete):
copy_of_dict = dict(dictionary_to_use) # creating clone/copy of the dictionary
if key_to_delete in copy_of_dict : # checking given key is present in the dictionary
del copy_of_dict [key_to_delete] # deleting the key from the dictionary
return copy_of_dict # returning the final dictionary
or you can also use dict.pop()
d = {"a": 1, "b": 2}
res = d.pop("c") # No `KeyError` here
print (res) # this line will not execute
or the better approach is
res = d.pop("c", "key not found")
print (res) # key not found
print (d) # {"a": 1, "b": 2}
res = d.pop("b", "key not found")
print (res) # 2
print (d) # {"a": 1}
Solution 1: with deleting
info = {'country': 'Iran'}
country = info.pop('country') if 'country' in info else None
Solution 2: without deleting
info = {'country': 'Iran'}
country = info.get('country') or None
Here's another variation using list comprehension:
original_d = {'a': None, 'b': 'Some'}
d = dict((k,v) for k, v in original_d.iteritems() if v)
# result should be {'b': 'Some'}
The approach is based on an answer from this post:
Efficient way to remove keys with empty strings from a dict
For Python 3 this is
original_d = {'a': None, 'b': 'Some'}
d = dict((k,v) for k, v in original_d.items() if v)
print(d)
species = {'HI': {'1': (1215.671, 0.41600000000000004),
'10': (919.351, 0.0012),
'1025': (1025.722, 0.0791),
'11': (918.129, 0.0009199999999999999),
'12': (917.181, 0.000723),
'1215': (1215.671, 0.41600000000000004),
'13': (916.429, 0.0005769999999999999),
'14': (915.824, 0.000468),
'15': (915.329, 0.00038500000000000003),
'CII': {'1036': (1036.3367, 0.11900000000000001), '1334': (1334.532, 0.129)}}
The following code will make a copy of dict species and delete items which are not in trans_HI
trans_HI=['1025','1215']
for transition in species['HI'].copy().keys():
if transition not in trans_HI:
species['HI'].pop(transition)
In Python 3, 'dict' object has no attribute 'remove'.
But with immutables package, can perform mutations that allow to apply changes to the Map object and create new (derived) Maps:
import immutables
map = immutables.Map(a=1, b=2)
map1 = map.delete('b')
print(map, map1)
# will print:
# <immutables.Map({'b': 2, 'a': 1})>
# <immutables.Map({'a': 1})>
can try my method. In one line.
yourList = [{'key':'key1','version':'1'},{'key':'key2','version':'2'},{'key':'key3','version':'3'}]
resultList = [{'key':dic['key']} for dic in yourList if 'key' in dic]
print(resultList)
Related
I have the horrible feeling this will be a duplicate, I tried my best to find the answer already.
I have a dictionary and a list, and I want to create a list of dictionaries, using the list to overwrite one of the key values, like this:
d={"a":1,"b":10}
c=[3,4,5]
arg=[]
for i in c:
e=d.copy()
e["a"]=i
arg.append(e)
this gives the desired result
arg
[{'a': 3, 'b': 10}, {'a': 4, 'b': 10}, {'a': 5, 'b': 10}]
but the code is ugly, especially with the copy command, and instead of one list I have 4 or 5 in my real example which leads to a huge nested loop. I feel sure there is a neater way with an iterator like
arg=[d *with* d[a]=i for i in c]
where I'm not sure what to put in the place of the "with".
Again, apologies if this is already answered.
IIUC, you could do:
d={"a":1,"b":10}
c=[3,4,5]
res = [{ **d, "a" : ci } for ci in c]
print(res)
Output
[{'a': 3, 'b': 10}, {'a': 4, 'b': 10}, {'a': 5, 'b': 10}]
The part:
"a" : ci
rewrites the value at the key "a" and **d unpacks the dictionary.
I would do it this way:
arg=[d.copy() for i in range(len(c))]
for i in range(len(arg)):
arg[i]['a']=c[i]
This code first creates a list of dictionaries with the length of c and then updates 'a' for each dictionary, with the respective itme of c
You could do it using a dictionary comprehension within a list comprehension, checking for key == 'a':
d = {"a":1,"b":10}
c = [3,4,5]
l = [{k: num if k == 'a' else v for k,v in d.items()} for num in c]
In Python 3.9 there is new method to create new dictionary with updated values and keep old dictionary without updates - using operator |
new_dict = old_dict | dict_with_updates
With list comprehension it will be
arg = [ d | {"a": i} for i in c]
Full example
d = {"a": 1, "b": 10}
c = [3, 4, 5]
arg = [ d | {"a": i} for i in c]
print(arg)
BTW: There is also |= to update existing dictionary
old_dict |= dict_with_updates
Doc: What’s New In Python 3.9
Lets say I have 2 dictionaries, a and b:
a = {"a": 2, "b": 4}
b = {"a": 5, "b": 2, "c": 10}
I want to find the common keys in both dictionaries, and then take the value of those keys from b to create a new one. Example:
c = intersect_keys(a, b)
# c = {"a": 5, "b": 2}
As you can understand, the keys which were not present in the first array were not used in the newly generated one. How can I do this in a fast way using Python?
Moreover, because we are always picking the values from the second array, would it be better to just make a into a list and then iterate over it and then the values from b? Thanks for any answers.
You can use a dictionary comprehension, in order to keep those keys in b that are also present in a:
{k:v for k,v in b.items() if k in a}
Output
{'a': 5, 'b': 2}
Try this?:
myKeys = set(a.keys()).intersection(b.keys())
c = {}
for key in myKeys:
c[key] = b[key]
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]:
d.pop(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.
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.
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.)
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]}
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]
print(d)
# {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.
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:
del[k]
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 = {
'k1':1,
'k2':2,
'k3':3,
'k4':4
}
print(my_dict)
for key, val in list(my_dict.items()):
if val == 2 or val == 4:
my_dict.pop(key)
print(my_dict)
Output:
{'k1': 1, 'k2': 2, 'k3': 3, 'k4': 4}
{'k1': 1, 'k3': 3}
This is just an example. Change it based on your case/requirements.
For Python 3:
{k:v for k,v in d.items() if v}
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]:
d.pop(key)
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():
with
for i in list(x)
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]:
pop_list.append(i)
for x in pop_list:
d.pop(x)
print (d)
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.
Nested null values
Let's say we have a dictionary with nested keys, some of which are null values:
dicti = {
"k0_l0":{
"k0_l1": {
"k0_l2": {
"k0_0":None,
"k1_1":1,
"k2_2":2.2
}
},
"k1_l1":None,
"k2_l1":"not none",
"k3_l1":[]
},
"k1_l0":"l0"
}
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]:
dicti.pop(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 '
'none'},
'k1_l0': 'l0'}
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():
print(key)
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()):
print(key)
if key == 'b':
del my_dict[key]
print(my_dict) # 👉️ {'a': 1, 'c': 3}
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
break
mydict.pop(keyToBeDeleted, None)
I have a dictionary
d = {'a':1, 'b':2, 'c':3}
I need to remove a key, say c and return the dictionary without that key in one function call
{'a':1, 'b':2}
d.pop('c') will return the key value - 3 - instead of the dictionary.
I am going to need one function solution if it exists, as this will go into comprehensions
How about this:
{i:d[i] for i in d if i!='c'}
It's called Dictionary Comprehensions and it's available since Python 2.7.
or if you are using Python older than 2.7:
dict((i,d[i]) for i in d if i!='c')
Why not roll your own? This will likely be faster than creating a new one using dictionary comprehensions:
def without(d, key):
new_d = d.copy()
new_d.pop(key)
return new_d
If you need an expression that does this (so you can use it in a lambda or comprehension) then you can use this little hack trick: create a tuple with the dictionary and the popped element, and then get the original item back out of the tuple:
(foo, foo.pop(x))[0]
For example:
ds = [{'a': 1, 'b': 2, 'c': 3}, {'a': 4, 'b': 5, 'c': 6}]
[(d, d.pop('c'))[0] for d in ds]
assert ds == [{'a': 1, 'b': 2}, {'a': 4, 'b': 5}]
Note that this actually modifies the original dictionary, so despite being a comprehension, it's not purely functional.
When you invoke pop the original dictionary is modified in place.
You can return that one from your function.
>>> a = {'foo': 1, 'bar': 2}
>>> a.pop('foo')
1
>>> a
{'bar': 2}
solution from me
item = dict({"A": 1, "B": 3, "C": 4})
print(item)
{'A': 1, 'B': 3, 'C': 4}
new_dict = (lambda d: d.pop('C') and d)(item)
print(new_dict)
{'A': 1, 'B': 3}
this will work,
(lambda dict_,key_:dict_.pop(key_,True) and dict_)({1:1},1)
EDIT
this will drop the key if exist in the dictionary and will return the dictionary without the key,value pair
in python there are functions that alter an object in place, and returns a value instead of the altered object, {}.pop function is an example.
we can use a lambda function as in the example, or more generic below
(lambda func:obj:(func(obj) and False) or obj)
to alter this behavior, and get a the expected behavior.
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]:
d.pop(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.
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.
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.)
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]}
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]
print(d)
# {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.
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:
del[k]
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 = {
'k1':1,
'k2':2,
'k3':3,
'k4':4
}
print(my_dict)
for key, val in list(my_dict.items()):
if val == 2 or val == 4:
my_dict.pop(key)
print(my_dict)
Output:
{'k1': 1, 'k2': 2, 'k3': 3, 'k4': 4}
{'k1': 1, 'k3': 3}
This is just an example. Change it based on your case/requirements.
For Python 3:
{k:v for k,v in d.items() if v}
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]:
d.pop(key)
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():
with
for i in list(x)
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]:
pop_list.append(i)
for x in pop_list:
d.pop(x)
print (d)
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.
Nested null values
Let's say we have a dictionary with nested keys, some of which are null values:
dicti = {
"k0_l0":{
"k0_l1": {
"k0_l2": {
"k0_0":None,
"k1_1":1,
"k2_2":2.2
}
},
"k1_l1":None,
"k2_l1":"not none",
"k3_l1":[]
},
"k1_l0":"l0"
}
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]:
dicti.pop(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 '
'none'},
'k1_l0': 'l0'}
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():
print(key)
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()):
print(key)
if key == 'b':
del my_dict[key]
print(my_dict) # 👉️ {'a': 1, 'c': 3}
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
break
mydict.pop(keyToBeDeleted, None)