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]
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}
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)
Iterate over a copy instead, such as the one returned by items():
for k, v in list(mydict.items()):
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.
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.
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 )
# {}
You can use a dictionary comprehension.
d = {k:d[k] for k in d if d[k] != val}
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]
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.
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])
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
}
Related
I want to remove a key from a dictionary if it is present. I currently use this code:
if 'key' in my_dict:
del my_dict['key']
Without the if statement, the code will raise KeyError if the key is not present. How can I handle this more simply?
See Delete an element from a dictionary for more general approaches to the problem of removing a key from a dict (including ones which produce a modified copy).
To delete a key regardless of whether it is in the dictionary, use the two-argument form of dict.pop():
my_dict.pop('key', None)
This will return my_dict[key] if key exists in the dictionary, and None otherwise. If the second parameter is not specified (i.e. my_dict.pop('key')) and key does not exist, a KeyError is raised.
To delete a key that is guaranteed to exist, you can also use
del my_dict['key']
This will raise a KeyError if the key is not in the dictionary.
Specifically to answer "is there a one line way of doing this?"
if 'key' in my_dict: del my_dict['key']
...well, you asked ;-)
You should consider, though, that this way of deleting an object from a dict is not atomic—it is possible that 'key' may be in my_dict during the if statement, but may be deleted before del is executed, in which case del will fail with a KeyError. Given this, it would be safest to either use dict.pop or something along the lines of
try:
del my_dict['key']
except KeyError:
pass
which, of course, is definitely not a one-liner.
It took me some time to figure out what exactly my_dict.pop("key", None) is doing. So I'll add this as an answer to save others googling time:
pop(key[, default])
If key is in the dictionary, remove it and return its value, else
return default. If default is not given and key is not in the
dictionary, a KeyError is raised.
Documentation
del my_dict[key] is slightly faster than my_dict.pop(key) for removing a key from a dictionary when the key exists
>>> import timeit
>>> setup = "d = {i: i for i in range(100000)}"
>>> timeit.timeit("del d[3]", setup=setup, number=1)
1.79e-06
>>> timeit.timeit("d.pop(3)", setup=setup, number=1)
2.09e-06
>>> timeit.timeit("d2 = {key: val for key, val in d.items() if key != 3}", setup=setup, number=1)
0.00786
But when the key doesn't exist if key in my_dict: del my_dict[key] is slightly faster than my_dict.pop(key, None). Both are at least three times faster than del in a try/except statement:
>>> timeit.timeit("if 'missing key' in d: del d['missing key']", setup=setup)
0.0229
>>> timeit.timeit("d.pop('missing key', None)", setup=setup)
0.0426
>>> try_except = """
... try:
... del d['missing key']
... except KeyError:
... pass
... """
>>> timeit.timeit(try_except, setup=setup)
0.133
If you need to remove a lot of keys from a dictionary in one line of code, I think using map() is quite succinct and Pythonic readable:
myDict = {'a':1,'b':2,'c':3,'d':4}
map(myDict.pop, ['a','c']) # The list of keys to remove
>>> myDict
{'b': 2, 'd': 4}
And if you need to catch errors where you pop a value that isn't in the dictionary, use lambda inside map() like this:
map(lambda x: myDict.pop(x,None), ['a', 'c', 'e'])
[1, 3, None] # pop returns
>>> myDict
{'b': 2, 'd': 4}
or in python3, you must use a list comprehension instead:
[myDict.pop(x, None) for x in ['a', 'c', 'e']]
It works. And 'e' did not cause an error, even though myDict did not have an 'e' key.
You can use a dictionary comprehension to create a new dictionary with that key removed:
>>> my_dict = {k: v for k, v in my_dict.items() if k != 'key'}
You can delete by conditions. No error if key doesn't exist.
You can use exception handling if you want to be very verbose:
try:
del dict[key]
except KeyError: pass
This is slower, however, than the pop() method, if the key doesn't exist.
my_dict.pop('key', None)
It won't matter for a few keys, but if you're doing this repeatedly, then the latter method is a better bet.
The fastest approach is this:
if 'key' in dict:
del myDict['key']
But this method is dangerous because if 'key' is removed in between the two lines, a KeyError will be raised.
We can delete a key from a Python dictionary by the some of the following approaches.
Using the del keyword; it's almost the same approach like you did though -
myDict = {'one': 100, 'two': 200, 'three': 300 }
print(myDict) # {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : del myDict['one']
print(myDict) # {'two': 200, 'three': 300}
Or
We can do like the following:
But one should keep in mind that, in this process actually it won't delete any key from the dictionary rather than making a specific key excluded from that dictionary. In addition, I observed that it returned a dictionary which was not ordered the same as myDict.
myDict = {'one': 100, 'two': 200, 'three': 300, 'four': 400, 'five': 500}
{key:value for key, value in myDict.items() if key != 'one'}
If we run it in the shell, it'll execute something like {'five': 500, 'four': 400, 'three': 300, 'two': 200} - notice that it's not the same ordered as myDict. Again if we try to print myDict, then we can see all keys including which we excluded from the dictionary by this approach. However, we can make a new dictionary by assigning the following statement into a variable:
var = {key:value for key, value in myDict.items() if key != 'one'}
Now if we try to print it, then it'll follow the parent order:
print(var) # {'two': 200, 'three': 300, 'four': 400, 'five': 500}
Or
Using the pop() method.
myDict = {'one': 100, 'two': 200, 'three': 300}
print(myDict)
if myDict.get('one') : myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
The difference between del and pop is that, using pop() method, we can actually store the key's value if needed, like the following:
myDict = {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : var = myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
print(var) # 100
Fork this gist for future reference, if you find this useful.
I prefer the immutable version
foo = {
1:1,
2:2,
3:3
}
removeKeys = [1,2]
def woKeys(dct, keyIter):
return {
k:v
for k,v in dct.items() if k not in keyIter
}
>>> print(woKeys(foo, removeKeys))
{3: 3}
>>> print(foo)
{1: 1, 2: 2, 3: 3}
Another way is by using items() + dict comprehension.
items() coupled with dict comprehension can also help us achieve the task of key-value pair deletion, but it has the drawback of not being an in place dict technique. Actually a new dict if created except for the key we don’t wish to include.
test_dict = {"sai" : 22, "kiran" : 21, "vinod" : 21, "sangam" : 21}
# Printing dictionary before removal
print ("dictionary before performing remove is : " + str(test_dict))
# Using items() + dict comprehension to remove a dict. pair
# removes vinod
new_dict = {key:val for key, val in test_dict.items() if key != 'vinod'}
# Printing dictionary after removal
print ("dictionary after remove is : " + str(new_dict))
Output:
dictionary before performing remove is : {'sai': 22, 'kiran': 21, 'vinod': 21, 'sangam': 21}
dictionary after remove is : {'sai': 22, 'kiran': 21, 'sangam': 21}
I want to remove a key from a dictionary if it is present. I currently use this code:
if 'key' in my_dict:
del my_dict['key']
Without the if statement, the code will raise KeyError if the key is not present. How can I handle this more simply?
See Delete an element from a dictionary for more general approaches to the problem of removing a key from a dict (including ones which produce a modified copy).
To delete a key regardless of whether it is in the dictionary, use the two-argument form of dict.pop():
my_dict.pop('key', None)
This will return my_dict[key] if key exists in the dictionary, and None otherwise. If the second parameter is not specified (i.e. my_dict.pop('key')) and key does not exist, a KeyError is raised.
To delete a key that is guaranteed to exist, you can also use
del my_dict['key']
This will raise a KeyError if the key is not in the dictionary.
Specifically to answer "is there a one line way of doing this?"
if 'key' in my_dict: del my_dict['key']
...well, you asked ;-)
You should consider, though, that this way of deleting an object from a dict is not atomic—it is possible that 'key' may be in my_dict during the if statement, but may be deleted before del is executed, in which case del will fail with a KeyError. Given this, it would be safest to either use dict.pop or something along the lines of
try:
del my_dict['key']
except KeyError:
pass
which, of course, is definitely not a one-liner.
It took me some time to figure out what exactly my_dict.pop("key", None) is doing. So I'll add this as an answer to save others googling time:
pop(key[, default])
If key is in the dictionary, remove it and return its value, else
return default. If default is not given and key is not in the
dictionary, a KeyError is raised.
Documentation
del my_dict[key] is slightly faster than my_dict.pop(key) for removing a key from a dictionary when the key exists
>>> import timeit
>>> setup = "d = {i: i for i in range(100000)}"
>>> timeit.timeit("del d[3]", setup=setup, number=1)
1.79e-06
>>> timeit.timeit("d.pop(3)", setup=setup, number=1)
2.09e-06
>>> timeit.timeit("d2 = {key: val for key, val in d.items() if key != 3}", setup=setup, number=1)
0.00786
But when the key doesn't exist if key in my_dict: del my_dict[key] is slightly faster than my_dict.pop(key, None). Both are at least three times faster than del in a try/except statement:
>>> timeit.timeit("if 'missing key' in d: del d['missing key']", setup=setup)
0.0229
>>> timeit.timeit("d.pop('missing key', None)", setup=setup)
0.0426
>>> try_except = """
... try:
... del d['missing key']
... except KeyError:
... pass
... """
>>> timeit.timeit(try_except, setup=setup)
0.133
If you need to remove a lot of keys from a dictionary in one line of code, I think using map() is quite succinct and Pythonic readable:
myDict = {'a':1,'b':2,'c':3,'d':4}
map(myDict.pop, ['a','c']) # The list of keys to remove
>>> myDict
{'b': 2, 'd': 4}
And if you need to catch errors where you pop a value that isn't in the dictionary, use lambda inside map() like this:
map(lambda x: myDict.pop(x,None), ['a', 'c', 'e'])
[1, 3, None] # pop returns
>>> myDict
{'b': 2, 'd': 4}
or in python3, you must use a list comprehension instead:
[myDict.pop(x, None) for x in ['a', 'c', 'e']]
It works. And 'e' did not cause an error, even though myDict did not have an 'e' key.
You can use a dictionary comprehension to create a new dictionary with that key removed:
>>> my_dict = {k: v for k, v in my_dict.items() if k != 'key'}
You can delete by conditions. No error if key doesn't exist.
You can use exception handling if you want to be very verbose:
try:
del dict[key]
except KeyError: pass
This is slower, however, than the pop() method, if the key doesn't exist.
my_dict.pop('key', None)
It won't matter for a few keys, but if you're doing this repeatedly, then the latter method is a better bet.
The fastest approach is this:
if 'key' in dict:
del myDict['key']
But this method is dangerous because if 'key' is removed in between the two lines, a KeyError will be raised.
We can delete a key from a Python dictionary by the some of the following approaches.
Using the del keyword; it's almost the same approach like you did though -
myDict = {'one': 100, 'two': 200, 'three': 300 }
print(myDict) # {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : del myDict['one']
print(myDict) # {'two': 200, 'three': 300}
Or
We can do like the following:
But one should keep in mind that, in this process actually it won't delete any key from the dictionary rather than making a specific key excluded from that dictionary. In addition, I observed that it returned a dictionary which was not ordered the same as myDict.
myDict = {'one': 100, 'two': 200, 'three': 300, 'four': 400, 'five': 500}
{key:value for key, value in myDict.items() if key != 'one'}
If we run it in the shell, it'll execute something like {'five': 500, 'four': 400, 'three': 300, 'two': 200} - notice that it's not the same ordered as myDict. Again if we try to print myDict, then we can see all keys including which we excluded from the dictionary by this approach. However, we can make a new dictionary by assigning the following statement into a variable:
var = {key:value for key, value in myDict.items() if key != 'one'}
Now if we try to print it, then it'll follow the parent order:
print(var) # {'two': 200, 'three': 300, 'four': 400, 'five': 500}
Or
Using the pop() method.
myDict = {'one': 100, 'two': 200, 'three': 300}
print(myDict)
if myDict.get('one') : myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
The difference between del and pop is that, using pop() method, we can actually store the key's value if needed, like the following:
myDict = {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : var = myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
print(var) # 100
Fork this gist for future reference, if you find this useful.
I prefer the immutable version
foo = {
1:1,
2:2,
3:3
}
removeKeys = [1,2]
def woKeys(dct, keyIter):
return {
k:v
for k,v in dct.items() if k not in keyIter
}
>>> print(woKeys(foo, removeKeys))
{3: 3}
>>> print(foo)
{1: 1, 2: 2, 3: 3}
Another way is by using items() + dict comprehension.
items() coupled with dict comprehension can also help us achieve the task of key-value pair deletion, but it has the drawback of not being an in place dict technique. Actually a new dict if created except for the key we don’t wish to include.
test_dict = {"sai" : 22, "kiran" : 21, "vinod" : 21, "sangam" : 21}
# Printing dictionary before removal
print ("dictionary before performing remove is : " + str(test_dict))
# Using items() + dict comprehension to remove a dict. pair
# removes vinod
new_dict = {key:val for key, val in test_dict.items() if key != 'vinod'}
# Printing dictionary after removal
print ("dictionary after remove is : " + str(new_dict))
Output:
dictionary before performing remove is : {'sai': 22, 'kiran': 21, 'vinod': 21, 'sangam': 21}
dictionary after remove is : {'sai': 22, 'kiran': 21, 'sangam': 21}
Do dictionaries allow for filtering based on key length? Altering dictionaries conditionally based on value seems straightforward, but what about doing the same with keys? i.e. what would it take to delete all dictionary keys that are not explicitly 4 characters long?
You could use a dictionary comprehension:
d = {'fooo': 1, 'bar': 2, 'foo': 3}
result = {k: v for k, v in d.items() if len(k) == 4}
print(result)
Output
{'fooo': 1}
As krflol mentioned, if you don't want to build a whole new dictionary but
rather just modify your old dictionary, a straightforward for loop will do:
my_dict = {"a": 1, "abcd": 4, "": 9, "xyzt": 16}
for k in list(my_dict):
if len(k) != 4:
del my_dict[k]
print(my_dict)
Another way could be using filter, like so :
my_dict = dict(filter(lambda k: len(k) != 4, my_dict.items()))
This can be accomplished as seen in the following:
new_dict = {k: v for k, v in old_dict.items() if len(k) == 4}
I want to remove a key from a dictionary if it is present. I currently use this code:
if key in my_dict:
del my_dict[key]
Without the if statement, the code will raise KeyError if the key is not present. How can I handle this more simply?
See Delete an element from a dictionary for more general approaches to the problem of removing a key from a dict (including ones which produce a modified copy).
To delete a key regardless of whether it is in the dictionary, use the two-argument form of dict.pop():
my_dict.pop('key', None)
This will return my_dict[key] if key exists in the dictionary, and None otherwise. If the second parameter is not specified (i.e. my_dict.pop('key')) and key does not exist, a KeyError is raised.
To delete a key that is guaranteed to exist, you can also use
del my_dict['key']
This will raise a KeyError if the key is not in the dictionary.
Specifically to answer "is there a one line way of doing this?"
if 'key' in my_dict: del my_dict['key']
...well, you asked ;-)
You should consider, though, that this way of deleting an object from a dict is not atomic—it is possible that 'key' may be in my_dict during the if statement, but may be deleted before del is executed, in which case del will fail with a KeyError. Given this, it would be safest to either use dict.pop or something along the lines of
try:
del my_dict['key']
except KeyError:
pass
which, of course, is definitely not a one-liner.
It took me some time to figure out what exactly my_dict.pop("key", None) is doing. So I'll add this as an answer to save others googling time:
pop(key[, default])
If key is in the dictionary, remove it and return its value, else
return default. If default is not given and key is not in the
dictionary, a KeyError is raised.
Documentation
del my_dict[key] is slightly faster than my_dict.pop(key) for removing a key from a dictionary when the key exists
>>> import timeit
>>> setup = "d = {i: i for i in range(100000)}"
>>> timeit.timeit("del d[3]", setup=setup, number=1)
1.79e-06
>>> timeit.timeit("d.pop(3)", setup=setup, number=1)
2.09e-06
>>> timeit.timeit("d2 = {key: val for key, val in d.items() if key != 3}", setup=setup, number=1)
0.00786
But when the key doesn't exist if key in my_dict: del my_dict[key] is slightly faster than my_dict.pop(key, None). Both are at least three times faster than del in a try/except statement:
>>> timeit.timeit("if 'missing key' in d: del d['missing key']", setup=setup)
0.0229
>>> timeit.timeit("d.pop('missing key', None)", setup=setup)
0.0426
>>> try_except = """
... try:
... del d['missing key']
... except KeyError:
... pass
... """
>>> timeit.timeit(try_except, setup=setup)
0.133
If you need to remove a lot of keys from a dictionary in one line of code, I think using map() is quite succinct and Pythonic readable:
myDict = {'a':1,'b':2,'c':3,'d':4}
map(myDict.pop, ['a','c']) # The list of keys to remove
>>> myDict
{'b': 2, 'd': 4}
And if you need to catch errors where you pop a value that isn't in the dictionary, use lambda inside map() like this:
map(lambda x: myDict.pop(x,None), ['a', 'c', 'e'])
[1, 3, None] # pop returns
>>> myDict
{'b': 2, 'd': 4}
or in python3, you must use a list comprehension instead:
[myDict.pop(x, None) for x in ['a', 'c', 'e']]
It works. And 'e' did not cause an error, even though myDict did not have an 'e' key.
You can use a dictionary comprehension to create a new dictionary with that key removed:
>>> my_dict = {k: v for k, v in my_dict.items() if k != 'key'}
You can delete by conditions. No error if key doesn't exist.
You can use exception handling if you want to be very verbose:
try:
del dict[key]
except KeyError: pass
This is slower, however, than the pop() method, if the key doesn't exist.
my_dict.pop('key', None)
It won't matter for a few keys, but if you're doing this repeatedly, then the latter method is a better bet.
The fastest approach is this:
if 'key' in dict:
del myDict['key']
But this method is dangerous because if 'key' is removed in between the two lines, a KeyError will be raised.
We can delete a key from a Python dictionary by the some of the following approaches.
Using the del keyword; it's almost the same approach like you did though -
myDict = {'one': 100, 'two': 200, 'three': 300 }
print(myDict) # {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : del myDict['one']
print(myDict) # {'two': 200, 'three': 300}
Or
We can do like the following:
But one should keep in mind that, in this process actually it won't delete any key from the dictionary rather than making a specific key excluded from that dictionary. In addition, I observed that it returned a dictionary which was not ordered the same as myDict.
myDict = {'one': 100, 'two': 200, 'three': 300, 'four': 400, 'five': 500}
{key:value for key, value in myDict.items() if key != 'one'}
If we run it in the shell, it'll execute something like {'five': 500, 'four': 400, 'three': 300, 'two': 200} - notice that it's not the same ordered as myDict. Again if we try to print myDict, then we can see all keys including which we excluded from the dictionary by this approach. However, we can make a new dictionary by assigning the following statement into a variable:
var = {key:value for key, value in myDict.items() if key != 'one'}
Now if we try to print it, then it'll follow the parent order:
print(var) # {'two': 200, 'three': 300, 'four': 400, 'five': 500}
Or
Using the pop() method.
myDict = {'one': 100, 'two': 200, 'three': 300}
print(myDict)
if myDict.get('one') : myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
The difference between del and pop is that, using pop() method, we can actually store the key's value if needed, like the following:
myDict = {'one': 100, 'two': 200, 'three': 300}
if myDict.get('one') : var = myDict.pop('one')
print(myDict) # {'two': 200, 'three': 300}
print(var) # 100
Fork this gist for future reference, if you find this useful.
I prefer the immutable version
foo = {
1:1,
2:2,
3:3
}
removeKeys = [1,2]
def woKeys(dct, keyIter):
return {
k:v
for k,v in dct.items() if k not in keyIter
}
>>> print(woKeys(foo, removeKeys))
{3: 3}
>>> print(foo)
{1: 1, 2: 2, 3: 3}
Another way is by using items() + dict comprehension.
items() coupled with dict comprehension can also help us achieve the task of key-value pair deletion, but it has the drawback of not being an in place dict technique. Actually a new dict if created except for the key we don’t wish to include.
test_dict = {"sai" : 22, "kiran" : 21, "vinod" : 21, "sangam" : 21}
# Printing dictionary before removal
print ("dictionary before performing remove is : " + str(test_dict))
# Using items() + dict comprehension to remove a dict. pair
# removes vinod
new_dict = {key:val for key, val in test_dict.items() if key != 'vinod'}
# Printing dictionary after removal
print ("dictionary after remove is : " + str(new_dict))
Output:
dictionary before performing remove is : {'sai': 22, 'kiran': 21, 'vinod': 21, 'sangam': 21}
dictionary after remove is : {'sai': 22, 'kiran': 21, 'sangam': 21}
Remove multiple keys
Marc Maxmeister's post discusses this but creates an unnecessary (imo) list while doing so. You can simply use a for-loop and throw away the popped values.
my_dict = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
lst = ['a', 'c', 'e']
for k in lst: my_dict.pop(k, None)
print(my_dict) # {'b': 2, 'd': 4}
or if you want to use map, then exhaust the map using a deque with max length 0.
from collections import deque
from itertools import repeat
deque(map(my_dict.pop, ['a', 'c', 'e'], repeat(None)), 0)
print(my_dict) # {'b': 2, 'd': 4}
Split dict into two
One case where dict.pop() may be useful is if you want to create a new dictionary with the popped key-value pairs, effectively splitting a dictionary into two in one for-loop.
new_dict = {k: v for k in lst if (v:=my_dict.pop(k, 'NULL')) != 'NULL'}
print(my_dict) # {'b': 2, 'd': 4}
print(new_dict) # {'a': 1, 'c': 3}
This question already has answers here:
Reverse / invert a dictionary mapping
(32 answers)
Closed 10 months ago.
I receive a dictionary as input, and would like to to return a dictionary whose keys will be the input's values and whose value will be the corresponding input keys. Values are unique.
For example, say my input is:
a = dict()
a['one']=1
a['two']=2
I would like my output to be:
{1: 'one', 2: 'two'}
To clarify I would like my result to be the equivalent of the following:
res = dict()
res[1] = 'one'
res[2] = 'two'
Any neat Pythonic way to achieve this?
Python 2:
res = dict((v,k) for k,v in a.iteritems())
Python 3 (thanks to #erik):
res = dict((v,k) for k,v in a.items())
new_dict = dict(zip(my_dict.values(), my_dict.keys()))
From Python 2.7 on, including 3.0+, there's an arguably shorter, more readable version:
>>> my_dict = {'x':1, 'y':2, 'z':3}
>>> {v: k for k, v in my_dict.items()}
{1: 'x', 2: 'y', 3: 'z'}
You can make use of dict comprehensions:
Python 3
res = {v: k for k, v in a.items()}
Python 2
res = {v: k for k, v in a.iteritems()}
Edited: For Python 3, use a.items() instead of a.iteritems(). Discussions about the differences between them can be found in iteritems in Python on SO.
In [1]: my_dict = {'x':1, 'y':2, 'z':3}
Python 3
In [2]: dict((value, key) for key, value in my_dict.items())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
Python 2
In [2]: dict((value, key) for key, value in my_dict.iteritems())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
The current leading answer assumes values are unique which is not always the case. What if values are not unique? You will loose information!
For example:
d = {'a':3, 'b': 2, 'c': 2}
{v:k for k,v in d.iteritems()}
returns {2: 'b', 3: 'a'}.
The information about 'c' was completely ignored.
Ideally it should had be something like {2: ['b','c'], 3: ['a']}. This is what the bottom implementation does.
Python 2.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.iteritems():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
Python 3.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.items():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
You could try:
Python 3
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.items())
d2
{'two': 2, 'one': 1}
Python 2
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.iteritems())
d2
{'two': 2, 'one': 1}
Beware that you cannot 'reverse' a dictionary if
More than one key shares the same value. For example {'one':1,'two':1}. The new dictionary can only have one item with key 1.
One or more of the values is unhashable. For example {'one':[1]}. [1] is a valid value but not a valid key.
See this thread on the python mailing list for a discussion on the subject.
res = dict(zip(a.values(), a.keys()))
new_dict = dict( (my_dict[k], k) for k in my_dict)
or even better, but only works in Python 3:
new_dict = { my_dict[k]: k for k in my_dict}
Another way to expand on Ilya Prokin's response is to actually use the reversed function.
dict(map(reversed, my_dict.items()))
In essence, your dictionary is iterated through (using .items()) where each item is a key/value pair, and those items are swapped with the reversed function. When this is passed to the dict constructor, it turns them into value/key pairs which is what you want.
Suggestion for an improvement for Javier answer :
dict(zip(d.values(),d))
Instead of d.keys() you can write just d, because if you go through dictionary with an iterator, it will return the keys of the relevant dictionary.
Ex. for this behavior :
d = {'a':1,'b':2}
for k in d:
k
'a'
'b'
Can be done easily with dictionary comprehension:
{d[i]:i for i in d}
dict(map(lambda x: x[::-1], YourDict.items()))
.items() returns a list of tuples of (key, value). map() goes through elements of the list and applies lambda x:[::-1] to each its element (tuple) to reverse it, so each tuple becomes (value, key) in the new list spitted out of map. Finally, dict() makes a dict from the new list.
Hanan's answer is the correct one as it covers more general case (the other answers are kind of misleading for someone unaware of the duplicate situation). An improvement to Hanan's answer is using setdefault:
mydict = {1:a, 2:a, 3:b}
result = {}
for i in mydict:
result.setdefault(mydict[i],[]).append(i)
print(result)
>>> result = {a:[1,2], b:[3]}
Using loop:-
newdict = {} #Will contain reversed key:value pairs.
for key, value in zip(my_dict.keys(), my_dict.values()):
# Operations on key/value can also be performed.
newdict[value] = key
If you're using Python3, it's slightly different:
res = dict((v,k) for k,v in a.items())
Adding an in-place solution:
>>> d = {1: 'one', 2: 'two', 3: 'three', 4: 'four'}
>>> for k in list(d.keys()):
... d[d.pop(k)] = k
...
>>> d
{'two': 2, 'one': 1, 'four': 4, 'three': 3}
In Python3, it is critical that you use list(d.keys()) because dict.keys returns a view of the keys. If you are using Python2, d.keys() is enough.
I find this version the most comprehensive one:
a = {1: 'one', 2: 'two'}
swapped_a = {value : key for key, value in a.items()}
print(swapped_a)
output :
{'one': 1, 'two': 2}
An alternative that is not quite as readable (in my opinion) as some of the other answers:
new_dict = dict(zip(*list(zip(*old_dict.items()))[::-1]))
where list(zip(*old_dict.items()))[::-1] gives a list of 2 tuples, old_dict's values and keys, respectively.