How can I add new keys to a dictionary?


How do I add a key to an existing dictionary? It doesn’t have an .add() method.

Asked By: lfaraone



dictionary[key] = value
Answered By: Jason Creighton

You create a new key/value pair on a dictionary by assigning a value to that key

d = {'key': 'value'}
print(d)  # {'key': 'value'}

d['mynewkey'] = 'mynewvalue'

print(d)  # {'key': 'value', 'mynewkey': 'mynewvalue'}

If the key doesn’t exist, it’s added and points to that value. If it exists, the current value it points to is overwritten.

Answered By: Paolo Bergantino

To add multiple keys simultaneously, use dict.update():

>>> x = {1:2}
>>> print(x)
{1: 2}

>>> d = {3:4, 5:6, 7:8}
>>> x.update(d)
>>> print(x)
{1: 2, 3: 4, 5: 6, 7: 8}

For adding a single key, the accepted answer has less computational overhead.

Answered By: user142911

I feel like consolidating info about Python dictionaries:

Creating an empty dictionary

data = {}
# OR
data = dict()

Creating a dictionary with initial values

data = {'a': 1, 'b': 2, 'c': 3}
# OR
data = dict(a=1, b=2, c=3)
# OR
data = {k: v for k, v in (('a', 1), ('b',2), ('c',3))}

Inserting/Updating a single value

data['a'] = 1  # Updates if 'a' exists, else adds 'a'
# OR
data.update({'a': 1})
# OR
# OR

Inserting/Updating multiple values

data.update({'c':3,'d':4})  # Updates 'c' and adds 'd'

Python 3.9+:

The update operator |= now works for dictionaries:

data |= {'c':3,'d':4}

Creating a merged dictionary without modifying originals

data3 = {}
data3.update(data)  # Modifies data3, not data
data3.update(data2)  # Modifies data3, not data2

Python 3.5+:

This uses a new feature called dictionary unpacking.

data = {**data1, **data2, **data3}

Python 3.9+:

The merge operator | now works for dictionaries:

data = data1 | {'c':3,'d':4}

Deleting items in dictionary

del data[key]  # Removes specific element in a dictionary
data.pop(key)  # Removes the key & returns the value
data.clear()  # Clears entire dictionary

Check if a key is already in dictionary

key in data

Iterate through pairs in a dictionary

for key in data: # Iterates just through the keys, ignoring the values
for key, value in d.items(): # Iterates through the pairs
for key in d.keys(): # Iterates just through key, ignoring the values
for value in d.values(): # Iterates just through value, ignoring the keys

Create a dictionary from two lists

data = dict(zip(list_with_keys, list_with_values))
Answered By: Yugal Jindle

If you want to add a dictionary within a dictionary you can do it this way.

Example: Add a new entry to your dictionary & sub dictionary

dictionary = {}
dictionary["new key"] = "some new entry" # add new dictionary entry
dictionary["dictionary_within_a_dictionary"] = {} # this is required by python
dictionary["dictionary_within_a_dictionary"]["sub_dict"] = {"other" : "dictionary"}
print (dictionary)


{'new key': 'some new entry', 'dictionary_within_a_dictionary': {'sub_dict': {'other': 'dictionarly'}}}

NOTE: Python requires that you first add a sub

dictionary["dictionary_within_a_dictionary"] = {}

before adding entries.

Answered By: Asher

The conventional syntax is d[key] = value, but if your keyboard is missing the square bracket keys you could also do:

d.__setitem__(key, value)

In fact, defining __getitem__ and __setitem__ methods is how you can make your own class support the square bracket syntax. See Dive Into Python, Classes That Act Like Dictionaries.

Answered By: Colonel Panic

You can create one:

class myDict(dict):

    def __init__(self):
        self = dict()

    def add(self, key, value):
        self[key] = value

## example

myd = myDict()


{'apples': 6, 'bananas': 3}
Answered By: kiriloff

This popular question addresses functional methods of merging dictionaries a and b.

Here are some of the more straightforward methods (tested in Python 3)…

c = dict( a, **b ) ## see also
c = dict( list(a.items()) + list(b.items()) )
c = dict( i for d in [a,b] for i in d.items() )

Note: The first method above only works if the keys in b are strings.

To add or modify a single element, the b dictionary would contain only that one element…

c = dict( a, **{'d':'dog'} ) ## returns a dictionary based on 'a'

This is equivalent to…

def functional_dict_add( dictionary, key, value ):
   temp = dictionary.copy()
   temp[key] = value
   return temp

c = functional_dict_add( a, 'd', 'dog' )
Answered By: Brent Bradburn

"Is it possible to add a key to a Python dictionary after it has been created? It doesn’t seem to have an .add() method."

Yes it is possible, and it does have a method that implements this, but you don’t want to use it directly.

To demonstrate how and how not to use it, let’s create an empty dict with the dict literal, {}:

my_dict = {}

Best Practice 1: Subscript notation

To update this dict with a single new key and value, you can use the subscript notation (see Mappings here) that provides for item assignment:

my_dict['new key'] = 'new value'

my_dict is now:

{'new key': 'new value'}

Best Practice 2: The update method – 2 ways

We can also update the dict with multiple values efficiently as well using the update method. We may be unnecessarily creating an extra dict here, so we hope our dict has already been created and came from or was used for another purpose:

my_dict.update({'key 2': 'value 2', 'key 3': 'value 3'})

my_dict is now:

{'key 2': 'value 2', 'key 3': 'value 3', 'new key': 'new value'}

Another efficient way of doing this with the update method is with keyword arguments, but since they have to be legitimate python words, you can’t have spaces or special symbols or start the name with a number, but many consider this a more readable way to create keys for a dict, and here we certainly avoid creating an extra unnecessary dict:

my_dict.update(foo='bar', foo2='baz')

and my_dict is now:

{'key 2': 'value 2', 'key 3': 'value 3', 'new key': 'new value', 
 'foo': 'bar', 'foo2': 'baz'}

So now we have covered three Pythonic ways of updating a dict.

Magic method, __setitem__, and why it should be avoided

There’s another way of updating a dict that you shouldn’t use, which uses the __setitem__ method. Here’s an example of how one might use the __setitem__ method to add a key-value pair to a dict, and a demonstration of the poor performance of using it:

>>> d = {}
>>> d.__setitem__('foo', 'bar')
>>> d
{'foo': 'bar'}

>>> def f():
...     d = {}
...     for i in xrange(100):
...         d['foo'] = i
>>> def g():
...     d = {}
...     for i in xrange(100):
...         d.__setitem__('foo', i)
>>> import timeit
>>> number = 100
>>> min(timeit.repeat(f, number=number))
>>> min(timeit.repeat(g, number=number))

So we see that using the subscript notation is actually much faster than using __setitem__. Doing the Pythonic thing, that is, using the language in the way it was intended to be used, usually is both more readable and computationally efficient.

There is also the strangely named, oddly behaved, and yet still handy dict.setdefault().


value = my_dict.setdefault(key, default)

basically just does this:

    value = my_dict[key]
except KeyError: # key not found
    value = my_dict[key] = default


>>> mydict = {'a':1, 'b':2, 'c':3}
>>> mydict.setdefault('d', 4)
4 # returns new value at mydict['d']
>>> print(mydict)
{'a':1, 'b':2, 'c':3, 'd':4} # a new key/value pair was indeed added
# but see what happens when trying it on an existing key...
>>> mydict.setdefault('a', 111)
1 # old value was returned
>>> print(mydict)
{'a':1, 'b':2, 'c':3, 'd':4} # existing key was ignored
Answered By: Michael Ekoka

Let’s pretend you want to live in the immutable world and do not want to modify the original but want to create a new dict that is the result of adding a new key to the original.

In Python 3.5+ you can do:

params = {'a': 1, 'b': 2}
new_params = {**params, **{'c': 3}}

The Python 2 equivalent is:

params = {'a': 1, 'b': 2}
new_params = dict(params, **{'c': 3})

After either of these:

params is still equal to {'a': 1, 'b': 2}


new_params is equal to {'a': 1, 'b': 2, 'c': 3}

There will be times when you don’t want to modify the original (you only want the result of adding to the original). I find this a refreshing alternative to the following:

params = {'a': 1, 'b': 2}
new_params = params.copy()
new_params['c'] = 3


params = {'a': 1, 'b': 2}
new_params = params.copy()
new_params.update({'c': 3})

Reference: What does `**` mean in the expression `dict(d1, **d2)`?

Answered By: campeterson

If you’re not joining two dictionaries, but adding new key-value pairs to a dictionary, then using the subscript notation seems like the best way.

import timeit

timeit.timeit('dictionary = {"karga": 1, "darga": 2}; dictionary.update({"aaa": 123123, "asd": 233})')
>> 0.49582505226135254

timeit.timeit('dictionary = {"karga": 1, "darga": 2}; dictionary["aaa"] = 123123; dictionary["asd"] = 233;')
>> 0.20782899856567383

However, if you’d like to add, for example, thousands of new key-value pairs, you should consider using the update() method.

Answered By: Burak Özdemir

First to check whether the key already exists:


Then you can add the new key and value.

Answered By: Agus Mathew

I think it would also be useful to point out Python’s collections module that consists of many useful dictionary subclasses and wrappers that simplify the addition and modification of data types in a dictionary, specifically defaultdict:

dict subclass that calls a factory function to supply missing values

This is particularly useful if you are working with dictionaries that always consist of the same data types or structures, for example a dictionary of lists.

>>> from collections import defaultdict
>>> example = defaultdict(int)
>>> example['key'] += 1
>>> example['key']
defaultdict(<class 'int'>, {'key': 1})

If the key does not yet exist, defaultdict assigns the value given (in our case 10) as the initial value to the dictionary (often used inside loops). This operation therefore does two things: it adds a new key to a dictionary (as per question), and assigns the value if the key doesn’t yet exist. With the standard dictionary, this would have raised an error as the += operation is trying to access a value that doesn’t yet exist:

>>> example = dict()
>>> example['key'] += 1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'key'

Without the use of defaultdict, the amount of code to add a new element would be much greater and perhaps looks something like:

# This type of code would often be inside a loop
if 'key' not in example:
    example['key'] = 0  # add key and initial value to dict; could also be a list
example['key'] += 1  # this is implementing a counter

defaultdict can also be used with complex data types such as list and set:

>>> example = defaultdict(list)
>>> example['key'].append(1)
>>> example
defaultdict(<class 'list'>, {'key': [1]})

Adding an element automatically initialises the list.

Answered By: m_____z

Here’s another way that I didn’t see here:

>>> foo = dict(a=1,b=2)
>>> foo
{'a': 1, 'b': 2}
>>> goo = dict(c=3,**foo)
>>> goo
{'c': 3, 'a': 1, 'b': 2}

You can use the dictionary constructor and implicit expansion to reconstruct a dictionary. Moreover, interestingly, this method can be used to control the positional order during dictionary construction (post Python 3.6). In fact, insertion order is guaranteed for Python 3.7 and above!

>>> foo = dict(a=1,b=2,c=3,d=4)
>>> new_dict = {k: v for k, v in list(foo.items())[:2]}
>>> new_dict
{'a': 1, 'b': 2}
>>> new_dict.update(newvalue=99)
>>> new_dict
{'a': 1, 'b': 2, 'newvalue': 99}
>>> new_dict.update({k: v for k, v in list(foo.items())[2:]})
>>> new_dict
{'a': 1, 'b': 2, 'newvalue': 99, 'c': 3, 'd': 4}

The above is using dictionary comprehension.

Answered By: ingyhere

Add a dictionary (key,value) class.

class myDict(dict):

    def __init__(self):
        self = dict()

    def add(self, key, value):
        #self[key] = value # add new key and value overwriting any exiting same key
        if self.get(key)!=None:
            print('key', key, 'already used') # report if key already used
        self.setdefault(key, value) # if key exit do nothing

## example

myd = myDict()
name = "fred"

print('n', myd)
print('n', myd)
myd.add('jack', 7)
print('n', myd)
myd.add(name, myd)
print('n', myd)
myd.add('apples', 23)
print('n', myd)
myd.add(name, 2)
Answered By: Susan

This question has already been answered ad nauseam, but since my (now deleted)
gained a lot of traction, here it is as an answer:

Adding new keys without updating the existing dict

If you are here trying to figure out how to add a key and return a new dictionary (without modifying the existing one), you can do this using the techniques below

Python >= 3.5

new_dict = {**mydict, 'new_key': new_val}

Python < 3.5

new_dict = dict(mydict, new_key=new_val)

Note that with this approach, your key will need to follow the rules of valid identifier names in Python.

Answered By: cs95

Adding keys to dictionary without using add

        # Inserting/Updating single value
        # subscript notation method
        d['mynewkey'] = 'mynewvalue' # Updates if 'a' exists, else adds 'a'
        # OR
        d.update({'mynewkey': 'mynewvalue'})
        # OR
        # OR
        print(d)  # {'key': 'value', 'mynewkey': 'mynewvalue'}
        # To add/update multiple keys simultaneously, use d.update():
        x = {3:4, 5:6, 7:8}
        print(d) # {'key': 'value', 'mynewkey': 'mynewvalue', 3: 4, 5: 6, 7: 8}
        # update operator |= now works for dictionaries:
        d |= {'c':3,'d':4}
        # Assigning new key value pair using dictionary unpacking.
        data1 = {4:6, 9:10, 17:20}
        data2 = {20:30, 32:48, 90:100}
        data3 = { 38:"value", 99:"notvalid"}
        d = {**data1, **data2, **data3}
        # The merge operator | now works for dictionaries:
        data = data1 | {'c':3,'d':4}
        # Create a dictionary from two lists
        data = dict(zip(list_with_keys, list_with_values))
Answered By: dataninsight

dico["new key"] = "value"

Answered By: Egidius

You can write a function for doing that, e.g.:

def dict_ins( d, key, value ):
    if key in d: raise Exception('This key is already in the dictionary.')
    d[key] = value
Answered By: Anton Shepelev
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