Related
I want to do something like this:
myList = [10, 20, 30]
yourList = myList.append(40)
Unfortunately, list append does not return the modified list.
So, how can I allow append to return the new list?
See also: Why do these list operations (methods) return None, rather than the resulting list?
Don't use append but concatenation instead:
yourList = myList + [40]
This returns a new list; myList will not be affected. If you need to have myList affected as well either use .append() anyway, then assign yourList separately from (a copy of) myList.
In python 3 you may create new list by unpacking old one and adding new element:
a = [1,2,3]
b = [*a,4] # b = [1,2,3,4]
when you do:
myList + [40]
You actually have 3 lists.
list.append is a built-in and therefore cannot be changed. But if you're willing to use something other than append, you could try +:
In [106]: myList = [10,20,30]
In [107]: yourList = myList + [40]
In [108]: print myList
[10, 20, 30]
In [109]: print yourList
[10, 20, 30, 40]
Of course, the downside to this is that a new list is created which takes a lot more time than append
Hope this helps
Try using itertools.chain(myList, [40]). That will return a generator as a sequence, rather than allocating a new list. Essentially, that returns all of the elements from the first iterable until it is exhausted, then proceeds to the next iterable, until all of the iterables are exhausted.
Unfortunately, none of the answers here solve exactly what was asked. Here is a simple approach:
lst = [1, 2, 3]
lst.append(4) or lst # the returned value here would be the OP's `yourList`
# [1, 2, 3, 4]
One may ask the real need of doing this, like when someone needs to improve RAM usage, do micro-benchmarks etc. that are, usually, useless. However, sometimes someone is really "asking what was asked" (I don't know if this is the case here) and the reality is more diverse than we can know of. So here is a (contrived because out-of-a-context) usage...
Instead of doing this:
dic = {"a": [1], "b": [2], "c": [3]}
key, val = "d", 4 # <- example
if key in dic:
dic[key].append(val)
else:
dic[key] = [val]
dic
# {'a': [1], 'b': [2], 'c': [3], 'd': [4]}
key, val = "b", 5 # <- example
if key in dic:
dic[key].append(val)
else:
dic[key] = [val]
dic
# {'a': [1], 'b': [2, 5], 'c': [3], 'd': [4]}
One can use the OR expression above in any place an expression is needed (instead of a statement):
key, val = "d", 4 # <- example
dic[key] = dic[key].append(val) or dic[key] if key in dic else [val]
# {'a': [1], 'b': [2], 'c': [3], 'd': [4]}
key, val = "b", 5 # <- example
dic[key] = dic[key].append(val) or dic[key] if key in dic else [val]
# {'a': [1], 'b': [2, 5], 'c': [3], 'd': [4]}
Or, equivalently, when there are no falsy values in the lists, one can try dic.get(key, <default value>) in some better way.
You can subclass the built-in list type and redefine the 'append' method. Or even better, create a new one which will do what you want it to do. Below is the code for a redefined 'append' method.
#!/usr/bin/env python
class MyList(list):
def append(self, element):
return MyList(self + [element])
def main():
l = MyList()
l1 = l.append(1)
l2 = l1.append(2)
l3 = l2.append(3)
print "Original list: %s, type %s" % (l, l.__class__.__name__)
print "List 1: %s, type %s" % (l1, l1.__class__.__name__)
print "List 2: %s, type %s" % (l2, l2.__class__.__name__)
print "List 3: %s, type %s" % (l3, l3.__class__.__name__)
if __name__ == '__main__':
main()
Hope that helps.
Just to expand on Storstamp's answer
You only need to do
myList.append(40)
It will append it to the original list,now you can return the variable containing the original list.
If you are working with very large lists this is the way to go.
You only need to do
myList.append(40)
It will append it to the original list, not return a new list.
I want to create a data structure for storing various possible paths through a plane with polygons scattered across it. I decided on using nested, multi-level dictionaries to save the various possible paths splitting at fixed points.
A possible instance of such a dictionary would be:
path_dictionary = {starting_coordinates:{new_fixpoint1:{new_fixpoint1_1:...}, new_fixpoint2:{new_fixpoint2_1:...}}}
Now I want to continue building up that structure with new paths from the last fixpoints, so I would have to edit the dictionary at various nesting levels. My plan was to provide a sorted keylist which contains all the fixpoints of the given path and I would have a function to add at to the last provided key.
To achieve this I would have to be able to access the dictionary with the keylist like this:
keylist = [starting_coordinates, new_fixpoint1, new_fixpoint1_1, new_fixpoint1_1_3, ...]
path_dictionary = {starting_coordinates:{new_fixpoint1:{new_fixpoint1_1:...}, new_fixpoint2:{new_fixpoint2_1:...}}}
path_dictionary [keylist [0]] [keylist [1]] [keylist [2]] [...] = additional_fixpoint
Question: How can I write to a variable nesting/depth level in the multi-level dictionary when I have a keylist of some length?
Any help is very much appreciated.
I was playing around with the idea of using multiple indexes, and a defaultdict. And this came out:
from collections import defaultdict
class LayeredDict(defaultdict):
def __getitem__(self, key):
if isinstance(key, (tuple, list)):
if len(key) == 1:
return self[key[0]]
return self[key[0]][key[1:]]
return super(LayeredDict, self).__getitem__(key)
def __setitem__(self, key, value):
if isinstance(key, (tuple, list)):
if len(key) == 1:
self[key[0]] = value
else:
self[key[0]][key[1:]] = value
else:
super(LayeredDict, self).__setitem__(key, value)
def __init__(self, *args, **kwargs):
super(LayeredDict, self).__init__(*args, **kwargs)
self.default_factory = type(self) # override default
I haven't fully tested it, but it should allow you to create any level of nested dictionaries, and index them with a tuple.
>>> x = LayeredDict()
>>> x['abc'] = 'blah'
>>> x['abc']
'blah'
>>> x[0, 8, 2] = 1.2345
>>> x[0, 8, 1] = 8.9
>>> x[0, 8, 'xyz'] = 10.1
>>> x[0, 8].keys()
[1, 2, 'xyz']
>>> x['abc', 1] = 5
*** TypeError: 'str' object does not support item assignment
Unfortunately expansion notation (or whatever it's called) isn't supported, but
you can just pass a list or tuple in as an index.
>>> keylist = (0, 8, 2)
>>> x[*keylist]
*** SyntaxError: invalid syntax (<stdin>, line 1)
>>> x[keylist]
1.2345
Also, the isinstance(key, (tuple, list)) condition means a tuple can't be used as a key.
You can certainly write accessors for such a nested dictionary:
def get(d,l):
return get(d[l[0]],l[1:]) if l else d
def set(d,l,v):
while len(l)>1:
d=d[l.pop(0)]
l,=l # verify list length of 1
d[l]=v
(Neither of these is efficient for long lists; faster versions would use a variable index rather than [1:] or pop(0).)
As for other approaches, there’s not nearly enough here to go on for picking one.
How do I add a key to an existing dictionary? It doesn't have an .add() method.
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.
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
data.update(dict(a=1))
# OR
data.update(a=1)
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))
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.
"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))
0.0020880699157714844
>>> min(timeit.repeat(g, number=number))
0.005071878433227539
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.
dictionary[key] = value
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)
Output:
{'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.
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.
You can create one:
class myDict(dict):
def __init__(self):
self = dict()
def add(self, key, value):
self[key] = value
## example
myd = myDict()
myd.add('apples',6)
myd.add('bananas',3)
print(myd)
Gives:
>>>
{'apples': 6, 'bananas': 3}
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 https://stackoverflow.com/q/2255878
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' )
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}
and
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
or
params = {'a': 1, 'b': 2}
new_params = params.copy()
new_params.update({'c': 3})
Reference: What does `**` mean in the expression `dict(d1, **d2)`?
There is also the strangely named, oddly behaved, and yet still handy dict.setdefault().
This
value = my_dict.setdefault(key, default)
basically just does this:
try:
value = my_dict[key]
except KeyError: # key not found
value = my_dict[key] = default
E.g.,
>>> 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
This question has already been answered ad nauseam, but since my
comment
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.
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.
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.
First to check whether the key already exists:
a={1:2,3:4}
a.get(1)
2
a.get(5)
None
Then you can add the new key and value.
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"
myd.add('apples',6)
print('\n', myd)
myd.add('bananas',3)
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)
print(myd)
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.
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
d.update(dict('mynewkey'='mynewvalue'))
# OR
d.update('mynewkey'='mynewvalue')
print(d) # {'key': 'value', 'mynewkey': 'mynewvalue'}
# To add/update multiple keys simultaneously, use d.update():
x = {3:4, 5:6, 7:8}
d.update(x)
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))
dico["new key"] = "value"
Just wrote some nasty code that iterates over a dict or a list in Python. I have a feeling this was not the best way to go about it.
The problem is that in order to iterate over a dict, this is the convention:
for key in dict_object:
dict_object[key] = 1
But modifying the object properties by key does not work if the same thing is done on a list:
# Throws an error because the value of key is the property value, not
# the list index:
for key in list_object:
list_object[key] = 1
The way I solved this problem was to write this nasty code:
if isinstance(obj, dict):
for key in obj:
do_loop_contents(obj, key)
elif isinstance(obj, list):
for i in xrange(0, len(obj)):
do_loop_contents(obj, i)
def do_loop_contents(obj, key):
obj[key] = 1
Is there a better way to do this?
Thanks!
I've never needed to do this, ever. But if I did, I'd probably do something like this:
seq_iter = x if isinstance(x, dict) else xrange(len(x))
For example, in function form:
>>> def seq_iter(obj):
... return obj if isinstance(obj, dict) else xrange(len(obj))
...
>>> x = [1,2,3]
>>> for i in seq_iter(x):
... x[i] = 99
...
>>> x
[99, 99, 99]
>>>
>>> x = {1: 2, 2:3, 3:4}
>>> for i in seq_iter(x):
... x[i] = 99
...
>>> x
{1: 99, 2: 99, 3: 99}
This is the correct way, but if for some reason you need to treat these two objects the same way, you can create an iterable that will return indexes / keys no matter what:
def common_iterable(obj):
if isinstance(obj, dict):
return obj
else:
return (index for index, value in enumerate(obj))
Which will behave in the way you wanted:
>>> d = {'a': 10, 'b': 20}
>>> l = [1,2,3,4]
>>> for index in common_iterable(d):
d[index] = 0
>>> d
{'a': 0, 'b': 0}
>>> for index in common_iterable(l):
l[index] = 0
>>> l
[0, 0, 0, 0]
Or probably more efficiently, using a generator:
def common_iterable(obj):
if isinstance(obj, dict):
for key in obj:
yield key
else:
for index, value in enumerate(obj):
yield index
To be pythonic and ducktype-y, and also to follow "ask for forgiveness not permission", you could do something like:
try:
iterator = obj.iteritems()
except AttributeError:
iterator = enumerate(obj)
for reference, value in iterator:
do_loop_contents(obj, reference)
Though if all you need is the key/index:
try:
references = obj.keys()
except AttributeError:
references = range(len(obj))
for reference in references:
do_loop_contents(obj, reference)
Or as a function:
def reference_and_value_iterator(iterable):
try:
return iterable.iteritems()
except AttributeError:
return enumerate(iterable)
for reference, value in reference_and_value_iterator(obj):
do_loop_contents(obj, reference)
Or for just the references:
def references(iterable):
try:
return iterable.keys()
except AttributeError:
return range(len(iterable))
for reference in references(obj):
do_loop_contents(obj, reference)
test_list = [2, 3, 4]
for i, entry in enumerate(test_list):
test_list[i] = entry * 2
print(test_list) # Gives: [4, 6, 8]
But you probably want a list comprehension:
test_list = [2, 3, 4]
test_list = [entry * 2 for entry in test_list]
print(test_list) # Gives: [4, 6, 8]
You probably just want to have a different code depending on if the object you are trying to change is a dict or a list.
if type(object)==type([]):
for key in range(len(object)):
object[key]=1
elif type(object)==type({}): #use 'else' if you know that object will be a dict if not a list
for key in object:
object[key]=1
I stumbled upon this post while searching for a better one, here's how I did it.
for row in [dict_or_list] if not type(dict_or_list) is list else dict_or_list:
for i,v in row.items():
print(i,v)
How do I add a key to an existing dictionary? It doesn't have an .add() method.
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.
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
data.update(dict(a=1))
# OR
data.update(a=1)
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))
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.
"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))
0.0020880699157714844
>>> min(timeit.repeat(g, number=number))
0.005071878433227539
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.
dictionary[key] = value
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)
Output:
{'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.
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.
You can create one:
class myDict(dict):
def __init__(self):
self = dict()
def add(self, key, value):
self[key] = value
## example
myd = myDict()
myd.add('apples',6)
myd.add('bananas',3)
print(myd)
Gives:
>>>
{'apples': 6, 'bananas': 3}
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 https://stackoverflow.com/q/2255878
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' )
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}
and
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
or
params = {'a': 1, 'b': 2}
new_params = params.copy()
new_params.update({'c': 3})
Reference: What does `**` mean in the expression `dict(d1, **d2)`?
There is also the strangely named, oddly behaved, and yet still handy dict.setdefault().
This
value = my_dict.setdefault(key, default)
basically just does this:
try:
value = my_dict[key]
except KeyError: # key not found
value = my_dict[key] = default
E.g.,
>>> 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
This question has already been answered ad nauseam, but since my
comment
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.
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.
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.
First to check whether the key already exists:
a={1:2,3:4}
a.get(1)
2
a.get(5)
None
Then you can add the new key and value.
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"
myd.add('apples',6)
print('\n', myd)
myd.add('bananas',3)
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)
print(myd)
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.
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
d.update(dict('mynewkey'='mynewvalue'))
# OR
d.update('mynewkey'='mynewvalue')
print(d) # {'key': 'value', 'mynewkey': 'mynewvalue'}
# To add/update multiple keys simultaneously, use d.update():
x = {3:4, 5:6, 7:8}
d.update(x)
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))
dico["new key"] = "value"