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I am iterating through a database column
I need to create a dictionary that updates certain values of certain keys if a criterion is met.
For example
The first iteration is: 'apples'
The dictionary should be {'apples': 1}
The second iteration is: 'peers'
The dictionary should be {'apples': 1, 'peers': 1}
The third iteration is: 'apples'
The dictionary should be {'apples': 2, 'peers': 1}
I apologise for the basic explanation. Its the best way ( I think ) to communicate what I want, because I don't know how to code this.
I need this to be in a dictionary because this operation is deep into a nested for loop structure
THE GOAL:
Is to get the iteration that appears most
DESIRED OUTCOME:
mostListed = 'apples'
I am new to python, if I am missing something obvious I am very open to learning
Using Counter() from collections:
>>> from collections import Counter
>>> l = ["apples", "pears", "apples"]
>>> Counter(l)
Counter({'apples': 2, 'pears': 1})
Making it work for your case goes for example like:
from collections import Counter
list_ = []
for item in ["first", "second", "third"]:
input_value = input(f"{item} iteration: ")
list_.append(input_value)
count = Counter(list_)
print(count) # output: Counter({'apples': 2, 'pears': 1})
print(count.most_common(1)) # output: [('apples', 2)]
Without defaultdict
You can use the following code:
d = {}
for iteration in ['first', 'second', 'third']:
value = input(f'The {iteration} iteration is:')
if value in d:
d[value] += 1
else:
d[value] = 1
print(d)
Output:
The first iteration is:apples
The second iteration is:peers
The third iteration is:apples
{'apples': 2, 'peers': 1}
Using defaultdict
You can create a defaultdict which default value is 0 as follows:
from _collections import defaultdict
d = defaultdict(lambda: 0)
for iteration in ['first', 'second', 'third']:
value = input(f'The {iteration} iteration is:')
d[value] += 1
print(dict(d))
Output
The first iteration is:apples
The second iteration is:peers
The third iteration is:apples
{'apples': 2, 'peers': 1}
Adding this to the already numerous answers for its clarity
from collections import Counter
values = ['apples', 'peers', 'apples']
Counter(values).most_common(1)
>>> [('apples', 2)]
Here is an example:
my_list = ['apples', 'apples', 'peers', 'apples', 'peers']
new_dict = {}
for i in my_list:
if i in new_dict:
new_dict[i] += 1
else:
new_dict[i] = 1
print(new_dict)
You can update the values of a key in a dictionary by doing
if 'apples' in dict:
dict['apples'] += 1
else:
dict['apples'] = 1
and you can find the key with the maximum value by something like this:
most_listed = max(dict, key=lambda k: dict[k])
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"
How do I remove duplicates and combine multiple lists into one like so:
function([["hello","me.txt"],["good","me.txt"],["good","money.txt"], ["rep", "money.txt"]]) should return exactly:
[["good", ["me.txt", "money.txt"]], ["hello", ["me.txt"]], ["rep", ["money.txt"]]]
The easiest one would be using defaultdict .
>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> for i,j in l:
d[i].append(j) #append value to the key
>>> d
=> defaultdict(<class 'list'>, {'hello': ['me.txt'], 'good': ['me.txt', 'money.txt'],
'rep': ['money.txt']})
#to get it in a list
>>> out = [ [key,d[key]] for key in d]
>>> out
=> [['hello', ['me.txt']], ['good', ['me.txt', 'money.txt']], ['rep', ['money.txt']]]
#driver values :
IN : l = [["hello","me.txt"],["good","me.txt"],["good","money.txt"], ["rep", "money.txt"]]
Try This ( no library needed ):
your_input_data = [ ["hello","me.txt"], ["good","me.txt"], ["good","me.txt"], ["good","money.txt"], ["rep", "money.txt"] ]
my_dict = {}
for box in your_input_data:
if box[0] in my_dict:
buffer_items = []
for items in box[1:]:
if items not in my_dict[box[0]]:
buffer_items.append(items)
remove_dup = list(set(buffer_items + my_dict[box[0]]))
my_dict[box[0]] = remove_dup
else:
buffer_items = []
for items in box[1:]:
buffer_items.append(items)
remove_dup = list(set(buffer_items))
my_dict[box[0]] = remove_dup
last_point = [[keys, values] for keys, values in my_dict.items()]
print(last_point)
Good Luck ...
You can do it with traditional dictionaries too.
In [30]: l1 = [["hello","me.txt"],["good","me.txt"],["good","money.txt"], ["rep", "money.txt"]]
In [31]: for i, j in l1:
...: if i not in d2:
...: d2[i] = j
...: else:
...: val = d2[i]
...: d2[i] = [val, j]
...:
In [32]: d2
Out[32]: {'good': ['me.txt', 'money.txt'], 'hello': 'me.txt', 'rep': 'money.txt'}
In [33]: out = [ [key,d1[key]] for key in d1]
In [34]: out
Out[34]:
[['rep', ['money.txt']],
['hello', ['me.txt']],
['good', ['me.txt', 'money.txt']]]
Let's first understand the actual problem :
Example Hint :
For these types of list problems there is a pattern :
So suppose you have a list :
a=[(2006,1),(2007,4),(2008,9),(2006,5)]
And you want to convert this to a dict as the first element of the tuple as key and second element of the tuple. something like :
{2008: [9], 2006: [5], 2007: [4]}
But there is a catch you also want that those keys which have different values but keys are same like (2006,1) and (2006,5) keys are same but values are different. you want that those values append with only one key so expected output :
{2008: [9], 2006: [1, 5], 2007: [4]}
for this type of problem we do something like this:
first create a new dict then we follow this pattern:
if item[0] not in new_dict:
new_dict[item[0]]=[item[1]]
else:
new_dict[item[0]].append(item[1])
So we first check if key is in new dict and if it already then add the value of duplicate key to its value:
full code:
a=[(2006,1),(2007,4),(2008,9),(2006,5)]
new_dict={}
for item in a:
if item[0] not in new_dict:
new_dict[item[0]]=[item[1]]
else:
new_dict[item[0]].append(item[1])
print(new_dict)
Your actual problem solution :
list_1=[["hello","me.txt"],["good","me.txt"],["good","money.txt"], ["rep", "money.txt"]]
no_dublicates={}
for item in list_1:
if item[0] not in no_dublicates:
no_dublicates[item[0]]=["".join(item[1:])]
else:
no_dublicates[item[0]].extend(item[1:])
list_result=[]
for key,value in no_dublicates.items():
list_result.append([key,value])
print(list_result)
output:
[['hello', ['me.txt']], ['rep', ['money.txt']], ['good', ['me.txt', 'money.txt']]]
yourList=[["hello","me.txt"],["good","me.txt"],["good","money.txt"], ["rep", "money.txt"]]
expectedList=[["good", ["me.txt", "money.txt"]], ["hello", ["me.txt"]], ["rep", ["money.txt"]]]
def getall(allsec, listKey, uniqlist):
if listKey not in uniqlist:
uniqlist.append(listKey)
return [listKey, [x[1] for x in allsec if x[0] == listKey]]
uniqlist=[]
result=sorted(list(filter(lambda x:x!=None, [getall(yourList,elem[0],uniqlist) for elem in yourList])))
print(result)
hope this helps
This can easily be solved using dict and sets.
def combine_duplicates(given_list):
data = {}
for element_1, element_2 in given_list:
data[element_1] = data.get(element_1, set()).add(element_2)
return [[k, list(v)] for k, v in data.items()]
Using Python to create a function that gives you the exact required output can be done as follows:
from collections import defaultdict
def function(data):
entries = defaultdict(list)
for k, v in data:
entries[k].append(v)
return sorted([k, v] for k, v in entries.items())
print(function([["hello","me.txt"],["good","me.txt"],["good","money.txt"], ["rep", "money.txt"]]))
The output is sorted before being returned as per your requirement. This would display the return from the function as:
[['good', ['me.txt', 'money.txt']], ['hello', ['me.txt']], ['rep', ['money.txt']]]
It also ensures that the keys are sorted. A dictionary is used to deal with the removal of duplicates (as keys need to be unique).
A defaultdict() is used to simplify the building of lists within the dictionary. The alternative would be to try and append a new value to an existing key, and if there is a KeyError exception, then add the new key instead as follows:
def function(data):
entries = {}
for k, v in data:
try:
entries[k].append(v)
except KeyError as e:
entries[k] = [v]
return sorted([k, v] for k, v in entries.items())
Create a empty array push the index 0 from childs arrays and join to convert all values to a string separate by space .
var your_input_data = [ ["hello","hi", "jel"], ["good"], ["good2","lo"], ["good3","lt","ahhahah"], ["rep", "nice","gr8", "job"] ];
var myprint = []
for(var i in your_input_data){
myprint.push(your_input_data[i][0]);
}
console.log(myprint.join(' '))
In python if I have a dictionary which has a single key value pair and if I don't know what the key might be, how can I get the value?
(and if I have a dict with more than 1 key, value pair, how can I return any one of the values without knowing any of the keys?)
You just have to use dict.values().
This will return a list containing all the values of your dictionary, without having to specify any key.
You may also be interested in:
.keys(): return a list containing the keys
.items(): return a list of tuples (key, value)
Note that in Python 3, returned value is not actually proper list but view object.
Other solution, using popitem and unpacking:
d = {"unknow_key": "value"}
_, v = d.popitem()
assert v == "value"
Further to Delgan's excellent answer, here is an example for Python 3 that demonstrates how to use the view object:
In Python 3 you can print the values, without knowing/using the keys, thus:
for item in my_dict:
print( list( item.values() )[0] )
Example:
cars = {'Toyota':['Camry','Turcel','Tundra','Tacoma'],'Ford':['Mustang','Capri','OrRepairDaily'],'Chev':['Malibu','Corvette']}
vals = list( cars.values() )
keyz = list( cars.keys() )
cnt = 0
for val in vals:
print('[_' + keyz[cnt] + '_]')
if len(val)>1:
for part in val:
print(part)
else:
print( val[0] )
cnt += 1
OUTPUT:
[_Toyota_]
Camry
Turcel
Tundra
Tacoma
[_Ford_]
Mustang
Capri
OrRepairDaily
[_Chev_]
Malibu
Corvette
That Py3 docs reference again:
https://docs.python.org/3.5/library/stdtypes.html#dict-views
Two more ways:
>>> d = {'k': 'v'}
>>> next(iter(d.values()))
'v'
>>> v, = d.values()
>>> v
'v'
One more way: looping with for/in through a dictionary we get the key(s) of the key-value pair(s), and with that, we get the value of the value.
>>>my_dict = {'a' : 25}
>>>for key in my_dict:
print(my_dict[key])
25
>>> my_other_dict = {'b': 33, 'c': 44}
>>> for key in my_other_dict:
print(my_other_dict[key])
33
44
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"