In ECMAScript6 you can use the spread operator to destructure an object like this
const {a, ...rest} = obj;
Which does a shallow copy of obj to rest without attribute a.
Is there a clean way of doing the same in python?
Python dict literals can use the unpacking operator, like { **obj, 'a': 3 } to make a shallow copy overriding some particular properties, but there is no special syntax to omit a property from the unpacking. Also, while python can destructure sequences and nested sequences, you cannot use a dictionary pattern on the left-hand-side of an destructuring assignment.
However, it is possible to do this a bit more verbosely using a dict comprehension:
rest = { k: v for k, v in obj.items() if k != 'a' }
But for readability, two lines is probably better:
rest = dict(obj)
rest.pop('a', None)
Or if you expect the key 'a' to always appear in the original dictionary, and you'd like a KeyError when it's missing, you can use del:
rest = dict(obj)
del rest['a']
I wouldn't call it "clean", but we can (ab)use the function keyword parameter mechanism:
>>> obj = {'b': 2, 'a': 1, 'c': 3}
>>> (lambda a=None, **rest: rest)(**obj)
{'b': 2, 'c': 3}
Or if the goal was to assign to two variables a and rest (I'm not sure since I don't do ECMAScript):
obj = {'b': 2, 'a': 1, 'c': 3}
a, rest = (lambda a=None, **rest: (a, rest))(**obj)
print('a: ', a)
print('rest:', rest)
Output:
a: 1
rest: {'b': 2, 'c': 3}
An possible answer that works, but may not be the prettiest is:
new_dict= rest.copy() # create copy to delete keys
try:
del new_dict["a"]
except KeyError:
print("The key 'a' was not found.")
I was looking for a way to "unpack" a dictionary in a generic way and found a relevant question (and answers) which explained various techniques (TL;DR: it is not too elegant).
That question, however, addresses the case where the keys of the dict are not known, the OP anted to have them added to the local namespace automatically.
My problem is possibly simpler: I get a dict from a function and would like to dissecate it on the fly, knowing the keys I will need (I may not need all of them every time). Right now I can only do
def myfunc():
return {'a': 1, 'b': 2, 'c': 3}
x = myfunc()
a = x['a']
my_b_so_that_the_name_differs_from_the_key = x['b']
# I do not need c this time
while I was looking for the equivalent of
def myotherfunc():
return 1, 2
a, b = myotherfunc()
but for a dict (which is what is returned by my function). I do not want to use the latter solution for several reasons, one of them being that it is not obvious which variable corresponds to which returned element (the first solution has at least the merit of being readable).
Is such operation available?
If you really must, you can use an operator.itemgetter() object to extract values for multiple keys as a tuple:
from operator import itemgetter
a, b = itemgetter('a', 'b')(myfunc())
This is still not pretty; I'd prefer the explicit and readable separate lines where you first assign the return value, then extract those values.
Demo:
>>> from operator import itemgetter
>>> def myfunc():
... return {'a': 1, 'b': 2, 'c': 3}
...
>>> itemgetter('a', 'b')(myfunc())
(1, 2)
>>> a, b = itemgetter('a', 'b')(myfunc())
>>> a
1
>>> b
2
You could also use map:
def myfunc():
return {'a': 1, 'b': 2, 'c': 3}
a,b = map(myfunc().get,["a","b"])
print(a,b)
In addition to the operator.itemgetter() method, you can also write your own myotherfunc(). It takes list of the required keys as an argument and returns a tuple of their corresponding value.
def myotherfunc(keys_list):
reference_dict = myfunc()
return tuple(reference_dict[key] for key in keys_list)
>>> a,b = myotherfunc(['a','b'])
>>> a
1
>>> b
2
>>> a,c = myotherfunc(['a','c'])
>>> a
1
>>> c
3
This question already has answers here:
How do I merge two dictionaries in a single expression in Python?
(43 answers)
Closed 10 years ago.
How do you calculate the union of two dict objects in Python, where a (key, value) pair is present in the result iff key is in either dict (unless there are duplicates)?
For example, the union of {'a' : 0, 'b' : 1} and {'c' : 2} is {'a' : 0, 'b' : 1, 'c' : 2}.
Preferably you can do this without modifying either input dict. Example of where this is useful: Get a dict of all variables currently in scope and their values
This question provides an idiom. You use one of the dicts as keyword arguments to the dict() constructor:
dict(y, **x)
Duplicates are resolved in favor of the value in x; for example
dict({'a' : 'y[a]'}, **{'a', 'x[a]'}) == {'a' : 'x[a]'}
You can also use update method of dict like
a = {'a' : 0, 'b' : 1}
b = {'c' : 2}
a.update(b)
print a
For a static dictionary, combining snapshots of other dicts:
As of Python 3.9, the binary "or" operator | has been defined to concatenate dictionaries. (A new, concrete dictionary is eagerly created):
>>> a = {"a":1}
>>> b = {"b":2}
>>> a|b
{'a': 1, 'b': 2}
Conversely, the |= augmented assignment has been implemented to mean the same as calling the update method:
>>> a = {"a":1}
>>> a |= {"b": 2}
>>> a
{'a': 1, 'b': 2}
For details, check PEP-584
Prior to Python 3.9, the simpler way to create a new dictionary is to create a new dictionary using the "star expansion" to add teh contents of each subctionary in place:
c = {**a, **b}
For dynamic dictionary combination, working as "view" to combined, live dicts:
If you need both dicts to remain independent, and updatable, you can create a single object that queries both dictionaries in its __getitem__ method (and implement get, __contains__ and other mapping method as you need them).
A minimalist example could be like this:
class UDict(object):
def __init__(self, d1, d2):
self.d1, self.d2 = d1, d2
def __getitem__(self, item):
if item in self.d1:
return self.d1[item]
return self.d2[item]
And it works:
>>> a = UDict({1:1}, {2:2})
>>> a[2]
2
>>> a[1]
1
>>> a[3]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 7, in __getitem__
KeyError: 3
>>>
NB: If one wants to lazily maintain a Union "view" of two
or more dictionaries, check collections.ChainMap in the standard library - as it has all dictionary methods and cover corner cases not
contemplated in the example above.
Two dictionaries
def union2(dict1, dict2):
return dict(list(dict1.items()) + list(dict2.items()))
n dictionaries
def union(*dicts):
return dict(itertools.chain.from_iterable(dct.items() for dct in dicts))
I've read the examples in python docs, but still can't figure out what this method means. Can somebody help? Here are two examples from the python docs
>>> from collections import defaultdict
>>> s = 'mississippi'
>>> d = defaultdict(int)
>>> for k in s:
... d[k] += 1
...
>>> d.items()
dict_items([('m', 1), ('i', 4), ('s', 4), ('p', 2)])
and
>>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
>>> d = defaultdict(list)
>>> for k, v in s:
... d[k].append(v)
...
>>> d.items()
[('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]
the parameters int and list are for what?
Usually, a Python dictionary throws a KeyError if you try to get an item with a key that is not currently in the dictionary. The defaultdict in contrast will simply create any items that you try to access (provided of course they do not exist yet). To create such a "default" item, it calls the function object that you pass to the constructor (more precisely, it's an arbitrary "callable" object, which includes function and type objects). For the first example, default items are created using int(), which will return the integer object 0. For the second example, default items are created using list(), which returns a new empty list object.
defaultdict means that if a key is not found in the dictionary, then instead of a KeyError being thrown, a new entry is created. The type of this new entry is given by the argument of defaultdict.
For example:
somedict = {}
print(somedict[3]) # KeyError
someddict = defaultdict(int)
print(someddict[3]) # print int(), thus 0
defaultdict
"The standard dictionary includes the method setdefault() for retrieving a value and establishing a default if the value does not exist. By contrast, defaultdict lets the caller specify the default(value to be returned) up front when the container is initialized."
as defined by Doug Hellmann in The Python Standard Library by Example
How to use defaultdict
Import defaultdict
>>> from collections import defaultdict
Initialize defaultdict
Initialize it by passing
callable as its first argument(mandatory)
>>> d_int = defaultdict(int)
>>> d_list = defaultdict(list)
>>> def foo():
... return 'default value'
...
>>> d_foo = defaultdict(foo)
>>> d_int
defaultdict(<type 'int'>, {})
>>> d_list
defaultdict(<type 'list'>, {})
>>> d_foo
defaultdict(<function foo at 0x7f34a0a69578>, {})
**kwargs as its second argument(optional)
>>> d_int = defaultdict(int, a=10, b=12, c=13)
>>> d_int
defaultdict(<type 'int'>, {'a': 10, 'c': 13, 'b': 12})
or
>>> kwargs = {'a':10,'b':12,'c':13}
>>> d_int = defaultdict(int, **kwargs)
>>> d_int
defaultdict(<type 'int'>, {'a': 10, 'c': 13, 'b': 12})
How does it works
As is a child class of standard dictionary, it can perform all the same functions.
But in case of passing an unknown key it returns the default value instead of error. For ex:
>>> d_int['a']
10
>>> d_int['d']
0
>>> d_int
defaultdict(<type 'int'>, {'a': 10, 'c': 13, 'b': 12, 'd': 0})
In case you want to change default value overwrite default_factory:
>>> d_int.default_factory = lambda: 1
>>> d_int['e']
1
>>> d_int
defaultdict(<function <lambda> at 0x7f34a0a91578>, {'a': 10, 'c': 13, 'b': 12, 'e': 1, 'd': 0})
or
>>> def foo():
... return 2
>>> d_int.default_factory = foo
>>> d_int['f']
2
>>> d_int
defaultdict(<function foo at 0x7f34a0a0a140>, {'a': 10, 'c': 13, 'b': 12, 'e': 1, 'd': 0, 'f': 2})
Examples in the Question
Example 1
As int has been passed as default_factory, any unknown key will return 0 by default.
Now as the string is passed in the loop, it will increase the count of those alphabets in d.
>>> s = 'mississippi'
>>> d = defaultdict(int)
>>> d.default_factory
<type 'int'>
>>> for k in s:
... d[k] += 1
>>> d.items()
[('i', 4), ('p', 2), ('s', 4), ('m', 1)]
>>> d
defaultdict(<type 'int'>, {'i': 4, 'p': 2, 's': 4, 'm': 1})
Example 2
As a list has been passed as default_factory, any unknown(non-existent) key will return [ ](ie. list) by default.
Now as the list of tuples is passed in the loop, it will append the value in the d[color]
>>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
>>> d = defaultdict(list)
>>> d.default_factory
<type 'list'>
>>> for k, v in s:
... d[k].append(v)
>>> d.items()
[('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]
>>> d
defaultdict(<type 'list'>, {'blue': [2, 4], 'red': [1], 'yellow': [1, 3]})
Dictionaries are a convenient way to store data for later retrieval by name (key). Keys must be unique, immutable objects, and are typically strings. The values in a dictionary can be anything. For many applications, the values are simple types such as integers and strings.
It gets more interesting when the values in a dictionary are collections (lists, dicts, etc.) In this case, the value (an empty list or dict) must be initialized the first time a given key is used. While this is relatively easy to do manually, the defaultdict type automates and simplifies these kinds of operations.
A defaultdict works exactly like a normal dict, but it is initialized with a function (“default factory”) that takes no arguments and provides the default value for a nonexistent key.
A defaultdict will never raise a KeyError. Any key that does not exist gets the value returned by the default factory.
from collections import defaultdict
ice_cream = defaultdict(lambda: 'Vanilla')
ice_cream['Sarah'] = 'Chunky Monkey'
ice_cream['Abdul'] = 'Butter Pecan'
print(ice_cream['Sarah'])
>>>Chunky Monkey
print(ice_cream['Joe'])
>>>Vanilla
Here is another example on How using defaultdict, we can reduce complexity
from collections import defaultdict
# Time complexity O(n^2)
def delete_nth_naive(array, n):
ans = []
for num in array:
if ans.count(num) < n:
ans.append(num)
return ans
# Time Complexity O(n), using hash tables.
def delete_nth(array,n):
result = []
counts = defaultdict(int)
for i in array:
if counts[i] < n:
result.append(i)
counts[i] += 1
return result
x = [1,2,3,1,2,1,2,3]
print(delete_nth(x, n=2))
print(delete_nth_naive(x, n=2))
In conclusion, whenever you need a dictionary, and each element’s value should start with a default value, use a defaultdict.
There is a great explanation of defaultdicts here: http://ludovf.net/blog/python-collections-defaultdict/
Basically, the parameters int and list are functions that you pass. Remember that Python accepts function names as arguments. int returns 0 by default and list returns an empty list when called with parentheses.
In normal dictionaries, if in your example I try calling d[a], I will get an error (KeyError), since only keys m, s, i and p exist and key a has not been initialized. But in a defaultdict, it takes a function name as an argument, when you try to use a key that has not been initialized, it simply calls the function you passed in and assigns its return value as the value of the new key.
The behavior of defaultdict can be easily mimicked using dict.setdefault instead of d[key] in every call.
In other words, the code:
from collections import defaultdict
d = defaultdict(list)
print(d['key']) # empty list []
d['key'].append(1) # adding constant 1 to the list
print(d['key']) # list containing the constant [1]
is equivalent to:
d = dict()
print(d.setdefault('key', list())) # empty list []
d.setdefault('key', list()).append(1) # adding constant 1 to the list
print(d.setdefault('key', list())) # list containing the constant [1]
The only difference is that, using defaultdict, the list constructor is called only once, and using dict.setdefault the list constructor is called more often (but the code may be rewriten to avoid this, if really needed).
Some may argue there is a performance consideration, but this topic is a minefield. This post shows there isn't a big performance gain in using defaultdict, for example.
IMO, defaultdict is a collection that adds more confusion than benefits to the code. Useless for me, but others may think different.
Since the question is about "how it works", some readers may want to see more nuts and bolts. Specifically, the method in question is the __missing__(key) method. See: https://docs.python.org/2/library/collections.html#defaultdict-objects .
More concretely, this answer shows how to make use of __missing__(key) in a practical way:
https://stackoverflow.com/a/17956989/1593924
To clarify what 'callable' means, here's an interactive session (from 2.7.6 but should work in v3 too):
>>> x = int
>>> x
<type 'int'>
>>> y = int(5)
>>> y
5
>>> z = x(5)
>>> z
5
>>> from collections import defaultdict
>>> dd = defaultdict(int)
>>> dd
defaultdict(<type 'int'>, {})
>>> dd = defaultdict(x)
>>> dd
defaultdict(<type 'int'>, {})
>>> dd['a']
0
>>> dd
defaultdict(<type 'int'>, {'a': 0})
That was the most typical use of defaultdict (except for the pointless use of the x variable). You can do the same thing with 0 as the explicit default value, but not with a simple value:
>>> dd2 = defaultdict(0)
Traceback (most recent call last):
File "<pyshell#7>", line 1, in <module>
dd2 = defaultdict(0)
TypeError: first argument must be callable
Instead, the following works because it passes in a simple function (it creates on the fly a nameless function which takes no arguments and always returns 0):
>>> dd2 = defaultdict(lambda: 0)
>>> dd2
defaultdict(<function <lambda> at 0x02C4C130>, {})
>>> dd2['a']
0
>>> dd2
defaultdict(<function <lambda> at 0x02C4C130>, {'a': 0})
>>>
And with a different default value:
>>> dd3 = defaultdict(lambda: 1)
>>> dd3
defaultdict(<function <lambda> at 0x02C4C170>, {})
>>> dd3['a']
1
>>> dd3
defaultdict(<function <lambda> at 0x02C4C170>, {'a': 1})
>>>
My own 2¢: you can also subclass defaultdict:
class MyDict(defaultdict):
def __missing__(self, key):
value = [None, None]
self[key] = value
return value
This could come in handy for very complex cases.
Well, defaultdict can also raise keyerror in the following case:
from collections import defaultdict
d = defaultdict()
print(d[3]) #raises keyerror
Always remember to give argument to the defaultdict like
d = defaultdict(int)
The defaultdict tool is a container in the collections class of Python. It's similar to the usual dictionary (dict) container, but it has one difference: The value fields' data type is specified upon initialization.
For example:
from collections import defaultdict
d = defaultdict(list)
d['python'].append("awesome")
d['something-else'].append("not relevant")
d['python'].append("language")
for i in d.items():
print i
This prints:
('python', ['awesome', 'language'])
('something-else', ['not relevant'])
In short:
defaultdict(int) - the argument int indicates that the values will be int type.
defaultdict(list) - the argument list indicates that the values will be list type.
I think its best used in place of a switch case statement. Imagine if we have a switch case statement as below:
option = 1
switch(option) {
case 1: print '1st option'
case 2: print '2nd option'
case 3: print '3rd option'
default: return 'No such option'
}
There is no switch case statements available in python. We can achieve the same by using defaultdict.
from collections import defaultdict
def default_value(): return "Default Value"
dd = defaultdict(default_value)
dd[1] = '1st option'
dd[2] = '2nd option'
dd[3] = '3rd option'
print(dd[4])
print(dd[5])
print(dd[3])
It prints:
Default Value
Default Value
3rd option
In the above snippet dd has no keys 4 or 5 and hence it prints out a default value which we have configured in a helper function. This is quite nicer than a raw dictionary where a KeyError is thrown if key is not present. From this it is evident that defaultdict more like a switch case statement where we can avoid a complicated if-elif-elif-else blocks.
One more good example that impressed me a lot from this site is:
>>> from collections import defaultdict
>>> food_list = 'spam spam spam spam spam spam eggs spam'.split()
>>> food_count = defaultdict(int) # default value of int is 0
>>> for food in food_list:
... food_count[food] += 1 # increment element's value by 1
...
defaultdict(<type 'int'>, {'eggs': 1, 'spam': 7})
>>>
If we try to access any items other than eggs and spam we will get a count of 0.
Without defaultdict, you can probably assign new values to unseen keys but you cannot modify it. For example:
import collections
d = collections.defaultdict(int)
for i in range(10):
d[i] += i
print(d)
# Output: defaultdict(<class 'int'>, {0: 0, 1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9})
import collections
d = {}
for i in range(10):
d[i] += i
print(d)
# Output: Traceback (most recent call last): File "python", line 4, in <module> KeyError: 0
The standard dictionary includes the method setdefault() for retrieving a value and establishing a default if the value does not exist. By contrast, defaultdict lets the caller specify the default up front when the container is initialized.
import collections
def default_factory():
return 'default value'
d = collections.defaultdict(default_factory, foo='bar')
print 'd:', d
print 'foo =>', d['foo']
print 'bar =>', d['bar']
This works well as long as it is appropriate for all keys to have the same default. It can be especially useful if the default is a type used for aggregating or accumulating values, such as a list, set, or even int. The standard library documentation includes several examples of using defaultdict this way.
$ python collections_defaultdict.py
d: defaultdict(<function default_factory at 0x100468c80>, {'foo': 'bar'})
foo => bar
bar => default value
#dictinary and defaultdict
normaldictionary=dict()
print(type(normaldictionary))
#print(normaldictionary["keynotexisit"])
#Above normal dictionary give an error as key not present
from collections import defaultdict
defaultdict1=defaultdict()
print(type(defaultdict1))
#print(defaultdict1['keynotexisit'])
######################################
from collections import defaultdict
default2=defaultdict(int)
print(default2['keynotexist'])
https://msatutorpy.medium.com/different-between-dictionary-and-defaultdictionary-cb215f682971
The documentation and the explanation are pretty much self-explanatory:
http://docs.python.org/library/collections.html#collections.defaultdict
The type function(int/str etc.) passed as an argument is used to initialize a default value for any given key where the key is not present in the dict.