The addition of collections.defaultdict in Python 2.5 greatly reduced the need for dict's setdefault method. This question is for our collective education:
What is setdefault still useful for, today in Python 2.6/2.7?
What popular use cases of setdefault were superseded with collections.defaultdict?
You could say defaultdict is useful for settings defaults before filling the dict and setdefault is useful for setting defaults while or after filling the dict.
Probably the most common use case: Grouping items (in unsorted data, else use itertools.groupby)
# really verbose
new = {}
for (key, value) in data:
if key in new:
new[key].append( value )
else:
new[key] = [value]
# easy with setdefault
new = {}
for (key, value) in data:
group = new.setdefault(key, []) # key might exist already
group.append( value )
# even simpler with defaultdict
from collections import defaultdict
new = defaultdict(list)
for (key, value) in data:
new[key].append( value ) # all keys have a default already
Sometimes you want to make sure that specific keys exist after creating a dict. defaultdict doesn't work in this case, because it only creates keys on explicit access. Think you use something HTTP-ish with many headers -- some are optional, but you want defaults for them:
headers = parse_headers( msg ) # parse the message, get a dict
# now add all the optional headers
for headername, defaultvalue in optional_headers:
headers.setdefault( headername, defaultvalue )
I commonly use setdefault for keyword argument dicts, such as in this function:
def notify(self, level, *pargs, **kwargs):
kwargs.setdefault("persist", level >= DANGER)
self.__defcon.set(level, **kwargs)
try:
kwargs.setdefault("name", self.client.player_entity().name)
except pytibia.PlayerEntityNotFound:
pass
return _notify(level, *pargs, **kwargs)
It's great for tweaking arguments in wrappers around functions that take keyword arguments.
defaultdict is great when the default value is static, like a new list, but not so much if it's dynamic.
For example, I need a dictionary to map strings to unique ints. defaultdict(int) will always use 0 for the default value. Likewise, defaultdict(intGen()) always produces 1.
Instead, I used a regular dict:
nextID = intGen()
myDict = {}
for lots of complicated stuff:
#stuff that generates unpredictable, possibly already seen str
strID = myDict.setdefault(myStr, nextID())
Note that dict.get(key, nextID()) is insufficient because I need to be able to refer to these values later as well.
intGen is a tiny class I build that automatically increments an int and returns its value:
class intGen:
def __init__(self):
self.i = 0
def __call__(self):
self.i += 1
return self.i
If someone has a way to do this with defaultdict I'd love to see it.
As most answers state setdefault or defaultdict would let you set a default value when a key doesn't exist. However, I would like to point out a small caveat with regard to the use cases of setdefault. When the Python interpreter executes setdefaultit will always evaluate the second argument to the function even if the key exists in the dictionary. For example:
In: d = {1:5, 2:6}
In: d
Out: {1: 5, 2: 6}
In: d.setdefault(2, 0)
Out: 6
In: d.setdefault(2, print('test'))
test
Out: 6
As you can see, print was also executed even though 2 already existed in the dictionary. This becomes particularly important if you are planning to use setdefault for example for an optimization like memoization. If you add a recursive function call as the second argument to setdefault, you wouldn't get any performance out of it as Python would always be calling the function recursively.
Since memoization was mentioned, a better alternative is to use functools.lru_cache decorator if you consider enhancing a function with memoization. lru_cache handles the caching requirements for a recursive function better.
I use setdefault() when I want a default value in an OrderedDict. There isn't a standard Python collection that does both, but there are ways to implement such a collection.
As Muhammad said, there are situations in which you only sometimes wish to set a default value. A great example of this is a data structure which is first populated, then queried.
Consider a trie. When adding a word, if a subnode is needed but not present, it must be created to extend the trie. When querying for the presence of a word, a missing subnode indicates that the word is not present and it should not be created.
A defaultdict cannot do this. Instead, a regular dict with the get and setdefault methods must be used.
Theoretically speaking, setdefault would still be handy if you sometimes want to set a default and sometimes not. In real life, I haven't come across such a use case.
However, an interesting use case comes up from the standard library (Python 2.6, _threadinglocal.py):
>>> mydata = local()
>>> mydata.__dict__
{'number': 42}
>>> mydata.__dict__.setdefault('widgets', [])
[]
>>> mydata.widgets
[]
I would say that using __dict__.setdefault is a pretty useful case.
Edit: As it happens, this is the only example in the standard library and it is in a comment. So may be it is not enough of a case to justify the existence of setdefault. Still, here is an explanation:
Objects store their attributes in the __dict__ attribute. As it happens, the __dict__ attribute is writeable at any time after the object creation. It is also a dictionary not a defaultdict. It is not sensible for objects in the general case to have __dict__ as a defaultdict because that would make each object having all legal identifiers as attributes. So I can't foresee any change to Python objects getting rid of __dict__.setdefault, apart from deleting it altogether if it was deemed not useful.
I rewrote the accepted answer and facile it for the newbies.
#break it down and understand it intuitively.
new = {}
for (key, value) in data:
if key not in new:
new[key] = [] # this is core of setdefault equals to new.setdefault(key, [])
new[key].append(value)
else:
new[key].append(value)
# easy with setdefault
new = {}
for (key, value) in data:
group = new.setdefault(key, []) # it is new[key] = []
group.append(value)
# even simpler with defaultdict
new = defaultdict(list)
for (key, value) in data:
new[key].append(value) # all keys have a default value of empty list []
Additionally,I categorized the methods as reference:
dict_methods_11 = {
'views':['keys', 'values', 'items'],
'add':['update','setdefault'],
'remove':['pop', 'popitem','clear'],
'retrieve':['get',],
'copy':['copy','fromkeys'],}
One drawback of defaultdict over dict (dict.setdefault) is that a defaultdict object creates a new item EVERYTIME non existing key is given (eg with ==, print). Also the defaultdict class is generally way less common then the dict class, its more difficult to serialize it IME.
P.S. IMO functions|methods not meant to mutate an object, should not mutate an object.
Here are some examples of setdefault to show its usefulness:
"""
d = {}
# To add a key->value pair, do the following:
d.setdefault(key, []).append(value)
# To retrieve a list of the values for a key
list_of_values = d[key]
# To remove a key->value pair is still easy, if
# you don't mind leaving empty lists behind when
# the last value for a given key is removed:
d[key].remove(value)
# Despite the empty lists, it's still possible to
# test for the existance of values easily:
if d.has_key(key) and d[key]:
pass # d has some values for key
# Note: Each value can exist multiple times!
"""
e = {}
print e
e.setdefault('Cars', []).append('Toyota')
print e
e.setdefault('Motorcycles', []).append('Yamaha')
print e
e.setdefault('Airplanes', []).append('Boeing')
print e
e.setdefault('Cars', []).append('Honda')
print e
e.setdefault('Cars', []).append('BMW')
print e
e.setdefault('Cars', []).append('Toyota')
print e
# NOTE: now e['Cars'] == ['Toyota', 'Honda', 'BMW', 'Toyota']
e['Cars'].remove('Toyota')
print e
# NOTE: it's still true that ('Toyota' in e['Cars'])
I use setdefault frequently when, get this, setting a default (!!!) in a dictionary; somewhat commonly the os.environ dictionary:
# Set the venv dir if it isn't already overridden:
os.environ.setdefault('VENV_DIR', '/my/default/path')
Less succinctly, this looks like this:
# Set the venv dir if it isn't already overridden:
if 'VENV_DIR' not in os.environ:
os.environ['VENV_DIR'] = '/my/default/path')
It's worth noting that you can also use the resulting variable:
venv_dir = os.environ.setdefault('VENV_DIR', '/my/default/path')
But that's less necessary than it was before defaultdicts existed.
Another use case that I don't think was mentioned above.
Sometimes you keep a cache dict of objects by their id where primary instance is in the cache and you want to set cache when missing.
return self.objects_by_id.setdefault(obj.id, obj)
That's useful when you always want to keep a single instance per distinct id no matter how you obtain an obj each time. For example when object attributes get updated in memory and saving to storage is deferred.
One very important use-case I just stumbled across: dict.setdefault() is great for multi-threaded code when you only want a single canonical object (as opposed to multiple objects that happen to be equal).
For example, the (Int)Flag Enum in Python 3.6.0 has a bug: if multiple threads are competing for a composite (Int)Flag member, there may end up being more than one:
from enum import IntFlag, auto
import threading
class TestFlag(IntFlag):
one = auto()
two = auto()
three = auto()
four = auto()
five = auto()
six = auto()
seven = auto()
eight = auto()
def __eq__(self, other):
return self is other
def __hash__(self):
return hash(self.value)
seen = set()
class cycle_enum(threading.Thread):
def run(self):
for i in range(256):
seen.add(TestFlag(i))
threads = []
for i in range(8):
threads.append(cycle_enum())
for t in threads:
t.start()
for t in threads:
t.join()
len(seen)
# 272 (should be 256)
The solution is to use setdefault() as the last step of saving the computed composite member -- if another has already been saved then it is used instead of the new one, guaranteeing unique Enum members.
In addition to what have been suggested, setdefault might be useful in situations where you don't want to modify a value that has been already set. For example, when you have duplicate numbers and you want to treat them as one group. In this case, if you encounter a repeated duplicate key which has been already set, you won't update the value of that key. You will keep the first encountered value. As if you are iterating/updating the repeated keys once only.
Here's a code example of recording the index for the keys/elements of a sorted list:
nums = [2,2,2,2,2]
d = {}
for idx, num in enumerate(sorted(nums)):
# This will be updated with the value/index of the of the last repeated key
# d[num] = idx # Result (sorted_indices): [4, 4, 4, 4, 4]
# In the case of setdefault, all encountered repeated keys won't update the key.
# However, only the first encountered key's index will be set
d.setdefault(num,idx) # Result (sorted_indices): [0, 0, 0, 0, 0]
sorted_indices = [d[i] for i in nums]
[Edit] Very wrong! The setdefault would always trigger long_computation, Python being eager.
Expanding on Tuttle's answer. For me the best use case is cache mechanism. Instead of:
if x not in memo:
memo[x]=long_computation(x)
return memo[x]
which consumes 3 lines and 2 or 3 lookups, I would happily write :
return memo.setdefault(x, long_computation(x))
I like the answer given here:
http://stupidpythonideas.blogspot.com/2013/08/defaultdict-vs-setdefault.html
In short, the decision (in non-performance-critical apps) should be made on the basis of how you want to handle lookup of empty keys downstream (viz. KeyError versus default value).
The different use case for setdefault() is when you don't want to overwrite the value of an already set key. defaultdict overwrites, while setdefault() does not. For nested dictionaries it is more often the case that you want to set a default only if the key is not set yet, because you don't want to remove the present sub dictionary. This is when you use setdefault().
Example with defaultdict:
>>> from collection import defaultdict()
>>> foo = defaultdict()
>>> foo['a'] = 4
>>> foo['a'] = 2
>>> print(foo)
defaultdict(None, {'a': 2})
setdefault doesn't overwrite:
>>> bar = dict()
>>> bar.setdefault('a', 4)
>>> bar.setdefault('a', 2)
>>> print(bar)
{'a': 4}
Another usecase for setdefault in CPython is that it is atomic in all cases, whereas defaultdict will not be atomic if you use a default value created from a lambda.
cache = {}
def get_user_roles(user_id):
if user_id in cache:
return cache[user_id]['roles']
cache.setdefault(user_id, {'lock': threading.Lock()})
with cache[user_id]['lock']:
roles = query_roles_from_database(user_id)
cache[user_id]['roles'] = roles
If two threads execute cache.setdefault at the same time, only one of them will be able to create the default value.
If instead you used a defaultdict:
cache = defaultdict(lambda: {'lock': threading.Lock()}
This would result in a race condition. In my example above, the first thread could create a default lock, and the second thread could create another default lock, and then each thread could lock its own default lock, instead of the desired outcome of each thread attempting to lock a single lock.
Conceptually, setdefault basically behaves like this (defaultdict also behaves like this if you use an empty list, empty dict, int, or other default value that is not user python code like a lambda):
gil = threading.Lock()
def setdefault(dict, key, value_func):
with gil:
if key not in dict:
return
value = value_func()
dict[key] = value
Conceptually, defaultdict basically behaves like this (only when using python code like a lambda - this is not true if you use an empty list):
gil = threading.Lock()
def __setitem__(dict, key, value_func):
with gil:
if key not in dict:
return
value = value_func()
with gil:
dict[key] = value
If I have a dictionary that is nested, and I pass in a string like "key1.key2.key3" which would translate to:
myDict["key1"]["key2"]["key3"]
What would be an elegant way to be able to have a method where I could pass on that string and it would translate to that key assignment? Something like
myDict.set_nested('key1.key2.key3', someValue)
Using only builtin stuff:
def set(my_dict, key_string, value):
"""Given `foo`, 'key1.key2.key3', 'something', set foo['key1']['key2']['key3'] = 'something'"""
# Start off pointing at the original dictionary that was passed in.
here = my_dict
# Turn the string of key names into a list of strings.
keys = key_string.split(".")
# For every key *before* the last one, we concentrate on navigating through the dictionary.
for key in keys[:-1]:
# Try to find here[key]. If it doesn't exist, create it with an empty dictionary. Then,
# update our `here` pointer to refer to the thing we just found (or created).
here = here.setdefault(key, {})
# Finally, set the final key to the given value
here[keys[-1]] = value
myDict = {}
set(myDict, "key1.key2.key3", "some_value")
assert myDict == {"key1": {"key2": {"key3": "some_value"}}}
This traverses myDict one key at a time, ensuring that each sub-key refers to a nested dictionary.
You could also solve this recursively, but then you risk RecursionError exceptions without any real benefit.
There are a number of existing modules that will already do this, or something very much like it. For example, the jmespath module will resolve jmespath expressions, so given:
>>> mydict={'key1': {'key2': {'key3': 'value'}}}
You can run:
>>> import jmespath
>>> jmespath.search('key1.key2.key3', mydict)
'value'
The jsonpointer module does something similar, although it likes / for a separator instead of ..
Given the number of pre-existing modules I would avoid trying to write your own code to do this.
EDIT: OP's clarification makes it clear that this answer isn't what he's looking for. I'm leaving it up here for people who find it by title.
I implemented a class that did this a while back... it should serve your purposes.
I achieved this by overriding the default getattr/setattr functions for an object.
Check it out! AndroxxTraxxon/cfgutils
This lets you do some code like the following...
from cfgutils import obj
a = obj({
"b": 123,
"c": "apple",
"d": {
"e": "nested dictionary value"
}
})
print(a.d.e)
>>> nested dictionary value
I am trying to create a multi-level dict with variable depth and with list and int type.
Data structure is like below
A
--B1
-----C1=1
-----C2=[1]
--B2=[3]
D
--E
----F
------G=4
In the case of above data structure, the last value can be an int or list.
If the above data structure has the only int then I can be easily achieved by using the below code:
from collections import defaultdict
f = lambda: defaultdict(f)
d = f()
d['A']['B1']['C1'] = 1
But as the last value has both list and int, it becomes a bit problematic for me.
Now we can insert data in a list using two ways.
d['A']['B1']['C2']= [1]
d['A']['B1']['C2'].append([2])
But when I am using only the append method it is causing the error.
Error is:
AttributeError: 'collections.defaultdict' object has no attribute 'append'
so Is there any way to use only the append method for a list?
There's no way you can use your current defaultdict-based structure to make d['A']['B1']['C2'].append(1) work properly if the 'C2' key doesn't already exist, since the data structure can't tell that the unknown key should correspond to a list rather than another layer of dictionary. It doesn't know what method you're going to call on the value it returns, so it can't know it shouldn't return a dictionary (like it did when it first looked up 'A' and 'B').
This isn't an issue for bare integers, since for those you're as assigning directly to a new key (and all the earlier levels are dictionaries). When you're assigning, the data structure isn't creating the value, you are, so you can use any type you want.
Now, if your keys are distinctive in some way, so that given a key like 'C2' you can know for sure that it should correspond to a list, you may have a chance. You can write your own dict subclass, defining a __missing__ method to handle lookups of keys that don't exist yet in your own special way:
def Tree(dict):
def __missing__(self, key):
if key_corresponds_to_list(key): # magic from somewhere
result = self[key] = []
else:
result = self[key] = Tree()
return result
# you might also want a custom __repr__
Here's an example run with a magic key function that makes any even-length key default to a list, while an odd-length key defaults to a dict:
> def key_corresponds_to_list(key):
return len(key) % 2 == 0
> t = Tree()
> t["A"]["B"]["C2"].append(1) # the default value for C2 is a list because it's even length
> t
{'A': {'B': {'C2': [1]}}}
> t["A"]["B"]["C10"]["D"] = 2 # C10's another layer of dict, since it's length is odd
> t
{'A': {'B': {'C10': {'D': 2}, 'C2': [1]}}} # it didn't matter what length D was though
You probably won't actually want to use a global function to control the class like this, I just did that as an example. If you go with this approach, I'd suggest putting the logic directly into the __missing__ method (or maybe passing a function as a parameter, like defaultdict does with its factory function).
This question already has answers here:
Is there a standard class for an infinitely nested defaultdict?
(6 answers)
Closed 9 years ago.
I'm creating a dictionary structure that is several levels deep. I'm trying to do something like the following:
dict = {}
dict['a']['b'] = True
At the moment the above fails because key 'a' does not exist. At the moment I have to check at every level of nesting and manually insert an empty dictionary. Is there some type of syntactic sugar to be able to do something like the above can produce:
{'a': {'b': True}}
Without having to create an empty dictionary at each level of nesting?
As others have said, use defaultdict. This is the idiom I prefer for arbitrarily-deep nesting of dictionaries:
def nested_dict():
return collections.defaultdict(nested_dict)
d = nested_dict()
d[1][2][3] = 'Hello, dictionary!'
print(d[1][2][3]) # Prints Hello, dictionary!
This also makes checking whether an element exists a little nicer, too, since you may no longer need to use get:
if not d[2][3][4][5]:
print('That element is empty!')
This has been edited to use a def rather than a lambda for pep8 compliance. The original lambda form looked like this below, which has the drawback of being called <lambda> everywhere instead of getting a proper function name.
>>> nested_dict = lambda: collections.defaultdict(nested_dict)
>>> d = nested_dict()
>>> d[1][2][3]
defaultdict(<function <lambda> at 0x037E7540>, {})
Use defaultdict.
Python: defaultdict of defaultdict?
Or you can do this, since dict() function can handle **kwargs:
http://docs.python.org/2/library/functions.html#func-dict
print dict(a=dict(b=True))
# {'a': {'b' : True}}
If the depth of your data structure is fixed (that is, you know in advance that you need mydict[a][b][c] but not mydict[a][b][c][d]), you can build a nested defaultdict structure using lambda expressions to create the inner structures:
two_level = defaultdict(dict)
three_level = defaultdict(lambda: defaultdict(dict))
four_level = defaultdict(lamda: defaultdict(lambda: defaultdict(dict)))