I need a dictionary that is automatically filled with a default value for each accessed key that is missing. I've found defaultdict and some other ways to achieve this, but the problem in my case is that I want the default value for each key to be specific to the key itself.
For example, with defaultdict I can achieve something like this:
from collections import defaultdict
d = defaultdict(lambda: 5)
> d[1] = 3
> d[1]
> 3
> d[2]
> 5
But what if I need the default value for each accessed missing key to be for example key + 5? Something like:
from collections import defaultdict
d = defaultdict(lambda key: key + 5) # <-- This does not work as defaultdict expects lambda function to be without any parameters
> d[1] = 3
> d[1]
> 3
> d[2]
> 7 <- Calculated from accessed key + 5 (2+5)
> d[5]
> 10 <- Calculated from accessed key + 5 (5+5)
Is there a clean, builtin way to achieve what I need? I know that I can create a subclass of dict and implement this specific functionality at __getitem__ level, but I want to avoid that if possible.
I couldn't find a solution in other answers, so sorry if it is still a duplicate.
I don't think there is a builtin way of doing this. However, instead of subclassing dict and change getitem, you can subclass defaultdict itself to tell __missing__() to call the default_factory with an argument (the key), rather than without args. It is still a custom class, not a builtin, but it should be quite efficient still.
from collections import defaultdict
class DefaultDict(defaultdict):
def __missing__(self, key):
return self.default_factory(key)
Then:
d = DefaultDict(lambda key: key + 5)
d[2]
# 7
You can also use a function like this
dic = {}
DEFAULT_VALUE = 5
def dict_get(item):
try:
return [dic[item]]
except:
dic[int(item)] = DEFAULT_VALUE + int(item)
return DEFAULT_VALUE + int(item)
print(dict_get(10))
Here is how to do it without creating any class. You can use dict.setdefault:
d = {}
d_default_factory = lambda key: key + 5
d.setdefault(2, d_default_factory(2))
# 7
Optionally, you could save the default factory as an entry in the dictionary, like so:
d = {'default_factory': lambda key: key + 5}
d.setdefault(2, d['default_factory'](2))
# 7
Related
I am trying to provide a function as the default argument for the dictionary's get function, like this
def run():
print "RUNNING"
test = {'store':1}
test.get('store', run())
However, when this is run, it displays the following output:
RUNNING
1
so my question is, as the title says, is there a way to provide a callable as the default value for the get method without it being called if the key exists?
Another option, assuming you don't intend to store falsy values in your dictionary:
test.get('store') or run()
In python, the or operator does not evaluate arguments that are not needed (it short-circuits)
If you do need to support falsy values, then you can use get_or_run(test, 'store', run) where:
def get_or_run(d, k, f):
sentinel = object() # guaranteed not to be in d
v = d.get(k, sentinel)
return f() if v is sentinel else v
See the discussion in the answers and comments of dict.get() method returns a pointer. You have to break it into two steps.
Your options are:
Use a defaultdict with the callable if you always want that value as the default, and want to store it in the dict.
Use a conditional expression:
item = test['store'] if 'store' in test else run()
Use try / except:
try:
item = test['store']
except KeyError:
item = run()
Use get:
item = test.get('store')
if item is None:
item = run()
And variations on those themes.
glglgl shows a way to subclass defaultdict, you can also just subclass dict for some situations:
def run():
print "RUNNING"
return 1
class dict_nokeyerror(dict):
def __missing__(self, key):
return run()
test = dict_nokeyerror()
print test['a']
# RUNNING
# 1
Subclassing only really makes sense if you always want the dict to have some nonstandard behavior; if you generally want it to behave like a normal dict and just want a lazy get in one place, use one of my methods 2-4.
I suppose you want to have the callable applied only if the key does not exist.
There are several approaches to do so.
One would be to use a defaultdict, which calls run() if key is missing.
from collections import defaultdict
def run():
print "RUNNING"
test = {'store':1}
test.get('store', run())
test = defaultdict(run, store=1) # provides a value for store
test['store'] # gets 1
test['runthatstuff'] # gets None
Another, rather ugly one, one would be to only save callables in the dict which return the apropriate value.
test = {'store': lambda:1}
test.get('store', run)() # -> 1
test.get('runrun', run)() # -> None, prints "RUNNING".
If you want to have the return value depend on the missing key, you have to subclass defaultdict:
class mydefaultdict(defaultdict):
def __missing__(self, key):
val = self[key] = self.default_factory(key)
return val
d = mydefaultdict(lambda k: k*k)
d[10] # yields 100
#mydefaultdict # decorators are fine
def d2(key):
return -key
d2[5] # yields -5
And if you want not to add this value to the dict for the next call, you have a
def __missing__(self, key): return self.default_factory(key)
instead which calls the default factory every time a key: value pair was not explicitly added.
If you only know what the callable is likely to be at he get call site you could subclass dict something like this
class MyDict(dict):
def get_callable(self,key,func,*args,**kwargs):
'''Like ordinary get but uses a callable to
generate the default value'''
if key not in self:
val = func(*args,**kwargs)
else:
val = self[key]
return val
This can then be used like so:-
>>> d = MyDict()
>>> d.get_callable(1,complex,2,3)
(2+3j)
>>> d[1] = 2
>>> d.get_callable(1,complex,2,3)
2
>>> def run(): print "run"
>>> repr(d.get_callable(1,run))
'2'
>>> repr(d.get_callable(2,run))
run
'None'
This is probably most useful when the callable is expensive to compute.
I have a util directory in my project with qt.py, general.py, geom.py, etc. In general.py I have a bunch of python tools like the one you need:
# Use whenever you need a lambda default
def dictGet(dict_, key, default):
if key not in dict_:
return default()
return dict_[key]
Add *args, **kwargs if you want to support calling default more than once with differing args:
def dictGet(dict_, key, default, *args, **kwargs):
if key not in dict_:
return default(*args, **kwargs)
return dict_[key]
Here's what I use:
def lazy_get(d, k, f):
return d[k] if k in d else f(k)
The fallback function f takes the key as an argument, e.g.
lazy_get({'a': 13}, 'a', lambda k: k) # --> 13
lazy_get({'a': 13}, 'b', lambda k: k) # --> 'b'
You would obviously use a more meaningful fallback function, but this illustrates the flexibility of lazy_get.
Here's what the function looks like with type annotation:
from typing import Callable, Mapping, TypeVar
K = TypeVar('K')
V = TypeVar('V')
def lazy_get(d: Mapping[K, V], k: K, f: Callable[[K], V]) -> V:
return d[k] if k in d else f(k)
Say I have a dict that contains only two values:
foo = {'bar':someThing, 'foobar':someThingElse}
and I am given in a function foo and a key which can be either bar or foobar:
is there a generic way to select the other element? This function should use the knowledge that $foo$ only contains only two elements.
The hard coded way would be
def funnyStuff(foo, key):
if key == 'bar':
myKey = 'foobar'
if key == 'foobar':
myKey = 'bar'
return foo[myKey]
(python2 only, and breaks when there are not exactly 2 key/value pairs)
foo.values()[not foo.keys().index(key)]
Only two elements? You don't need a dict then.
foo = {'bar':'someThing', 'foobar':'someThingElse'}
def selectOther(data, key):
# structure unpacking is almost like pattern matching!
[(k1, v1), (k2, v2)] = data.items()
if k1 == key:
return v2
elif k2 == key:
return v1
else:
raise KeyError("Your %r is not among data keys!" % key)
You could use list comprehension:
def funnyStuff(foo, key):
return foo[[x for x in foo if x != key][0]]
How about something like this:
foo = {'bar':'someThing', 'foobar':'someThingElse'}
def get_other(foo, key):
return foo.get([k for k in foo.keys() if k != key][0])
print get_other('bar')
Output:
'someThingElse'
Although this works, it is inefficient since we create a list from another list, just to get to the 0th item!
foo = {'bar':'someThing', 'foobar':'someThingElse'}
def funnyStuff(foo, key):
mykey = ""
key_list = foo.keys()
if key == key_list[0]:
mykey = key_list[1]
else:
mykey = key_list[0]
return mykey
print foo[funnyStuff(foo, 'foobar')]
Depending on what you are trying to do this idea could be useful, make a new mapping from your keys to boolean, so that it is easy to negate them:
keys2bool = dict(zip(foo.keys(), [True, False]))
bool2values = dict(zip([True, False], foo.values()))
>>> bool2values[not keys2bool['bar']]
'someThingElse'
I am trying to determine if a specific key and value pair exist in a dictionary; however, if I use the contains or has-key method, it only checks for the key. I need it to check both the key and the specific value. Some background:
We have a total of 4 dictionaries: one for A, B, CompareList, and ChangeList. Once A is initialized, I put A's contents into CompareList (I would compare them directly; but A and B are double hash tables. And I've tried all of the methods here; but none of them work for me). So once we put A into CompareList, I compare it with the ObjectAttributes dictionary in B to see if anything changed. So for example, B may have the key,value pairs shape:circle and fill:no. If CompareList had shape:circle and fill:yes, then I want only fill:yes to be ChangeList. The problem lies in the "if attributes.getName() not in self.CompareList:" line. Here is the code; I am running it on Python 2.7.8. Thanks in advance for any help!!
class ObjectSemanticNetwork:
def __init__(self):
self.ObjectNames = {}
self.ObjectAttributes = {}
def setName(self, name):
self.ObjectNames[name] = self.ObjectAttributes
def setData(self, name, attribute):
self.ObjectAttributes[name] = attribute
def checkData(self, key):
print(key)
for key, value in self.ObjectAttributes.iteritems():
print(key)
print(value)
print("\n")
class Agent:
(self):
self.CompareList = {}
self.ChangeListAB = {}
self.ChangeListCD = {}
def addToCompareList(self, name, value):
self.CompareList[name] = value
def addToChangeListAB(self, name, value):
self.ChangeListAB[name] = value
def addToChangeListCD(self, name, value):
self.ChangeListCD[name] = value
def CheckList(self, List, ListName):
print '-------------------------',ListName,'--------------------------------'
for key, value in List.iteritems():
print(key)
print(value)
def Solve(self,problem):
OSNAB = ObjectSemanticNetwork()
for object in problem.getFigures().get("A").getObjects():
for attributes in object.getAttributes():
self.addToCompareList(attributes.getName(), attributes.getValue())
OSNAB.ObjectNames["A"] = OSNAB.setData(attributes.getName(), attributes.getValue())
#OSNAB.checkData("A")
self.CheckList(self.CompareList,"CompareList")
for object in problem.getFigures().get("B").getObjects():
for attributes in object.getAttributes():
if attributes.getName() not in self.CompareList:
self.addToChangeListAB(attributes.getName(), attributes.getValue())
OSNAB.ObjectNames["B"] = OSNAB.setData(attributes.getName(), attributes.getValue())
# OSNAB.checkData("B")
self.CheckList(self.ChangeListAB,"ChangeList")
OSNCD = ObjectSemanticNetwork()
for object in problem.getFigures().get("C").getObjects():
for attributes in object.getAttributes():
OSNCD.ObjectNames["C"] = OSNCD.setData(attributes.getName(), attributes.getValue())
# OSNCD.checkData("C")
for object in problem.getFigures().get("1").getObjects():
for attributes in object.getAttributes():
OSNCD.ObjectNames["D"] = OSNCD.setData(attributes.getName(), attributes.getValue())
# OSNCD.checkData("D")
return "6"
Use
if key in d and d[key] == value:
Or (only in Python 3)
if (key, value) in d.items():
In Python 3 d.items() returns a Dictionary view object, which supports fast membership testing. In Python 2 d.items() returns a list, which is both slow to create and slow to to test membership. Python 2.7 is a special case where you can use d.viewitems() and get the same thing that you get with d.items() in Python 3.
Edit: In a comment you indicate that for performance reasons you prefer checkKeyValuePairExistence over key in d and d[key] == value. Below are some timings showing that checkKeyValuePairExistence is always slower (by about 2x on my system when the key-value pair is present 16x when it is not). I also tested larger and smaller dictionaries and found little variation in the timings.
>>> import random
>>> from timeit import timeit
>>> def checkKeyValuePairExistence(dic, key, value):
... try:
... return dic[key] == value
... except KeyError:
... return False
...
>>> d = {random.randint(0, 100000):random.randint(0, 100000) for i in range(1000)}
>>> setup = 'from __main__ import k, d, v, checkKeyValuePairExistence'
>>> test_try_except = 'checkKeyValuePairExistence(d, k, v)'
>>> test_k_in_d_and = 'k in d and d[k] == v'
>>> k, v = random.choice(d.items()) # to test if found
>>> timeit(test_try_except, setup=setup)
0.1984054392365806
>>> timeit(test_k_in_d_and, setup=setup)
0.10442071140778353
>>> k = -1 # test if not found
>>> timeit(test_try_except, setup=setup)
1.2896073903002616
>>> timeit(test_k_in_d_and, setup=setup)
0.07827843747497809
How about this function:
def checkKeyValuePairExistence(dic, key, value):
try:
return dic[key] == value
except KeyError:
return False
If you are using another type of dictionary other then the one python offers (I'm sorry, I couldnt understand from your post if you are using it or not) then let me know and i'll try to give your another solution
Why not just do this:
a = {1:'a', 2:'b'}
b = (1, 'a')
print b in a.iteritems() # prints True
One minor annoyance with dict.setdefault is that it always evaluates its second argument (when given, of course), even when the first the first argument is already a key in the dictionary.
For example:
import random
def noisy_default():
ret = random.randint(0, 10000000)
print 'noisy_default: returning %d' % ret
return ret
d = dict()
print d.setdefault(1, noisy_default())
print d.setdefault(1, noisy_default())
This produces ouptut like the following:
noisy_default: returning 4063267
4063267
noisy_default: returning 628989
4063267
As the last line confirms, the second execution of noisy_default is unnecessary, since by this point the key 1 is already present in d (with value 4063267).
Is it possible to implement a subclass of dict whose setdefault method evaluates its second argument lazily?
EDIT:
Below is an implementation inspired by BrenBarn's comment and Pavel Anossov's answer. While at it, I went ahead and implemented a lazy version of get as well, since the underlying idea is essentially the same.
class LazyDict(dict):
def get(self, key, thunk=None):
return (self[key] if key in self else
thunk() if callable(thunk) else
thunk)
def setdefault(self, key, thunk=None):
return (self[key] if key in self else
dict.setdefault(self, key,
thunk() if callable(thunk) else
thunk))
Now, the snippet
d = LazyDict()
print d.setdefault(1, noisy_default)
print d.setdefault(1, noisy_default)
produces output like this:
noisy_default: returning 5025427
5025427
5025427
Notice that the second argument to d.setdefault above is now a callable, not a function call.
When the second argument to LazyDict.get or LazyDict.setdefault is not a callable, they behave the same way as the corresponding dict methods.
If one wants to pass a callable as the default value itself (i.e., not meant to be called), or if the callable to be called requires arguments, prepend lambda: to the appropriate argument. E.g.:
d1.setdefault('div', lambda: div_callback)
d2.setdefault('foo', lambda: bar('frobozz'))
Those who don't like the idea of overriding get and setdefault, and/or the resulting need to test for callability, etc., can use this version instead:
class LazyButHonestDict(dict):
def lazyget(self, key, thunk=lambda: None):
return self[key] if key in self else thunk()
def lazysetdefault(self, key, thunk=lambda: None):
return (self[key] if key in self else
self.setdefault(key, thunk()))
This can be accomplished with defaultdict, too. It is instantiated with a callable which is then called when a nonexisting element is accessed.
from collections import defaultdict
d = defaultdict(noisy_default)
d[1] # noise
d[1] # no noise
The caveat with defaultdict is that the callable gets no arguments, so you can not derive the default value from the key as you could with dict.setdefault. This can be mitigated by overriding __missing__ in a subclass:
from collections import defaultdict
class defaultdict2(defaultdict):
def __missing__(self, key):
value = self.default_factory(key)
self[key] = value
return value
def noisy_default_with_key(key):
print key
return key + 1
d = defaultdict2(noisy_default_with_key)
d[1] # prints 1, sets 2, returns 2
d[1] # does not print anything, does not set anything, returns 2
For more information, see the collections module.
You can do that in a one-liner using a ternary operator:
value = cache[key] if key in cache else cache.setdefault(key, func(key))
If you are sure that the cache will never store falsy values, you can simplify it a little bit:
value = cache.get(key) or cache.setdefault(key, func(key))
No, evaluation of arguments happens before the call. You can implement a setdefault-like function that takes a callable as its second argument and calls it only if it is needed.
There seems to be no one-liner that doesn't require an extra class or extra lookups. For the record, here is a easy (even not concise) way of achieving that without either of them.
try:
value = dct[key]
except KeyError:
value = noisy_default()
dct[key] = value
return value
Consider the following:
p1=1;
p2=5;
p3=7;
highest=max(p1,p2,p3).
The max function would return 7. I am looking to create a similar function, which would return "p3". I have created a small function (by simple comparisons) for the above example, shown below. however I am having trouble when the number of arguments go up.
def highest(p1,p2,p3):
if (p1>p2) and (p1>p3):
return "p1"
if (p2>p1) and (p2>p3):
return "p2"
if (p3>p1) and (p3>p1):
return "p3"
Is there a simpler way to do this>
Update: Paul Hankin pointed out that max() took a key function, which I didn't know. So:
>>> def argmax(**kw):
... return max(kw, key=kw.get)
...
>>> argmax(foo=3, bar=5, frotz=1, kaka=-3)
'bar'
Other solutions for completeness:
In Python 2.7 and 3.x you can use dictionary comprehensions.
>>> def argmax(**kw):
... wk = {v:k for k,v in kw.items()}
... return wk[max(wk)]
...
>>> argmax(foo=3, bar=5, frotz=1, kaka=-3)
'bar'
Dictionary comprehensions are neat. :)
In earlier versions of Python you can do this:
>>> def argmax(**kw):
... wk = dict([(v,k) for k,v in kw.items()])
... return wk[max(wk)]
...
>>> argmax(foo=3, bar=5, frotz=1, kaka=-3)
'bar'
Which will work in anything after Python 2.2 or so.
There is no way to get the name of the variable that had the highest value in the caller (because it might be a number or a complex expression), but by using keyword arguments exclusively, you can get the name of the parameter. Something like this:
def argmax(**kwargs):
mx = -1e+400 # overflows to -Inf
amx = None
for k, v in kwargs.iteritems():
if v > mx:
mx = v
amx = k
return amx
Works like this:
>>> argmax(a=1,b=2,c=3)
'c'
but the catch is, it doesn't work if any of the arguments is positional:
>>> argmax(1,2,3)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: argmax() takes exactly 0 arguments (3 given)
Depending on what you're doing, this construct might be more useful: (hat tip to http://lemire.me/blog/archives/2008/12/17/fast-argmax-in-python/ )
>>> a = [9,99,999]
>>> a.index(max(a))
2
The only way to keep this anything close to extensible is to take a list as argument and return the index of its highest element. You can put a p in front and begin counting at 1 if you really want to.
def highest(x):
return 'p' + repr(x.index(max(x))+1)
Obviously it dose not handle variable length arguments. If you want variable length argument then that's a different issue. If you have 10 arguments then just add them in the definition and it will return the correct argument name (not necessarily started with 'p'). But the catch is the number of arguments (3 or 5 or 10 or whatever else) is not variable. You need to know how many arguments you require.
def highest(p1,p2,p3,p4,p5):
d = locals()
keys = d.keys()
max_key = keys[0]
max_val = d[max_key]
for i in range(1,len(keys)):
key = keys[i]
val = d[key]
if val > max_val:
max_val = val
max_key = key
return max_key
print highest(3,2,5,10,1)
print highest(1,5,2,2,3)
print highest(5,2,5,1,11)
print highest(3,2,1,1,2)