I want to convert certain values of a dictionary as multiple args and pass it to a method. The key of the dict needs to be the name of the variable passed to the method.
eg:
myDict={'one':1,'two':2,'three':3,'four':4}
#call myMethod as
myMethod(one=1,two=2)
def myMeth(self,*args):
do somthing with args
You can use the unpacking notation. If you have a method like this
def function(arg1,arg2,arg3):
# do something
and a dictionary dct = {'arg1':3,'arg2':3,'arg3':6}, you can call the function like
function(**dct)
which is equivalent to doing function(arg1=3,arg2=3,arg3=6).
Notice the double stars. This means to take the dictionary and pass its values as named parameters where the keys are the names. A single star would unpack a list, passing its values in order as unnamed parameters. These can be combined.
See section 4.7.4 of the python documentation for more detail. Here is another article discussing these as well.
Use ** for passing or accepting dictionary arguments. The following code
d={'foo': 'bar','test': 123}
def my_method(**kwargs):
print kwargs
my_method(**d)
would print the contents of d, {'foo': 'bar','test': 123}
EDIT: #Matthew 's answer sums it up nicely, via the comparison of *args.
Related
For this function
def eat_dog(name, should_digest=True):
print "ate dog named %s. Digested, too? %" % (name, str(should_digest))
I want to, external to the function, read its arguments and any default values attached. So for this specific example, I want to know that name has no default value (i.e. that it is a required argument) and that True is the default value for should_digest.
I'm aware of inspect.getargspec(), which does give me information about arguments and default values, but I see no connection between the two:
ArgSpec(args=['name', 'should_digest'], varargs=None, keywords=None, defaults=(True,))
From this output how can I tell that True (in the defaults tuple) is the default value for should_digest?
Additionally, I'm aware of the "ask for forgiveness" model of approaching a problem, but unfortunately output from that error won't tell me the name of the missing argument:
>>> eat_dog()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: eat_dog() takes at least 1 argument (0 given)
To give context (why I want to do this), I'm exposing functions in a module over a JSON API. If the caller omits certain function arguments, I want to return a specific error that names the specific function argument that was omitted. If a client omits an argument, but there's a default provided in the function signature, I want to use that default.
Python3.x
In a python3.x world, you should probably use a Signature object:
import inspect
def get_default_args(func):
signature = inspect.signature(func)
return {
k: v.default
for k, v in signature.parameters.items()
if v.default is not inspect.Parameter.empty
}
Python2.x (old answer)
The args/defaults can be combined as:
import inspect
a = inspect.getargspec(eat_dog)
zip(a.args[-len(a.defaults):],a.defaults)
Here a.args[-len(a.defaults):] are the arguments with defaults values and obviously a.defaults are the corresponding default values.
You could even pass the output of zip to the dict constructor and create a mapping suitable for keyword unpacking.
looking at the docs, this solution will only work on python2.6 or newer since I assume that inspect.getargspec returns a named tuple. Earlier versions returned a regular tuple, but it would be very easy to modify accordingly. Here's a version which works with older (and newer) versions:
import inspect
def get_default_args(func):
"""
returns a dictionary of arg_name:default_values for the input function
"""
args, varargs, keywords, defaults = inspect.getargspec(func)
return dict(zip(args[-len(defaults):], defaults))
Come to think of it:
return dict(zip(reversed(args), reversed(defaults)))
would also work and may be more intuitive to some people.
Depending on exactly what you need, you might not need the inspect module since you can check the __defaults__ attribute of the function:
>>> eat_dog.__defaults__
(True,)
>>> eat_dog.__code__.co_argcount
2
>>> eat_dog.__code__.co_varnames
('name', 'should_digest')
>>>
>>> eat_dog.__kwdefaults__
>>> eat_dog.__code__.co_kwonlyargcount
0
You can use inspect module with its getargspec function:
inspect.getargspec(func)
Get the names and default values of a Python function’s arguments. A tuple of four things is returned: (args, varargs, keywords, defaults). args is a list of the argument names (it may contain nested lists). varargs and keywords are the names of the * and ** arguments or None. defaults is a tuple of default argument values or None if there are no default arguments; if this tuple has n elements, they correspond to the last n elements listed in args.
See mgilson's answer for exact code on how to retrieve argument names and their default values.
To those looking for a version to grab a specific default parameter with mgilson's answer.
value = signature(my_func).parameters['param_name'].default
Here's a full working version, done in Python 3.8.2
from inspect import signature
def my_func(a, b, c, param_name='apple'):
pass
value = signature(my_func).parameters['param_name'].default
print(value == 'apple') # True
to take care of keyword-only args (and because defaults and kwonlydefaults can be None):
spec = inspect.getfullargspec(func)
defaults = dict(zip(spec.args[::-1], (spec.defaults or ())[::-1]))
defaults.update(spec.kwonlydefaults or {})
You can get this via some of the __dunder__ vars as mentioned by other posts. Putting that into a simple helper function can get you a dictionary of default values.
.__code__.co_varnames: A tuple of all input variables
.__defaults__: A tuple of the default values
It is worth noting that this tuple only incudes the default provided variables which must always be positioned last in the function arguments
You can use these two items to match the last n variables in the .__code__.co_varnames with all the items in the .__defaults__
EDIT Thanks to #griloHBG - Added if statement to prevent exceptions when no defaults are specified.
def my_fn(a, b=2, c='a'):
pass
def get_defaults(fn):
if fn.__defaults__==None:
return {}
return dict(zip(
fn.__code__.co_varnames[-len(fn.__defaults__):],
fn.__defaults__
))
print(get_defaults(my_fn))
Should give:
{'b': 2, 'c': 'a'}
In python, all the arguments with default value come after the arguments without default value. So the mapping should start from the end till you exhaust the default value list. Hence the logic:
dict(zip(reversed(args), reversed(defaults)))
gives the correctly mapped defaults.
I want to create a method, which can dynamically add keyword arguments to another function that I intend to call within this method, something like:
def func(query1: str, query2: str):
func2(query1, query2)
where, the query1, query2 in the func are strings of the type "a=3", "b=4", the func should essentially take these strings and then call them in a keyword argument way like:
func2(a=3, b=4)
The a, b here are not fixed, they may vary, I mean to say I want to be able to take the a and b from the LHS of the string "a=3"
So my function takes the string "a=3" and parses the "a" as the keyword for the other function and the 3 as the value and calls the other function like func2(a=3)
I want to know if this kind of implementation is even possible or if there are any workarounds for achieving the same goal.
The reason why I want this is because I have a django application in which I want to create a model where I can store some filter rules and apply them each time that view is called, so I want to allow the user to select a bunch of filters and then I want to store them in the database and call those rules to get a queryset specific to that user's preferences.
def some_view(request):
# Get the filter rule strings from the model
queryset = some_model.objects.filter(<Pass all the rules here from those strings>)
return Response
edit: Add more details on the use case
Not sure I really understand what you want. You could translate those query strings to a dict, then use that to call the other function with **kwargs:
def f(a, b):
print(a, b)
def g(*queries):
return f(**dict((q.split("=", 1) for q in queries)))
>>> g("a=4", "b='string with = in the middle'")
4 'string with = in the middle'
Note: just split("=") might fail if the parameter is a string containing =; thus you might want to use split("=", 1) instead.
However, this passes all the arguments as raw strings; you might want to eval, or rather ast.literal_eval to the values, but that may also pose some risks. Something like this:
import ast
def g(*queries):
return f(**{k: ast.literal_eval(v)
for k, v in (q.split("=", 1) for q in queries)})
def accept(**kwargs):
pass
If I defined accept and I expect it be called by passing a param which is dict. Are the asterisks necessary for all dict params?
What if I do things like:
def accept(dict):
pass
dict = {...}
accept(dict)
Specifically speaking, I would like to implement a update method for a class, which keeps a dict as a container. Just like the dict.update method, it takes a dict as a param and modify the content of the container. In this specific case should I use the kwargs or not?
** in function parameter collects all keyword arguments as a dictionary.
>>> def accept(**kwargs): # isinstance(kwargs, dict) == True
... pass
...
Call using keyword arguments:
>>> accept(a=1, b=2)
Call using ** operator:
>>> d = {'a': 1, 'b': 2}
>>> accept(**d)
>>> accept(d)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: accept() takes exactly 0 arguments (1 given)
See Python tutorial - Keyword argument and Unpacking Argument Lists.
BTW, don't use dict as variable name. It shadows builtin function/type dict.
See f below. The function f has two parameters, a positional one called name and a keyword argument message. They are local variable in the frame of the function call.
When you do f("John", **{"foo": "123", "message": "Hello World"}), the function f will unpack the dictionary into local variable by its key/value pair. In the end you have three local varaibles: name, foo=123, and message=Hello World.
The purpose of **kwargs, double asterisks is for uknown keyword arguments.
Contrast this:
def f(name, message=None):
if message:
return name + message
return name
Here I am telling user if you ever want to call f, you can pass a keyword argument message. This is the only kwarg I will ever accept and expect to receive if there is such one. If you try f("John", foo="Hello world") you will get unexpected keyword argument.
**kwargs is useful if you don't know ahead of time what keyword arguments you want to receive (very common for dispatching to lower-level functions/methods), then you use it.
def f(name, message=None, **kwargs):
if message:
return name + message
return name
In the second example, you can do f("John", **{"foo": "Hello Foo"}) while omitting message. You can also do f("John", **{"foo": "Hello Foo", "message": "Hello Message"}).
Can I ignore it?
As you see yes you can ignore it. Here f("John", **{"foo": "Hello Foo", "message": "Hello Message"}) I still only use message and ignore everything else.
Don't use **kwargs unless you need to be careless about the inputs.
What if my input is a dictionary?
If your function simply takes the dictionary and modifies the dictionary, NOT using individaul key, then just pass the dictionary. There is no reason to make a dictionary item into variables.
But here are two main usages of **kwargs.
Supposed I have a class and I want to create attributes on the fly. I can use setattr to set class attributes from input.
class Foo(object):
def __init__(**kwargs):
for key, value in kwargs.items():
setattr(self, key, value)
If I do Foo(**{"a": 1, "b": 2}) I will get Foo.a and Foo.b at the end.
This is particularly useful when you have to deal with legacy code. However, there is a big security concern. Imagine you own a MongoDB instance and this is a container for writing into a database. Imagine this dict is a request form object from user. The user can shovel ANYTHING and you simply save it in the database like that? That's a security hole. Make sure you validate (use a loop).
The second common usage of kwargs is that you don't know things ahead of times which I have covered (it's actually sort of the first common usage anyway).
If you want to pass a dictionary as input to a function, you can simply do it like this
def my_function1(input_dict):
print input_dict
d = {"Month": 1, "Year": 2}
my_function1(d) # {'Month': 1, 'Year': 2}
This is straight forward. Lets see the **kwargs method. kwargs stands for keyword arguments. So, you need to actually pass the parameters as key-value pairs, like this
def my_function2(**kwargs):
print kwargs
my_function2(Month = 1, Year = 2)
But if you have a dictionary and if you want to pass that as a parameter to my_function2, it can be done with unpacking, like this
my_function2(**d)
see what I have:
> bar = [(u'code__regex', u'^[^J]'), (u'active__exact', u'0'), (u'type__id__exact', u'E01')]
There are the fields/values that I want to use to filter the model Foo.
> #want to have this equivalent:
> Foo.objects.filter(bar)
Thanks!
Foo.objects.filter(**dict(bar))
This isn't a django issue, this is a python issue. You want to pass the keyword pairs as keyword arguments (kwargs) to the filter. Your bar is perfect as a kwarg set, so the dict(bar) converts it to the dictionary, and the ** prefix informs the python parser that the dictionary is to be interpreted as keyword arguments by the receiver.
Stack overflow entry Understanding kwargs in Python covers this in more detail.
You didn't say if you want it in AND or OR. If AND is ok then you can simply convert that list into a dictionary and pass it to filter:
Foo.objects.filter(**dict(bar))
Does python have the ability to create dynamic keywords?
For example:
qset.filter(min_price__usd__range=(min_price, max_price))
I want to be able to change the usd part based on a selected currency.
Yes, It does. Use **kwargs in a function definition.
Example:
def f(**kwargs):
print kwargs.keys()
f(a=2, b="b") # -> ['a', 'b']
f(**{'d'+'e': 1}) # -> ['de']
But why do you need that?
If I understand what you're asking correctly,
qset.filter(**{
'min_price_' + selected_currency + '_range' :
(min_price, max_price)})
does what you need.
You can easily do this by declaring your function like this:
def filter(**kwargs):
your function will now be passed a dictionary called kwargs that contains the keywords and values passed to your function. Note that, syntactically, the word kwargs is meaningless; the ** is what causes the dynamic keyword behavior.
You can also do the reverse. If you are calling a function, and you have a dictionary that corresponds to the arguments, you can do
someFunction(**theDictionary)
There is also the lesser used *foo variant, which causes you to receive an array of arguments. This is similar to normal C vararg arrays.
Yes, sort of.
In your filter method you can declare a wildcard variable that collects all the unknown keyword arguments. Your method might look like this:
def filter(self, **kwargs):
for key,value in kwargs:
if key.startswith('min_price__') and key.endswith('__range'):
currency = key.replace('min_price__', '').replace('__range','')
rate = self.current_conversion_rates[currency]
self.setCurrencyRange(value[0]*rate, value[1]*rate)