Develop Reference

passing **kwargs to a de-referencing function

Hope the title is conveying the correct information.
My problem is that I don't understand why call kwarg_function(some_func, a=1, b=2, c=3) fails. I would have thought that as 'c' isn't referenced with some_func() it would simply be ignored. Can anyone explain why 'c' isn't simply ignored.
def kwarg_function(function, **kwargs):
print(kwargs)
function(**kwargs)
def some_func(a, b):
print(f"type: {type(a)} values: {a}")
print(f"type: {type(b)} values: {b}")
kwarg_function(some_func, a=1, b=2) # called successfully
kwarg_function(some_func, a=1, b=2, c=3) # fails with unexpected keyword arg 'c'

Think of the ** as "unpack what's on my right side as keyword arguments" in this case.
def foo(a,b,**kwargs):
# This will take any number of arguments provided that starting from the 3rd one they are keyword args
# Those are equivalent (and working) calls
foo(1,2, x = 7)
foo(1,2, **{"x":7})
# Those will fail
foo(1,2,7)
foo(1,2, {"x":7})
The function you declared expects 2 arguments
def some_func(a, b):
And you are calling it with three under the hood, because this:
kwarg_function(some_func, a=1, b=2, c=3) # fails with unexpected keyword arg 'c'
Does this (inside kwarg_function body):
funtion(a=1,b=2,c=3)

In python, * and ** are for unpacking iterables. They don't consider what's are in them, and just unpack whatever you pass in them.
You can find more info about it in this link.
So, when you pass a=1, b=2, c=3, ... as kwargs to your kwargs_function, you will get them as kwargs param, regardless of what you have passed.
And then, when you pass **kwargs to another function, all of your data would be passed to your another function, regardless of what's in that.
If you want your some_func be more flexible with your data and accept whatever you pass to it, you can add **kwargs param to it too:
def some_func(a, b, **kwargs):
print(f"type: {type(a)} values: {a}")
print(f"type: {type(b)} values: {b}")

The rationale of `functools.partial` behavior

I'm wondering what the story -- whether sound design or inherited legacy -- is behind these functools.partial and inspect.signature facts (talking python 3.8 here).
Set up:
from functools import partial
from inspect import signature
def bar(a, b):
return a / b
All starts well with the following, which seems compliant with curry-standards.
We're fixing a to 3 positionally, a disappears from the signature and it's value is indeed bound to 3:
f = partial(bar, 3)
assert str(signature(f)) == '(b)'
assert f(6) == 0.5 == f(b=6)
If we try to specify an alternate value for a, f won't tell us that we got an unexpected keyword, but rather that it got multiple values for argument a:
f(a=2, b=6) # TypeError: bar() got multiple values for argument 'a'
f(c=2, b=6) # TypeError: bar() got an unexpected keyword argument 'c'
But now if we fix b=3 through a keyword, b is not removed from the signature, it's kind changes to keyword-only, and we can still use it (overwrite the default, as a normal default, which we couldn't do with a in the previous case):
f = partial(bar, b=3)
assert str(signature(f)) == '(a, *, b=3)'
assert f(6) == 2.0 == f(6, b=3)
assert f(6, b=1) == 6.0
Why such asymmetry?
It gets even stranger, we can do this:
f = partial(bar, a=3)
assert str(signature(f)) == '(*, a=3, b)' # whaaa?! non-default argument follows default argument?
Fine: For keyword-only arguments, there can be no confusing of what parameter a default is assigned to, but I still wonder what design-thinking or constraints are behind these choices.

Using partial with a Positional Argument
f = partial(bar, 3)
By design, upon calling a function, positional arguments are assigned first. Then logically, 3 should be assigned to a with partial. It makes sense to remove it from the signature as there is no way to assign anything to it again!
when you have f(a=2, b=6), you are actually doing
bar(3, a=2, b=6)
when you have f(2, 2), you are actually doing
bar (3, 2, 2)
We never get rid of 3
For the new partial function:
We can't give a a different value with another positional argument
We can't use the keyword a to assign a different value to it as it is already "filled"
If there is a parameter with the same name as the keyword, then the argument value is assigned to that parameter slot. However, if the parameter slot is already filled, then that is an error.
I recommend reading the function calling behavior section of pep-3102 to get a better grasp of this matter.
Using partial with a Keyword Argument
f = partial(bar, b=3)
This is a different use case. We are applying a keyword argument to bar.
You are functionally turning
def bar(a, b):
...
into
def f(a, *, b=3):
...
where b becomes a keyword-only argument
instead of
def f(a, b=3):
...
inspect.signature correctly reflects a design decision of partial. The keyword arguments passed to partial are designed to append additional positional arguments (source).
Note that this behavior does not necessarily override the keyword arguments supplied with f = partial(bar, b=3), i.e., b=3 will be applied regardless of whether you supply the second positional argument or not (and there will be a TypeError if you do so). This is different from a positional argument with a default value.
>>> f(1, 2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: f() takes 1 positional argument but 2 were given
where f(1, 2) is equivalent to bar(1, 2, b=3)
The only way to override it is with a keyword argument
>>> f(2, b=2)
An argument that can only be assigned with a keyword but positionally? This is a keyword-only argument. Thus (a, *, b=3) instead of (a, b=3).
The Rationale of Non-default Argument follows Default Argument
f = partial(bar, a=3)
assert str(signature(f)) == '(*, a=3, b)' # whaaa?! non-default argument follows default argument?
You can't do def bar(a=3, b). a and b are so called positional-or-keyword arguments.
You can do def bar(*, a=3, b). a and b are keyword-only arguments.
Even though semantically, a has a default value and thus it is optional, we can't leave it unassigned because b, which is a positional-or-keyword argument needs to be assigned a value if we want to use b positionally. If we do not supply a value for a, we have to use b as a keyword argument.
Checkmate! There is no way for b to be a positional-or-keyword argument as we intended.
The PEP for positonal-only arguments also kind of shows the rationale behind it.
This also has something to do with the aforementioned "function calling behavior".
partial != Currying & Implementation Details
partial by its implementation wraps the original function while storing the fixed arguments you passed to it.
IT IS NOT IMPLEMENTED WITH CURRYING. It is rather partial application instead of currying in the sense of functional programming. partial is essentially applying the fixed arguments first, then the arguments you called with the wrapper:
def __call__(self, /, *args, **keywords):
keywords = {**self.keywords, **keywords}
return self.func(*self.args, *args, **keywords)
This explains f(a=2, b=6) # TypeError: bar() got multiple values for argument 'a'.
See also: Why is partial called partial instead of curry
Under the Hood of inspect
The outputs of inspect is another story.
inspect itself is a tool that produces user-friendly outputs. For partial() in particular (and partialmethod(), similarly), it follows the wrapped function while taking the fixed parameters into account:
if isinstance(obj, functools.partial):
wrapped_sig = _get_signature_of(obj.func)
return _signature_get_partial(wrapped_sig, obj)
Do note that it is not inspect.signature's goal to show you the actual signature of the wrapped function in the AST.
def _signature_get_partial(wrapped_sig, partial, extra_args=()):
"""Private helper to calculate how 'wrapped_sig' signature will
look like after applying a 'functools.partial' object (or alike)
on it.
"""
...
So we have a nice and ideal signature for f = partial(bar, 3)
but get f(a=2, b=6) # TypeError: bar() got multiple values for argument 'a' in reality.
Follow-up
If you want currying so badly, how do you implement it in Python, in the way which gives you the expected TypeError?

When you provide positional or keyword arguments to partial, the new function is constructed
f = partial(bar, 3)
f(a=2, b=6) # TypeError: bar() got multiple values for argument 'a'
f(c=2, b=6) # TypeError: bar() got an unexpected keyword argument 'c'
This is actually consistent with the idea of partial, which is that arguments are passed to the wrapped function with the addition of positional and keyword arguments passed to partial
These cases behave as expected:
bar(3, a=2, b=6) # TypeError: bar() got multiple values for argument 'a'
bar(3, c=2, b=6) # TypeError: bar() got an unexpected keyword argument 'c'
But now if we fix b=3 through a keyword, b is not removed from the signature,
f = partial(bar, b=3)
assert str(signature(f)) == '(a, *, b=3)'
assert f(6) == 2.0 == f(6, b=3)
assert f(6, b=1) == 6.0
This case is different from the above because in the previous case, a positional argument was provided to partial, not a keyword argument. When positional arguments are provided to partial, then it makes sense to remove them from the signature. Arguments provided as keywords are not removed from the signature.
So far, there is no inconsistency or asymmetry.
f = partial(bar, a=3)
assert str(signature(f)) == '(*, a=3, b)' # whaaa?! non-default argument follows default argument?
The signature here makes sense and is the expectation for partial(bar, a=3) -- it works the same as def f(*, a=3, b): ... and is the correct signature in this case. Note that when you provide a=3 to partial in this case, a becomes a keyword-only argument, as does b.
This is because when a positional argument is provided as a keyword, all following arguments must be specified keyword arguments.
sig = signature(f)
sig.parameters['a'].kind # <_ParameterKind.KEYWORD_ONLY: 3>
inspect.getfullargspec(f)
# FullArgSpec(args=[], varargs=None, varkw=None, defaults=None, kwonlyargs=['a', 'b'], kwonlydefaults={'a': 3}, annotations={})

How to use *args and self in Python constructor

I need a Python method to have access to self for instance variables and also be able to take any number of arguments. I basically want a method foo that can be called via
foo(a, b, c)
or
foo()
In the class, I think the constructor would be
def foo(self, *args):
Is this correct? Also, fyi, I am new to Python (if you can't tell).

You just have to add it after the self parameter:
class YourClass:
def foo(self, *args):
print(args)
def bar(self, *args, **kwargs):
print(args)
print(kwargs)
def baz(self, **kwargs):
print(kwargs)
I have also added a method in which you also add **kwargs, and the case in which you add both *args and **kwargs.
Examples
>>> o = YourClass()
>>> o.foo()
()
>>> o.foo(1)
(1,)
>>> o.foo(1, 2)
(1, 2)

def foo(self, *args):
Yes, that is correct.

You declared the method correctly. You can also use double asterisks to accept keyword arguments.
Reference: Expressions
A double asterisk ** denotes dictionary unpacking. Its operand must be a mapping. Each mapping item is added to the new dictionary. Later values replace values already set by earlier key/datum pairs and earlier dictionary unpackings.
....
An asterisk * denotes iterable unpacking. Its operand must be an iterable. The iterable is expanded into a sequence of items, which are included in the new tuple, list, or set, at the site of the unpacking.
Args will be a tuple. To access the values you will have to iterate or use positional arguments, ie: args[0]

TypeError: fun() takes at least 2 arguments (3 given)

This is my code:
def fun(x, y, b=None, c=None):
print(x,' ',y,' ',b,' ',c)
I am calling it as fun(1, b=2, c=4) and getting error TypeError: fun() takes at least 2 arguments (3 given).
I know this error is because of incorrect number of positional and keyword arguments.
Instead of this, I want whenever I call my function with incorrect no. of arguments, it should tell me which argument is provided.
For example: for above case it should say something like "argument y is nor provided".
Can I write a decorator for this purpose?

fun as defined need to get between 2 and 4 argument, as it has two mandatory arguments and two optional arguments. You did not provide one of the two mandatory ones:
fun(1, b=2, c=4) # What about the argument y?
You need to call it using one of the next forms:
fun(1, 2)
fun(1, 2, b=3)
fun(1, 2, c=4)
fun(1, 2, b=3, c=4)
If you want notification about insufficient arguments, you can use args and kwargs:
def fun(*args, **kwargs):
if len(args) < 2:
print("Need at least two arguments!"); return
if len(args) > 2 or len(kwargs) > 2:
print("Too much arguments supplied!"); return
x, y = args
a, b = kwargs.get('a', None), kwargs.get('b', None)
print(x, y, a, b)

I want to handle this error and prompt error like it requires these (names) positional arguments. Is it possible to write a decorator for this?
I did a bit of research and came across the inspect module. Perhaps something along these lines will suffice? Right now I'm catching TypeError and printing a message, but you may prefer throwing a new TypeError that contains the message instead.
import inspect
from functools import wraps
def inspect_signature(f):
signature = inspect.signature(f)
#wraps(f)
def decorator(*args, **kwargs):
try:
f(*args, **kwargs)
except TypeError:
print('Failed to call "{}" with signature {}. Provided args={} and kwargs={}.'.format(
f.__name__, signature, args, kwargs))
return decorator
#inspect_signature
def func(foo, bar):
print('Called successfully with foo={}, bar={}'.format(foo, bar))
pass
if __name__ == '__main__':
func(foo='a', bar='b')
func(foo='a')
func('a', 'b', 'c')
Output
Called successfully with foo=a, bar=b
Failed to call "func" with signature (foo, bar). Provided args=() and kwargs={'foo': 'a'}.
Failed to call "func" with signature (foo, bar). Provided args=('a', 'b', 'c') and kwargs={}.

*args treated as single argument

I'm trying to make a function designed to call another function multiple times:
def iterator(iterations, function, *args):
#called as:
iterator(5, my_function, arg1, arg2, arg3)
Note that the number of arguments here is variable: could 1, could be 2, could be 10.
fill them in based on the function that is being called.
def iterator(iterations, function, *args):
for i in range(iteration):
temp = function(args)
return temp
The problem here is:
TypeError: my_function() takes exactly 4 arguments (1 given)
And this is because (arg1, arg2, arg3, arg4) are being treated as a single argument.
How do I get around this?

By using the same syntax when applying the args sequence:
temp = function(*args)
The *args syntax here is closely related to the *args function parameter syntax; instead of capturing an arbitrary number of arguments, using *args in a call expands the sequence to separate arguments.
You may be interested to know that there is a **kwargs syntax too, to capture and apply keyword arguments:
def iterator(iterations, function, *args, **kwargs):
for i in range(iteration):
temp = function(*args, **kwargs)
return temp

Try this, unpacking the argument list (a.k.a. splatting it):
function(*args)
From the example in the documentation, you'll see that this is what you need:
range(3, 6) # ok
range([3, 6]) # won't work
range(*[3, 6]) # it works!