I have a Python function which takes several arguments. Some of these arguments could be omitted in some scenarios.
def some_function (self, a, b, c, d = None, e = None, f = None, g = None, h = None):
#code
The arguments d through h are strings which each have different meanings. It is important that I can choose which optional parameters to pass in any combination. For example, (a, b, C, d, e), or (a, b, C, g, h), or (a, b, C, d, e, f, or all of them (these are my choices).
It would be great if I could overload the function - but I read that Python does not support overloading. I tried to insert some of the required int arguments in the list - and got an argument mismatch error.
Right now I am sending empty strings in place of the first few missing arguments as placeholders. I would like to be able to call a function just using actual values.
Is there any way to do this? Could I pass a list instead of the argument list?
Right now the prototype using ctypes looks something like:
_fdll.some_function.argtypes = [c_void_p, c_char_p, c_int, c_char_p, c_char_p, c_char_p, c_char_p, c_char_p]
Just use the *args parameter, which allows you to pass as many arguments as you want after your a,b,c. You would have to add some logic to map args->c,d,e,f but its a "way" of overloading.
def myfunc(a,b, *args, **kwargs):
for ar in args:
print ar
myfunc(a,b,c,d,e,f)
And it will print values of c,d,e,f
Similarly you could use the kwargs argument and then you could name your parameters.
def myfunc(a,b, *args, **kwargs):
c = kwargs.get('c', None)
d = kwargs.get('d', None)
#etc
myfunc(a,b, c='nick', d='dog', ...)
And then kwargs would have a dictionary of all the parameters that are key valued after a,b
Try calling it like: obj.some_function( '1', 2, '3', g="foo", h="bar" ). After the required positional arguments, you can specify specific optional arguments by name.
It is very easy just do this
def foo(a = None):
print(a)
Instead of None you can type anything that should be in place if there was no argument for example if you will not write the value of the parameter like this foo() then it will print None because no argument is given and if you will GIVE it an argument like foo("hello world") then it will print hello world... oh well I just forgot to tell y'all that these types of parameters i.e optional parameters, need to be behind all the other parameters. This means that, let's take the previous function and add another parameter b
def foo(a = None, b):
print(a)
Now if you'll execute your python file it is going to raise an exception saying that Non-default arguments follow default arguments,
SyntaxError: non-default argument follows default argument
so you gotta put the optional or non-default argument after the arguments which are required
which means
def foo (a, b=None): ... #This one is right
def foo(b=None, a): ... #and this isn't
Required parameters first, optional parameters after. Optional parameters always with a =None.
Easy and fast example:
def example_function(param1, param2, param3=None, param4=None):
pass
# Doesn't work, param2 missing
example_function("hello")
# Works
example_function("hello", "bye")
# Works. Both the same
example_function("hello", "bye", "hey")
example_function("hello", "bye", param3="hey")
# Works. Both the same
example_function("hello", "bye", "hey", "foo")
example_function("hello", "bye", param3="hey", param4="foo")
Check this:
from typing import Optional
def foo(a: str, b: Optional[str] = None) -> str or None:
pass
To get a better sense of what's possible when passing parameters it's really helpful to refer to the various options: positional-or-keyword (arg or arg="default_value"), positional-only (before /, in the parameter list), keyword-only (after *, in the parameter list), var-positional (typically *args) or var-keyword (typically **kwargs). See the Python documentation for an excellent summary; the various other answers to the question make use of most of these variations.
Since you always have parameters a, b, c in your example and you appear to call them in a positional manner, you could make this more explicit by adding /,,
def some_function (self, a, b, c, /, d = None, e = None, f = None, g = None, h = None):
#code
To make Avión's answer work for vector argument inputs;
def test(M,v=None):
try:
if (v==None).all() == False:
print('argument passed')
return M + v
except:
print('no argument passed')
return M
Where M is some matrix and v some vector. Both test(M) and test(M,v) produce errors when I attempted to use if statements without using 'try/ except' statements.
As mentioned by cem, upgrading to python 3.10 would allow the union (x|y) (or the Optional[...])functionality which might open some doors for alternative methods, but I'm using Anaconda spyder so I think I have to wait for a new release to use python 3.10.
Related
I'm creating a wrapper for a function with functools.wraps. My wrapper has the effect of overriding a default parameter (and it doesn't do anything else):
def add(*, a=1, b=2):
"Add numbers"
return a + b
#functools.wraps(add)
def my_add(**kwargs):
kwargs.setdefault('b', 3)
return add(**kwargs)
This my_add definition behaves the same as
#functools.wraps(add)
def my_add(*, a=1, b=3):
return add(a=a, b=b)
except that I didn't have to manually type out the parameter list.
However, when I run help(my_add), I see the help string for add, which has the wrong function name and the wrong default argument for the parameter b:
add(*, a=1, b=2)
Add numbers
How can I override the function name and the default argument in this help() output?
(Or, is there a different way to define my_add, using for example some magic function my_add = magic(add, func_name='my_add', kwarg_defaults={'b': 3}) that will do what I want?)
Let me try and explain what happens.
When you call the help functions, this is going to request information about your function using the inspect module. Therefore you have to change the function signature, in order to change the default argument.
Now this is not something that is advised, or often preferred, but who cares about that right? The provided solution is considered hacky and probably won't work for all versions of Python. Therefore you might want to reconsider how important the help function is... Any way let's start with some explanation on how it was done, followed by the code and test case.
Copying functions
Now the first thing we will do is copy the entire function, this is because I only want to change the signature of the new function and not the original function. This decouples the new my_add signature (and default values) from the original add function.
See:
How to create a copy of a python function
How can I make a deepcopy of a function in Python?
For ideas of how to do this (I will show my version in a bit).
Copying / updating signature
The next step is to get a copy of the function signature, for that this post was very useful. Except for the part where we have to adjust the signature parameters to match the new keyword default arguments.
For that we have to change the value of a mappingproxy, which we can see when running the debugger on the return value of inspect.signature(g). Now so far this can only be done by changing the private variables (the values with leading underscores _private). Therefore this solution will be considered hacky and is not guaranteed to withstand possible updates. That said, let's see the solution!
Full code
import inspect
import types
import functools
def update_func(f, func_name='', update_kwargs: dict = None):
"""Based on http://stackoverflow.com/a/6528148/190597 (Glenn Maynard)"""
g = types.FunctionType(
code=f.__code__,
globals=f.__globals__.copy(),
name=f.__name__,
argdefs=f.__defaults__,
closure=f.__closure__
)
g = functools.update_wrapper(g, f)
g.__signature__ = inspect.signature(g)
g.__kwdefaults__ = f.__kwdefaults__.copy()
# Adjust your arguments
for key, value in (update_kwargs or {}).items():
g.__kwdefaults__[key] = value
g.__signature__.parameters[key]._default = value
g.__name__ = func_name or g.__name__
return g
def add(*, a=1, b=2):
"Add numbers"
return a + b
my_add = update_func(add, func_name="my_add", update_kwargs=dict(b=3))
Example
if __name__ == '__main__':
a = 2
print("*" * 50, f"\nMy add\n", )
help(my_add)
print("*" * 50, f"\nOriginal add\n", )
help(add)
print("*" * 50, f"\nResults:"
f"\n\tMy add : a = {a}, return = {my_add(a=a)}"
f"\n\tOriginal add: a = {a}, return = {add(a=a)}")
Output
**************************************************
My add
Help on function my_add in module __main__:
my_add(*, a=1, b=3)
Add numbers
**************************************************
Original add
Help on function add in module __main__:
add(*, a=1, b=2)
Add numbers
**************************************************
Results:
My add : a = 2, return = 5
Original add: a = 2, return = 4
Usages
f: is the function that you want to update
func_name: is optionally the new name of the function (if empty, keeps the old name)
update_kwargs: is a dictionary containing the key and value of the default arguments that you want to update.
Notes
The solution is using copy variables to make full copies of dictionaries, such that there is no impact on the original add function.
The _default value is a private variable, and can be changed in future releases of python.
The following code (gratefully gleaned from elsewhere in stackoverflow):
import inspect
def func(first, second, third):
frame = inspect.currentframe()
# print(f'frame: {inspect.getargvalues(frame)}')
args, _, _, values = inspect.getargvalues(frame)
print(f'function name {inspect.getframeinfo(frame)[2]}')
for i in args:
print(f'{i} = {values[i]}')
return [(i, values[i]) for i in args]
a = 'Test'
b = 'This'
c = 'Now'
func(a, b, c)
prints:
function name func
first = Test
second = This
third = Now
I would like the code to:
Print a = Test, b = This, c = Now instead of first = Test.. etc
According to No. 1 immediately above, print the arguments names passed to
*args. So instead of def func(first, second, third): it should be def func(*args): whose expected output should be:
function name func
a = Test
b = This
b = Now
I am deliberately avoiding using **kwargs (which would solve the problem easily but will require significant change in my code which may not be an option in the immediate term).
P.S. For clarification purposes, I am not looking for a way to get the name of the function (as this is already achieved with the code above) or for a way to get the name of the arguments when the arguments are explicitly named at the function definition. I am only stuck when the arguments are passed via *args (in which case the number or name of arguments cannot be determined until the time the function is being called and the name and the number of arguments are being passed to it)
I am new to Python kindly help!
In python we can do this:
def myFun1(one = '1', two = '2'):
...
Then we can call the function and pass the arguments by their name:
myFun1(two = 'two', one = 'one')
Also, we can do this:
def myFun2(**kwargs):
print kwargs.get('one', 'nothing here')
myFun2(one='one')
So I was wondering if it is possible to combine both methods like:
def myFun3(name, lname, **other_info):
...
myFun3(lname='Someone', name='myName', city='cityName', otherInfo='blah')
In general what combinations can we do?
Thanks and sorry for my silly question.
The general idea is:
def func(arg1, arg2, ..., kwarg1=default, kwarg2=default, ..., *args, **kwargs):
...
You can use as many of those as you want. The * and ** will 'soak up' any remaining values not otherwise accounted for.
Positional arguments (provided without defaults) can't be given by keyword, and non-default arguments can't follow default arguments.
Note Python 3 also adds the ability to specify keyword-only arguments by having them after *:
def func(arg1, arg2, *args, kwonlyarg=default):
...
You can also use * alone (def func(a1, a2, *, kw=d):) which means that no arguments are captured, but anything after is keyword-only.
So, if you are in 3.x, you could produce the behaviour you want with:
def myFun3(*, name, lname, **other_info):
...
Which would allow calling with name and lname as keyword-only.
Note this is an unusual interface, which may be annoying to the user - I would only use it in very specific use cases.
In 2.x, you would need to manually make this by parsing **kwargs.
You can add your named arguments along with kwargs. If the keys are available in the calling function It will taken to your named argument otherwise it will be taken by the kwargs dictionary.
def add(a=1, b=2,**c):
res = a+b
for items in c:
res = res + c[items]
print(res)
add(2,3)
5
add(b=4, a =3)
7
add(a =1,b=2,c=3,d=4)
10
It's possible at least for Python 2.7. Keyword arguments get assigned to positional parameters by name, so you can do
In [34]: def func(name, lname, **kwargs):
print 'name='+name, 'lname='+lname
print kwargs
....:
In [35]: func(lname='lname_val', name='name_val', city='cityName', otherInfo='blah')
name=name_val lname=lname_val
{'city': 'cityName', 'otherInfo': 'blah'}
Official docs state it that way:
"If keyword arguments are present, they are first converted to positional arguments, as follows. First, a list of unfilled slots is created for the formal parameters. If there are N positional arguments, they are placed in the first N slots. Next, for each keyword argument, the identifier is used to determine the corresponding slot (if the identifier is the same as the first formal parameter name, the first slot is used, and so on). If the slot is already filled, a TypeError exception is raised. Otherwise, the value of the argument is placed in the slot, filling it (even if the expression is None, it fills the slot)."
https://docs.python.org/2/reference/expressions.html#calls
I am trying to understand if it is a good idea or not to pass as parameter the python equivalent of null; which I believe is None.
Example: You have a function that accepts n parameters; in one case I need just the first and second parameters, so instead of writing a long function definition with args and kwargs, and manipulate them, I can just pass null to one of the parameters.
def myfunct(a, b, c[optional], d[optional], e, f....n):
[do something]
if d=="y":
[do something but use only a and b]
Execution:
myfunct(a, b, c, d, .....n) #OK!
myfunct(a, b, None, "y", None,....n) #OK?
This theoretically should not raise an error, since null is a value I believe (this is not C++), although I am not sure if this is a correct way to do things.
The function knows that there is a condition when one of the parameters is a specific value, and in that case, it won't ask for any other parameter but 1; so the risk of using null should be practically 0.
Is this acceptable or am I potentially causing issues down the road, using this approach?
There's nothing wrong with using None to mean "I am not supplying this argument".
You can check for None in your code:
if c is None:
# do something
if d is not None:
# do something else
One recommendation I would make is to have None be the default argument for any optional arguments:
def myfunct(a, b, e, f, c=None, d=None):
# do something
myfunct(A, B, E, F)
In python we can do this:
def myFun1(one = '1', two = '2'):
...
Then we can call the function and pass the arguments by their name:
myFun1(two = 'two', one = 'one')
Also, we can do this:
def myFun2(**kwargs):
print kwargs.get('one', 'nothing here')
myFun2(one='one')
So I was wondering if it is possible to combine both methods like:
def myFun3(name, lname, **other_info):
...
myFun3(lname='Someone', name='myName', city='cityName', otherInfo='blah')
In general what combinations can we do?
Thanks and sorry for my silly question.
The general idea is:
def func(arg1, arg2, ..., kwarg1=default, kwarg2=default, ..., *args, **kwargs):
...
You can use as many of those as you want. The * and ** will 'soak up' any remaining values not otherwise accounted for.
Positional arguments (provided without defaults) can't be given by keyword, and non-default arguments can't follow default arguments.
Note Python 3 also adds the ability to specify keyword-only arguments by having them after *:
def func(arg1, arg2, *args, kwonlyarg=default):
...
You can also use * alone (def func(a1, a2, *, kw=d):) which means that no arguments are captured, but anything after is keyword-only.
So, if you are in 3.x, you could produce the behaviour you want with:
def myFun3(*, name, lname, **other_info):
...
Which would allow calling with name and lname as keyword-only.
Note this is an unusual interface, which may be annoying to the user - I would only use it in very specific use cases.
In 2.x, you would need to manually make this by parsing **kwargs.
You can add your named arguments along with kwargs. If the keys are available in the calling function It will taken to your named argument otherwise it will be taken by the kwargs dictionary.
def add(a=1, b=2,**c):
res = a+b
for items in c:
res = res + c[items]
print(res)
add(2,3)
5
add(b=4, a =3)
7
add(a =1,b=2,c=3,d=4)
10
It's possible at least for Python 2.7. Keyword arguments get assigned to positional parameters by name, so you can do
In [34]: def func(name, lname, **kwargs):
print 'name='+name, 'lname='+lname
print kwargs
....:
In [35]: func(lname='lname_val', name='name_val', city='cityName', otherInfo='blah')
name=name_val lname=lname_val
{'city': 'cityName', 'otherInfo': 'blah'}
Official docs state it that way:
"If keyword arguments are present, they are first converted to positional arguments, as follows. First, a list of unfilled slots is created for the formal parameters. If there are N positional arguments, they are placed in the first N slots. Next, for each keyword argument, the identifier is used to determine the corresponding slot (if the identifier is the same as the first formal parameter name, the first slot is used, and so on). If the slot is already filled, a TypeError exception is raised. Otherwise, the value of the argument is placed in the slot, filling it (even if the expression is None, it fills the slot)."
https://docs.python.org/2/reference/expressions.html#calls