Python: Assigning function along with parameters to a variable

Question

def test(name):
    print "name:", name

func = test
func("testing") # it works, as I know that the function test accepts one parameter.

My question is, what if "test" has varying number of arguments depending on the scenario and how "func" knows how many number of arguments to pass and what are those arguments name.

Sorry, if I am not clear. This would give more clear picture on the scenario.

I have a function dispatcher.

testcase_obj  = testcase() # A object of a class    
if command.startswith("test1"):    
    output = exec_test1()    
elif command.startswith("do_test"):    
    output = exec_do_test(testcase_obj)

Now, I want to wrap a function whenever user sends an option while executing the script. I changed above code as:

testcase_obj  = testcase() # A object of a class    
if command.startswith("test1"):    
    func = exec_test1() # Mistake, this should be func = exec_test1
elif command.startswith("do_test"):    
    func = exec_do_test(testcase_obj) # I don't know how assign exec_do_test along
                                      # with its parameter to 'func'. I don't want to
                                      # to call exec_to_test.

if option_given:    
    func = wrapper_func(func)    
    output = func() # At this point I don't how many parameters that "func" takes.

You'll need to give more specifics about what you're trying to do. It doesn't make sense to try to call a "mystery function" without knowing what arguments it accepts. Even if you can programmatically figure out how many arguments it takes and what their names are, how would you decide what values to pass for them without reading the documentation and knowing what they do? — BrenBarn
– BrenBarn, Commented Oct 23, 2012 at 6:19
Your edited question still doesn't really explain how you hope to make use of the wrapped function. Do you know what exec_test1() and exec_do_test(testcase_obj) return? If you do, then you know what func will be in each case, and how many arguments to pass in each case. If you don't, knowing how many arguments they take won't help you, because you still won't know what values you can meaningfully pass for those arguments. — BrenBarn
– BrenBarn, Commented Oct 23, 2012 at 7:03

nneonneo · Accepted Answer · 2012-10-23 06:19:32Z

5

After saying func = test, func becomes just another name for test. So, you call func exactly the same way as test, and if you give func the wrong number of arguments you'll get a TypeError as if you had called test incorrectly.

For more information about the difference between variables in other languages and names in Python, see Code like a Pythonista.

answered Oct 23, 2012 at 6:19

nneonneo

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2 Comments

rajpy Over a year ago

If "test" takes variable number of arguments and it is assigned to "func". I want to know how many arguments that "func" has. Introspection(dir(func)) will not show how many arguments that "func" can take.

nneonneo Over a year ago

Well, dir(test) wouldn't tell you either. Remember, they are names for the same exact thing. You probably want help(func)...

nehz · Accepted Answer · 2012-10-23 06:32:59Z

3

Try the inspect module

import inspect
inspect.getargspec(func).args

will give:

['name']

answered Oct 23, 2012 at 6:32

nehz

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Comments

sharcashmo · Accepted Answer · 2012-10-23 06:19:36Z

1

It will be the same.

func is just an alias to test, not a function calling test

answered Oct 23, 2012 at 6:19

sharcashmo

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Comments

Burhan Khalid · Accepted Answer · 2012-10-23 07:02:13Z

1

If "test" takes variable number of arguments and it is assigned to "func". I want to know how many arguments that "func" has. Introspection(dir(func)) will not show how many arguments that "func" can take.

func is not a function. It is just an alias pointing to the function that is called test. So there is no way func can take a different number of arguments than test because func is not a function, just a name pointing to one. You can verify this:

>>> def test(arg1):
...    print 'test was given ',arg1
...
>>> func = test
>>> test.func_name
'test'
>>> func.func_name
'test'
>>> id(func)
3075004876L
>>> id(test)
3075004876L
>>> inspect.getargspec(func).args
['arg1']
>>> inspect.getargspec(test).args
['arg1']

answered Oct 23, 2012 at 7:02

Burhan Khalid

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2 Comments

glglgl Over a year ago

func and test are absolutely equivalent. Both are names pointing to a function. So nothing special about func.

rajpy Over a year ago

Thanks for the answer. Is it possible to assign 'test' along with its signature 'arg1' to 'func'? So that if I call func(), it should translate to the call test(arg1)

sawan gupta · Accepted Answer · 2012-10-23 06:39:20Z

0

Yes there is a way if you give default values to your function.

def test(name="default",hi=0):
    print "name:", name,hi

func = test
func("testing")
func("testing",6) 
func(0)

answered Oct 23, 2012 at 6:39

sawan gupta

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