How to pass a method that receives parameters as a parameter of another function in Python

Question

I know this is valid:

def printValue():
    print 'This is the printValue() method'

def callPrintValue(methodName):
    methodName()
    print 'This is the callPrintValue() method'

but is there a way to pass a method that receives parameters as a parameter of another function?

Doing this is not possible:

def printValue(value):
    print 'This is the printValue() method. The value is %s'%(value)

def callPrintValue(methodName):
    methodName()
    print 'This is the callPrintValue() method'

This is the stack trace i get:

This is the printValue() method. The value is dsdsd
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in callPrintValue
TypeError: 'NoneType' object is not callable

score 8 · Answer 1 · answered Jul 16 '12 at 12:59

Some people find lambda ugly, but it is a useful tool in cases like this. Rather than modifying the signature of callPrintValue(), you can use lambda to quickly define a new function that binds the arguments to printValue(). Whether you really want to do this depends on many factors, and it may be that adding an *args parameter as others have suggested is preferable. Still, this is an option worth considering. The following works with no modifications to your current code:

>>> callPrintValue(lambda: printValue('"Hello, I am a value"'))
This is the printValue() method. The value is "Hello, I am a value"
This is the callPrintValue() method

I agree that `lambda` may be what is called for here (+1). In fact, this is how it is often done in GUI programming (with `tkinter` for example) as there you don't have any way to modify the function doing the callback. — mgilson, Jul 16 '12 at 13:01

score 6 · Accepted Answer · answered Jul 16 '12 at 12:53

def printValue(value):
    print 'This is the printValue() method. The value is %s'%(value)

def callPrintValue(methodName, *args):
    methodName(*args)
    print 'This is the callPrintValue() method'

Then you can call it like so:

callPrintValue(printValue, "Value to pass to printValue")

This allows you to pass in an arbitrary number of arguments, and all of them are passed to the function that you call in callPrintValue

score 4 · Answer 3 · answered Jul 16 '12 at 12:53

4

I guess you can do this


def callPrintValue(methodName, *args):
    methodName(*args)
    print 'This is the callPrintValue() method'

make the call


callPrintValue(printValue, "abc")

answered Jul 16 '12 at 12:53

dolaameng

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score 3 · Answer 4 · answered Jul 16 '12 at 12:54

You want to use tuple unpacking:

def print_value(*values):
    print values

def call_print_value(func,args=None):
    func(*args)

call_print_value(print_value,args=('this','works')) #prints ('this', 'works')

From an API point of view, I prefer to keep the arguments that are passed on as a separate keyword. (then it's a little more explicit which arguments are being used by print_value and which ones are being used by call_print_value). Also note that in python, it is customary to have function (and method) names as name_with_underscores. CamelCase is generally used for class names.

score 2 · Answer 5 · edited May 23 '17 at 12:04

2

As a follow up to the answers provided already, you may want to check out the the following questions on stackoverflow for a better understanding of *args and/or **kwargs and lambda in python.

edited May 23 '17 at 12:04

Community

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answered Jul 16 '12 at 13:02

Kulbir Saini

3,926
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How to pass a method that receives parameters as a parameter of another function in Python

5 Answers5