How to make a function with a function as a parameter

Question

This is a function I've defined:

import sys

def hello():
    print("hello")

class parser():

    def parseSubcommand(self, name, function):
        if name == sys.argv[1]:
            result = function()
            return(result)


    def findArgument(self, name, function):
        print("dummy") #not started working on findArgument yet

But when I try to call it like this:

parser().parseSubcommand(name="hello", function="hello")

I'm getting the error

function() TypeError: 'str' object is not callable

Then you are passing a string as the "function" argument. You need to post the rest of the code that shows how test is being called. — Daniel Roseman, May 25 '15 at 10:22
possible duplicate of [python: convert string into variable name](http://stackoverflow.com/questions/30028055/python-convert-string-into-variable-name) — Aleksandr Kovalev, May 25 '15 at 10:31
you still cannot call a string, use a dict mapping string name to function — Padraic Cunningham, May 25 '15 at 10:32

DeJaVo · Answer 1 · 2015-05-25T10:26:47.790

1

You can do the following:

def function1():
    return 'hello world'

def function2(function_to_run):
    result = function_to_run()
    return result

function2(function1)

edited May 25 '15 at 10:26

answered May 25 '15 at 10:20

DeJaVo

3,091
2
17
32

Please mark it as answer for other will have the solution in case encountered in your question. – DeJaVo May 25 '15 at 10:23
We do not use mixedCase in Python for function names. PEP8 quote: *"Function names should be lowercase, with words separated by underscores as necessary to improve readability."*. – Dr. Jan-Philip Gehrcke May 25 '15 at 10:25
why would you not just `return function_to_run()`? – Padraic Cunningham May 25 '15 at 10:26
@PadraicCunningham I like to write explicit. but you can do `return functionToRun()` , it is the same – DeJaVo May 25 '15 at 10:29

Padraic Cunningham · Answer 2 · 2015-05-25T11:05:36.150

You need to map the functions to the string representation of their names:

def hello():
    print("hello")

class parser():    
    funcs = {"hello": hello}
    def parseSubcommand(self, name):
         return self.funcs.get(name,lambda: print("Invalid name"))()



parser().parseSubcommand(name="hello")
parser().parseSubcommand(name="foo")
hello
Invalid name

Which if you are using sys.argv[1] you will need as you will always be getting strings.

If you want to take args using argparse for python3 or optaprse for python2 might be a better idea.

SuperBiasedMan · Accepted Answer · 2015-05-25T10:55:10.817

1

hello is not the same as "hello"

"hello" is a string that can only be used as a string. It can't refer to variables unless used as a dictionary key to access a specific reference.

You want to use hello because that's the actual name of your function. Variable names are never accessed by string.

parser().parseSubcommand(name="hello", function=hello)

If you need to pass in the function names as a string, then you need to reference them in a dictionary, like this:

functionNames = {"hello":hello}

edited May 25 '15 at 10:55

answered May 25 '15 at 10:40

SuperBiasedMan

9,814
10
45
73

2

If the OP is using sys.argv to take args then this won't work – Padraic Cunningham May 25 '15 at 10:46
@PadraicCunningham That's true, I didn't pay attention to the `sys.argv` part. @Blaubeersirup You'll need to use a dictionary to use this properly with `sys.argv` – SuperBiasedMan May 25 '15 at 10:52
1

The op may not be using it at all it is hard to tell from the code, if they are taking arguments using optparse would probably be a better than trying to make their own parser – Padraic Cunningham May 25 '15 at 10:57

How to make a function with a function as a parameter

3 Answers3