Passing a function name as an argument in a function

python

I am trying to pass the name of a function into another function as an argument but I get an error: TypeError: 'str' object is not callable. Here is a simplified example of the problem:

def doIt(a, func, y, z):
    result = z
    result = func(a, y, result)
    return result

def dork1(arg1, arg2, arg3):
    thing = (arg1 + arg2) / arg3
    return thing

def dork2(arg1, arg2, arg3):
    thing = arg1 + (arg2 / arg3)
    return thing

When I call doIt like so:

var = 'dork1'
ned = doIt(3, var, 4, 9)
print (ned)

I get:

Traceback (most recent call last):
   File "<pyshell#9>", line 1, in <module>
     ned = doIt(3, var, 4, 9)
   File "<pyshell#2>", line 3, in doIt
     result = func(a, y, result)
TypeError: 'str' object is not callable

If you want to pass the function's name, as you said and you're doing, of course you can't call it -- why would one "call a name"? It's meaningless.

If you want to call it, pass the function itself, that is, most emphatically not

var = 'dork1'

but rather

var = dork1

without quotes!

Edit: the OP wonders in a comment (!) how to get a function object given the function name (as a string). As it happens I just showed how to do that in a tutorial I taught at OSCON (from which I'm just back) -- get the slides from here and see page 47, "Lazy-loading callbacks":

class LazyCallable(object):
  def __init__(self, name):
    self.n, self.f = name, None
  def __call__(self, *a, **k):
    if self.f is None:
      modn, funcn = self.n.rsplit('.', 1)
      if modn not in sys.modules:
        __import__(modn)
      self.f = getattr(sys.modules[modn],
                       funcn)
    self.f(*a, **k)

So you could pass LazyCallable('somemodule.dork1') and live happily ever after. If you don't need to deal with the module of course (what a weird architecture that must imply!-) it's easy to adjust this code.

Don't pass the name of a function.

Pass the function.

fun = dork1
ned = doIt(3, fun, 4, 9)
print (ned)

I was thrilled to find this and I don't know if this was answered. My solution to this is as follows:

def doIt(func, *args):
   func_dict = {'dork1':dork1,'dork2':dork2}
   result = func_dict.get(func)(*args)
   return result

def dork1(var1, var2, var3):
   thing = (float(var1 + var2) / var3)
   return thing

def dork2(var1, var2, var3):
   thing = float(var1) + (float(var2) / var3)
   return thing

This can be run as follows:

func = 'dork2'
ned = doIt(func,3, 4, 9)
print ned

var = 'dork1'
ned = doIt(3, var, 4, 9)
print (ned)

In this example, var is a string. The doIt function "calls" its second argument (for which you pass var). Pass a function instead.

Passing a function name as an argument in a function

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