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)

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.

If you have to pass it as a string you can call it with:

globals()['dork1'](arg1,arg2,arg3)

Functions are first-class objects in python. Do this:

var = dork1

If you must pass a string, such as user input, then:

globals()[var]

will look up the function object.

You probably shouldn't do this, but you can get the function using eval()

for example, to use len,

eval("len")(["list that len is called on"])