在python中将函数作为参数传递

Question

I am trying to get my head around this one.

So here is what I want to do: I have three functions. say, foo,bar and foo_bar

def foo_bar(function):
   for i in range(20):
      function # execute function
def foo(someargs):
   print "hello from foo"

def bar(someargs):
   print " hello from bar"

and when i do foo_bar(foo) # how to specify arguments to foo??

I am expecting that I see "hello from foo" 20 times ?

But since I am not seeing that.. I clearly dont understand this well enough?

Answer 1

You're not calling the function. You call a function called function by following the name function with brackets () , and inside those brackets you put any arguments you need:

function(function)

I've passed function as a parameter to itself because your foo and bar take arguments, but do nothing with them.

Answer 2

This should do basically what you want:

def foo_bar(function):
   for i in range(20):
      function(i) # execute function
def foo(someargs):
   print "hello from foo"

def bar(someargs):
   print " hello from bar"

foo_bar(foo)
foo_bar(bar)

Answer 3

The docs have a section on this here . You most likely want to construct foo_bar as

def foo_bar(function, *func_args):
    for i in range(20):
        function(func_args)

def foo(a):
    print "hello %s"%a

You can then call it as

foo_bar(foo, 'from foo!')

Answer 4

I would make foo_bar return a function that accept the parameters to pass to the internal function:

def foo_bar(fn, n=20):
    def _inner(*args, **kwargs):
        for i in range(n):
            fn(*args, **kwargs)
    return _inner

and call it like:

foo_bar(foo)('some_param')