Is there a way to loop through and execute all of the functions in a Python class?

Question

I have

class Foo():
    function bar():
        pass

    function foobar():
        pass

Rather than executing each function one by one as follows:

x = Foo()
x.bar()
x.foobar()

is there a built-in way to loop through and execute each function in the sequence in which they are written in the class?

Answer 1

No. You can access Foo.__dict__, and call each value in turn (catching errors for non-callable members), but the order is not preserved.

for callable in Foo.__dict__.values():
    try:
        callable()    
    except TypeError:
        pass

This assumes none of the functions take parameters, as in your example.

Answer 2

Since Python stores the methods (and other attributes) of a class in a dictionary, which is fundamentally unordered, this is impossible.

If you don't care about order, use the class's __dict__:

x = Foo()
results = []
for name, method in Foo.__dict__.iteritems():
    if callable(method):
        results.append(method(x))

This also works if the function takes extra parameters - just put them after the instance of the class.

Answer 3

def assignOrder(order):
  def do_assignment(to_func):
    to_func.order = order
    return to_func
  return do_assignment

class Foo():

  @assignOrder(1)
  def bar(self):
    print "bar"

  @assignOrder(2)
  def foo(self):
    print "foo"

  #don't decorate functions you don't want called
  def __init__(self):
    #don't call this one either!
    self.egg = 2

x = Foo()
functions = sorted(
             #get a list of fields that have the order set
             [getattr(x, field) for field in dir(x)
              if hasattr(getattr(x, field), "order")
             ],
             #sort them by their order
             key = (lambda field: field.order)
            )
for func in functions:
  func()

That funny @assignOrder(1) line above def bar(self) triggers this to happen:

Foo.bar = assignOrder(1)(Foo.bar)

assignOrder(1) returns a function that takes another function, changes it (adding the field order and setting it to 1) and returns it. This function is then called on the function it decorates (its order field gets thus set); the result replaces the original function.

It's a fancier, more readable and more maintainable way of saying:

  def bar(self):
    print "bar"
  Foo.bar.order = 1

Answer 4

So long as you're only interested in Python 3.x (and from the empty parentheses in your class statement I'll guess you might be), then there is actually a simple way to do this without decorators: Python 3 allows you to provide your own dictionary like object to use while the class is defined.

The following code is from PEP3115 except for the last couple of lines which I added to print out the methods in order:

# The custom dictionary
class member_table(dict):
  def __init__(self):
     self.member_names = []

  def __setitem__(self, key, value):
     # if the key is not already defined, add to the
     # list of keys.
     if key not in self:
        self.member_names.append(key)

     # Call superclass
     dict.__setitem__(self, key, value)

# The metaclass
class OrderedClass(type):

   # The prepare function
   @classmethod
   def __prepare__(metacls, name, bases): # No keywords in this case
      return member_table()

   # The metaclass invocation
   def __new__(cls, name, bases, classdict):
      # Note that we replace the classdict with a regular
      # dict before passing it to the superclass, so that we
      # don't continue to record member names after the class
      # has been created.
      result = type.__new__(cls, name, bases, dict(classdict))
      result.member_names = classdict.member_names
      return result

class MyClass(metaclass=OrderedClass):
  # method1 goes in array element 0
  def method1(self):
     pass

  # method2 goes in array element 1
  def method2(self):
     pass

x = MyClass()
print([name for name in x.member_names if hasattr(getattr(x, name), '__call__')])