Scope of lambda functions and their parameters?

Question

I need a callback function that is almost exactly the same for a series of gui events. The function will behave slightly differently depending on which event has called it.

Answer 1

the soluiton to lambda is more lambda

In [0]: funcs = [(lambda j: (lambda: j))(i) for i in ('do', 're', 'mi')]

In [1]: funcs
Out[1]: 
[<function __main__.<lambda>>,
 <function __main__.<lambda>>,
 <function __main__.<lambda>>]

In [2]: [f() for f in funcs]
Out[2]: ['do', 're', 'mi']

the outer lambda is used to bind the current value of i to j at the

each time the outer lambda is called it makes an instance of the inner lambda with j bound to the current value of i as i's value

Answer 2

Python does uses references of course, but it does not matter in this context.

When you define a lambda (or a function, since this is the exact same behavior), it does not evaluate the lambda expression before runtime:

# defining that function is perfectly fine
def broken():
    print undefined_var

broken() # but calling it will raise a NameError

Even more surprising than your lambda example:

i = 'bar'
def foo():
    print i

foo() # bar

i = 'banana'

foo() # you would expect 'bar' here? well it prints 'banana'

In short, think dynamic: nothing is evaluated before interpretation, that's why your code uses the latest value of m.

When it looks for m in the lambda execution, m is taken from the topmost scope, which means that, as others pointed out; you can circumvent that problem by adding another scope:

def factory(x):
    return lambda: callback(x)

for m in ('do', 're', 'mi'):
    funcList.append(factory(m))

Here, when the lambda is called, it looks in the lambda' definition scope for a x. This x is a local variable defined in factory's body. Because of this, the value used on lambda execution will be the value that was passed as a parameter during the call to factory. And doremi!

As a note, I could have defined factory as factory(m) [replace x by m], the behavior is the same. I used a different name for clarity :)

You might find that Andrej Bauer got similar lambda problems. What's interesting on that blog is the comments, where you'll learn more about python closure :)

Answer 3

Yes, that's a problem of scope, it binds to the outer m, whether you are using a lambda or a local function. Instead, use a functor:

class Func1(object):
    def __init__(self, callback, message):
        self.callback = callback
        self.message = message
    def __call__(self):
        return self.callback(self.message)
funcList.append(Func1(callback, m))

Answer 4

The problem here is the m variable (a reference) being taken from the surrounding scope. Only parameters are held in the lambda scope.

To solve this you have to create another scope for lambda:

def callback(msg):
    print msg

def callback_factory(m):
    return lambda: callback(m)

funcList=[]
for m in ('do', 're', 'mi'):
    funcList.append(callback_factory(m))
for f in funcList:
    f()

In the example above, lambda also uses the surounding scope to find m, but this time it's callback_factory scope which is created once per every callback_factory call.

Or with functools.partial:

from functools import partial

def callback(msg):
    print msg

funcList=[partial(callback, m) for m in ('do', 're', 'mi')]
for f in funcList:
    f()