Scope of Lambda Functions and Their Parameters

Scope of lambda functions and their parameters?

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()

How external variables can be passed into the scope of lambda functions?

I am not sure why you are having the problem. In a minimal example like this:

(lambda x: (lambda y: y+x)(3))(2)

You get the expected 5, so there must be something more to what you are doing.

What is the scope of a lambda variable in C#?

The brackets give the clue - the lambda variable is captured in the scope of where it's declared:

.Where(o => ... olist.Where(o1 => ...).Max(o1 => ...))
  //  |----------------------------------------------| scope of o
  //                       |---------|                 scope of first o1
  //                                      |---------|  scope of second o1

Note that there's no overlap for the two o1 variables, but they both overlap (or shadow) the o variable and hence can't use the same name.

Outer scope variable in lambda function

In this [lambda d: for d in [1,2]] what you have is

"Running d from 1 to 2, give me a function returning the value of d at the moment when the function is called."

After that list of two functions is constructed, you call [f() for f in that list]. At the moment the function is called, d in [1,2] has already completed, and d has its final value, 2. So both functions return 2.

One way to get a different result is to copy d to a variable for each lambda function you create. The easiest way to do that is:

[f() for f in [lambda x=d: x for d in [1,2]]]

Here each lambda has its own default value for x, which is d at the time the function is created.

Lambda function and variable scope

SPLAT.

for control in self.controls():
  self.connect(control, SIGNAL('clicked()'), lambda control=control:
    self.button(control.objectName()))

Python enclosing scope variables with lambda function

Even though i takes multiple values over time, there is effectively only one variable, i. The content of i is being changed during the loop. But the closures captures variables, not values. Nothing is evaluated inside the lambda until you call it. At the time you call the function, you access the current value of i, which happens to be the last one.

As for why i=i solves the problem, this is explained for example in The Hitchhiker's guide to Python (Common Gotchas):

Python’s default arguments are evaluated once when the function is defined, not each time the function is called (like it is in say, Ruby). This means that if you use a mutable default argument and mutate it, you will and have mutated that object for all future calls to the function as well.

And so, each fresh binding that occurs inside the closure you create (and happen to be named i just like the one outside) has its default value being computed when creating the closure. Consequently, you have the "right" value in place, ready to be used when the closure is called.

What do lambda function closures capture?

What do the closures capture exactly?

Closures in Python use lexical scoping: they remember the name and scope of the closed-over variable where it is created. However, they are still late binding: the name is looked up when the code in the closure is used, not when the closure is created. Since all the functions in your example are created in the same scope and use the same variable name, they always refer to the same variable.

There are at least two ways to get early binding instead:

1. The most concise, but not strictly equivalent way is the one recommended by Adrien Plisson. Create a lambda with an extra argument, and set the extra argument's default value to the object you want preserved.

2. More verbosely but also more robustly, we can create a new scope for each created lambda:

   >>> adders = [0,1,2,3]
   >>> for i in [0,1,2,3]:
   ...     adders[i] = (lambda b: lambda a: b + a)(i)
   ...     
   >>> adders[1](3)
   4
   >>> adders[2](3)
   5
   

   The scope here is created using a new function (another lambda, for brevity), which binds its argument, and passing the value you want to bind as the argument. In real code, though, you most likely will have an ordinary function instead of the lambda to create the new scope:

   def createAdder(x):
       return lambda y: y + x
   adders = [createAdder(i) for i in range(4)]
   

Lambda Scope Clarification

In Example 1, x is defined in the local scope of the lamdba expression and is not visible to the third line

In Example 2, now you've declared two variables named "x" at the same declaration scope (visibility is different)

With a lambda or anonymous method, it "captures" the scope at which it is running. If you have a local x in the same scope as the lambda definition, then it "captures" that x to pull into what the lambda can access--thus resulting in two definitions of "x". What you declare in the lambda doesn't get captured in the other direction so it isn't visible outside the lambda.

In Example 3, Now you're not using a variable that is local only the lambda outside the lambda, and not naming something the same at the same declaration scope.

Using lambdas in local scope in c++

Lambdas with empty brackets are equivalent with function pointers and will be copied by value in the same way. Answer yes.

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