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python lambda closures

What do (lambda) function closures capture?

发布于 2020-03-31 22:53:29

Recently I started playing around with Python and I came around something peculiar in the way closures work. Consider the following code:

adders=[0,1,2,3]

for i in [0,1,2,3]:
   adders[i]=lambda a: i+a

print adders[1](3)

It builds a simple array of functions that take a single input and return that input added by a number. The functions are constructed in for loop where the iterator i runs from 0 to 3. For each of these numbers a lambda function is created which captures i and adds it to the function's input. The last line calls the second lambda function with 3 as a parameter. To my surprise the output was 6.

I expected a 4. My reasoning was: in Python everything is an object and thus every variable is essential a pointer to it. When creating the lambda closures for i, I expected it to store a pointer to the integer object currently pointed to by i. That means that when i assigned a new integer object it shouldn't effect the previously created closures. Sadly, inspecting the adders array within a debugger shows that it does. All lambda functions refer to the last value of i, 3, which results in adders[1](3) returning 6.

Which make me wonder about the following:

  • What do the closures capture exactly?
  • What is the most elegant way to convince the lambda functions to capture the current value of i in a way that will not be affected when i changes its value?
Questioner
Boaz
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23
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♦ 2017-05-23 18:31

Your second question has been answered, but as for your first:

what does the closure capture exactly?

Scoping in Python is dynamic and lexical. A closure will always remember the name and scope of the variable, not the object it's pointing to. 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.

EDIT: Regarding your other question of how to overcome this, there are two ways that come to mind:

  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. A little more verbose but less hacky would be to create a new scope each time you create the 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 (a 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)]
Boaz 2010-02-19 18:25:14

Max, if you add an answer for my other (simpler question), I can mark this as an accepted answer. Thx!

Claudiu 2010-06-29 23:21:09

Python has static scoping, not dynamic scoping.. it's just all variables are references, so when you set a variable to a new object, the variable itself (the reference) has the same location, but it points to something else. the same thing happens in Scheme if you set!. see here for what dynamic scope really is: voidspace.org.uk/python/articles/code_blocks.shtml .

Crashworks 2011-09-20 10:15:25

Option 2 resembles what functional languages would call a "Curried function."

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