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dolaameng
dolaameng

Reputation: 1437

How does a lambda function refer to its parameters in python?

I am new in Python. My task was quite simple -- I need a list of functions that I can use to do things in batch. So I toyed it with some examples like

fs = [lambda x: x + i for i in xrange(10)]

Surprisingly, the call of 

[f(0) for f in fs]

gave me the result like [9, 9, 9, 9, 9, 9, 9, 9, 9, 9]. It was not what I expected as I'd like the variable i has different values in different functions.

So My question is:

  1. Is the variable i in lambda global or local?

  2. Does python has the same concept like 'closure' in javascript? I mean does each lambda here holds a reference to the i variable or they just hold a copy of the value of i in each?

  3. What should I do if I'd like the output to be [0, 1, .....9] in this case?

Upvotes: 6

Views: 1201

Answers (4)

ncoghlan
ncoghlan

Reputation: 41506

The problem you are running into here is the distinction between "early binding" and "late binding".

When Python looks up a variable from an outer scope (i in this case) it uses late binding. That means it sees the value of that variable at the time the function is called, rather than the value at the time the function is defined.

So, in your example code, all 10 lambda functions see the final value assigned to the i variable by the looping process: 9.

Greg's answer shows one way to force early binding behaviour (i.e. create an additional closure and call it immediately while still inside the loop).

Another commonly used approach to forcing early binding semantics is the "default argument hack", which binds the variable as a default argument at function definition time:

>>> fs = [(lambda x, _i=i: x + _i) for i in xrange(10)]
>>> [f(0) for f in fs]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Either method works. Greg's has the advantage of not messing with the returned function's signature, the default argument hack is faster and is significantly more readable than adding an additional closure level when defining named functions rather than using lambda expressions.

Upvotes: 6

jon_darkstar
jon_darkstar

Reputation: 16768

i is local and there are indeed closures in python.

I believe your confusion is that you assigh fs to a list of identical function. 

>>> fs = [lambda x: x + i for i in xrange(10)]
>>> fs
[<function <lambda> at 0x02C6E930>, <function <lambda> at 0x02C6E970>, <function <lambda> at 0x02C6E9B0>, <function <lambda> at 0x02C6E9F0>, <function <lambda> at 0x02C6EA30>, <function <lambda> at 0x02C6EA70>, <function <lambda> at 0x02C6EAB0>, <function <lambda> at 0x02C6EAF0>, <function <lambda> at 0x02C6EB30>, <function <lambda> at 0x02C6EB70>]
>>> fs[0](0)
9
>>> fs[0](100)
109
>>> fs[5](0)
9
>>> fs[5](100)
109

I think a single function returning a list would be more appropriate.

>>> fs3 = lambda x: [x + i for i in xrange(10)]
>>> fs3
<function <lambda> at 0x02C6EC70>
>>> fs3(0)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Upvotes: 1

Rafe Kettler
Rafe Kettler

Reputation: 76955

  1. The variable i is local to the list comprehension, but it's available to the lambda because the lambda is in its scope.
  2. Yes, lambdas are closures. Variable binding might not always work the way you want it to, but they're closures. You shouldn't rely on them heavily, though.
  3. You'd just want to loop over xrange(10). You could do this with lambdas (see the other answer), but you wouldn't want to. Lambdas should be used pretty sparingly.

The reason that the lambda behaves this way is because for each loop i is rebound. Since i isn't local to the lambda, it changes too, and the last value it holds is 9. So all you're doing is 0 + 9 10 times.

Upvotes: 4

Greg Hewgill
Greg Hewgill

Reputation: 993461

It looks a bit messy, but you can get what you want by doing something like this:

>>> fs = [(lambda y: lambda x: x + y)(i) for i in xrange(10)]
>>> [f(0) for f in fs]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Normally Python supports the "closure" concept similar to what you're used to in Javascript. However, for this particular case of a lambda expression inside a list comprehension, it seems as though i is only bound once and takes on each value in succession, leaving each returned function to act as though i is 9. The above hack explicitly passes each value of i into a lambda that returns another lambda, using the captured value of y.

Upvotes: 10

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