Generating functions inside loop with lambda expression in python

Question

If I make two lists of functions:

def makeFun(i): return lambda: i
a = [makeFun(i) for i in range(10)]
b = [lambda: i for i in range(10)]

why the lists a and b are not equal?

For example:

>>> a[2]()
2
>>> b[2]()
9

Answer 1

Lambdas in python share the variable scope they're created in. In your first case, the scope of the lambda is makeFun's. In your second case, it's the global i, which is 9 because it's a leftover from the loop.

That's what I understand of it anyway...

Answer 2

Technically, the lambda expression is closed over the i that's visible in the global scope, which is last set to 9. It's the same i being referred to in all 10 lambdas. For example,

i = 13
print b[3]()

In the makeFun function, the lambda closes on the i that's defined when the function is invoked. Those are ten different is.

Answer 3

Nice catch. The lambda in the list comprehension is seeing the same local i every time.

You can rewrite it as:

a = []
for i in range(10):
    a.append(makefun(i))

b = []
for i in range(10):
    b.append(lambda: i)

with the same result.

Answer 4

One set of functions (a) operates on the argument passed and the other (b) operates on a global variable which is then set to 9. Check the disassembly:

>>> import dis
>>> dis.dis(a[2])
  1           0 LOAD_DEREF               0 (i)
              3 RETURN_VALUE
>>> dis.dis(b[2])
  1           0 LOAD_GLOBAL              0 (i)
              3 RETURN_VALUE
>>>

Answer 5

As others have stated, scoping is the problem. Note that you can solve this by adding an extra argument to the lambda expression and assigning it a default value:

>> def makeFun(i): return lambda: i
... 
>>> a = [makeFun(i) for i in range(10)]
>>> b = [lambda: i for i in range(10)]
>>> c = [lambda i=i: i for i in range(10)]  # <-- Observe the use of i=i
>>> a[2](), b[2](), c[2]()
(2, 9, 2)

The result is that i is now explicitly placed in a scope confined to the lambda expression.