Generating functions inside loop with lambda expression in python

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 do lists a and b not behave in the save way?

For example:

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

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.

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.

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
>>>

To add some clarity (at least in my mind)

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

a uses makeFun(i) which is a function with an argument.

b uses lambda: i which is a function without arguments. The i it uses is very different from the previous

To make a and b equal we can make both functions to use no arguments:

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

Now both functions use the global i

>>> a[2]()
9
>>> b[2]()
9
>>> i=13
>>> a[2]()
13
>>> b[2]()
13

Or (more useful) make both use one argument:

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

I deliberately changed i with x where the variable is local. Now:

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

Success !

Generating functions inside loop with lambda expression in python

Related

Recent Posts