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)]

Answer 1

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 !