python map function with min argument and two lists

Question

Im having trouble figuring out why this map function is producing the output it's producing. The code is as follows:

    L1 = [1, 28, 36]
    L2 = [2, 57, 9]
    print map(min, L1, L2)

output is: [1, 28, 9]

I understand it took the min values from the first list, but why did it not take the 2 from the second, and instead took the 9. Appreciate any feedback, thank you.

Answer 1

The result is made up from

[min(L1[0], L2[0]), min(L1[1], L2[1]), min(L1[2], L2[2])]

so min is being called on each pair of values to construct the new list

Answer 2

map(min, L1, L2) means roughly this:

[min(L1[0], L2[0]), min(L1[1], L2[1]), min(L1[2], L2[2])]

So the min of [1,2] (first element of each list) is 1, min of [28,57] is 28, and min of [36,9] is 9.

You probably wanted map(min, [L1, L2]) instead.

Answer 3

The statement:

map(min, L1, L2)

compares each elements of the two lists with the same index.

Thus,
It performs:

list = []
list.append(min(1,2)) #1
list.append(min(28,57)) #28
list.append(min(36,9)) #9
print list

leading to the output [1, 28, 9]

Answer 4

Let's take a look at the document:

map(function, iterable, ...)

Return an iterator that applies function to every item of iterable, yielding the results. If additional iterable arguments are passed, function must take that many arguments and is applied to the items from all iterables in parallel. ...

Here, in parallel means items of same index from different iterable are passed to function for each index.