Question

I have two lists:

a,b=[1,2],[33,44]

I want to explore both their minimum. But

>>> min(a,b)

returns [1, 2] as min()

With more than one argument, return the smallest of the arguments.

Same happens if I use map() as
map(min,a,b)

is mostly equivalent to:

[f(x1, x2) for x1, x2 in zip(sequence1, sequence2)]

as already stated in this answer.

>>> map(min,[a,b])
[1, 33]

gives me what I want but I don’t really understand why. Can someone explain?

Answer 1

The answer is in Python map documentation:

Apply function to every item of iterable and return a list of 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.

When you call:

map(min, a, b)

You are actually passing two iterables to map. This successively calls min(1, 33) and min(2, 44), thus returning [1, 2].

However, in:

map(min, [a, b])

There is a single iterable, and map calls min on each element of the sequence:

• First calling min([1, 2]) which yields 1
• Then calling min([33, 44]) which yields 33

The result, as expected, is [1, 33].