How to convert list to tuple and search for max and min

I am currently learning python coding and I come across this qns on a learning site:

Create a function that takes a sequence of inputs (may be a list, a tuple, or just a bunch of inputs). The function should return the minimum and the maximum of the list.

This are some of the test values that are using:

minmax(5,4) 4,5

minmax(5,4,8,3,5,7,4) 3,8

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minmax([5,4,6]) 4,6

minmax(5.1,4.2,68.34,129.1,-90.3) -90.3,129.1

And I had tried doing it this way but when the parameter is a list, I can't seems to convert it into a tuple and find the max and min. Here is what I had tried:

def minmax(*a):

    b = tuple(a)

    minNum = min(b)

    maxNum = max(b)

    c = (minNum, maxNum)

    return c

When a list is taken in, the return result is ([5, 4, 6], [5, 4, 6])

def minmax(*a):

    if len(a) == 1:  # Single (list, tuple, or scalar) argument
        try:
            return minmax(*a[0])  # Expansion of sequence elements, if possible
        except TypeError:  # Case of minmax(42)
            pass  # The general code below handles this directly

    return (min(a), max(a))

>>> minmax(3, 5, 1, 10)
(1, 10)
>>> minmax([3, 5, 1, 10])
(1, 10)
>>> minmax((42, 123, -12))
(-12, 123)
>>> minmax(42)
42

This works in more cases than the built-in min() and max(), which do not work on a single scalar argument (min(42)).

>>> min(42)
TypeError: 'int' object is not iterable

It is however possible to write a simpler version that behaves like the built-in min() and max() (see my other answer, for instance).

This works by forcing min() to be given strictly more than 1 element, except in the special case of minmax(42), which calls min((42,)).

To be able to handle different ways of passing in arguments, you need to have some conditions to handle each case.

>>> def minmax(*a):
...     if len(a) == 1 and hasattr(a[0], '__getitem__'):
...         # handle a single sequence passed in
...         return min(a[0]), max(a[0])
...     # handle values passed in
...     return min(a), max(a)
...
>>> minmax(5, 4)
(4, 5)
>>> minmax(5, 4, 8, 3, 5, 7, 4)
(3, 8) 
>>> minmax([5, 4, 6])
(4, 6)
>>> minmax(5.1, 4.2, 68.34, 129.1, -90.3)
(-90.3, 129.1)

A simpler but slightly less powerful solution that mirrors my other solution is:

def minmax(*a):
    if len(a) == 1:  # Single (list or tuple) argument
        return (min(a[0]), max(a[0]))
    return minmax(a)  # Single tuple argument given to minmax()

This version forces min() to be given a single (list or tuple) argument.

It behaves like the built-in min() and max(): min(42) and minmax(42) both raise an exception:

>>> minmax(3, 5, 1, 10)
(1, 10)
>>> minmax([3, 5, 1, 10])
(1, 10)
>>> minmax((42, 123, -12))
(-12, 123)
>>> minmax(42)
TypeError: 'int' object is not iterable
>>> min(42)
TypeError: 'int' object is not iterable