CODEWITHSUNDEEP
python
John
John

Reputation: 3

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

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])

Upvotes: 0

Views: 973

Answers (3)

Eric O. Lebigot
Eric O. Lebigot

Reputation: 94565

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,)).

Upvotes: 2

Eric O. Lebigot
Eric O. Lebigot

Reputation: 94565

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

Upvotes: 0

Jeff Mercado
Jeff Mercado

Reputation: 134491

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)

Upvotes: 1

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