如何在元组列表中查找用户定义的最小值和最大值

Question

I have a list of tuples with 2 integers each:

a_list=[(20, 1), (16, 0), (21, 0), (20, 0), (24, 0), (25, 1)]

What I am looking for is a way to find and return the tuple that has the smallest second item with the biggest first item. In the above list, it will be the tuple (24, 0) .

Answer 1

Your example is trivial enough for the following approach to work

>>> a_list=[(20, 1), (16, 0), (21, 0), (20, 0), (24, 0), (25, 1)]
>>> max(a_list, key=lambda e:(-e[-1],e[0]))
(24, 0)

It assumes the fact that

1. Tuples are compared lexicographic-ally based on indexes

   See doc :

   Sequence types also support comparisons. In particular, tuples and lists are compared lexicographically by comparing corresponding elements. This means that to compare equal, every element must compare equal and the two sequences must be of the same type and have the same length

2. Ordering of numbers revereses on change of sign

Answer 2

This works:

>>> a_list = [(20, 1), (16, 0), (21, 0), (20, 0), (24, 0), (25, 1)]
>>> a = min(x[1] for x in a_list)
>>> max(b for b in a_list if b[1] == a)
(24, 0)
>>>

Answer 3

Assuming that my understanding is correct, then here's what you need to do.

Find the minimum second value:

lowest = min(a_list, key=lambda t: t[1])[1]

Then find all the items that have that as the second item

having_lowest = filter(lambda t: t[1] == lowest, a_list)

Then find the one with the highest first number

having_max = max(having_lowest, key=lambda t: t[0])