优雅的方法从元组列表中提取元组，具有最小的元素值

Question

I am having list of tuple from which I want the tuple with the minimum value at index 1 . For example, if my list is like:

a =[('a', 2), ('ee', 3), ('mm', 4), ('x', 1)]

I want the returned tuple to be ('x', 1) .

Currently I am using sorted function to get this result like:

min=sorted(a, key=lambda t: t[1])[0]

Is there better ways to do it? Maybe with min function?

Answer 1

You may use min() function with key parameter in order to find the tuple with minimum value in the list. There is no need to sort the list. Hence, your min call should be like:

>>> min(a, key=lambda t: t[1])
('x', 1)

Even better to use operator.itemgetter() instead of lambda expression ; as itemgetter are comparatively faster. In this case, the call to min function should be like:

>>> from operator import itemgetter

>>> min(a, key=itemgetter(1))
('x', 1)

Note: min() is a in-built function in Python. You should not be using it as a variable name.

Answer 2

This will also work:

min([x[::-1] for x in a])[::-1]
# ('x', 1)