Python sorted() or its alternative to get/return resulting key lambda value by which it was sorted-CodePudding

Suppose I have a dictionary/list/anything. I am sorting this set according to some function (not directly dependent on the value, square is given as an example):

dict_a = {('a', 3), ('b', 2), ('c', 1)}

def temp(x):
    return x**2

print(sorted(dict_a, key=lambda item:temp(item[1])))

Is there any way/syntax to get/return the value of the key obtained from a call to lambda (by which the resulting list was sorted), in order not to calculate it twice later, e.g.:

[(1, ('c', 1)), (4, ('b', 2)), (9, ('a', 3))]

rather than

[('c', 1), ('b', 2), ('a', 3)]

Preferably in a single line of code and only standard libraries. Does Python have capability to do that? Thanks!

CodePudding user response：

You could already map it before sorting:

>>> sorted(map(lambda x: (temp(x[1]), x), dict_a))
[(1, ('c', 1)), (4, ('b', 2)), (9, ('a', 3))]
>>>

Or with a list comprehension:

>>> sorted([(temp(x[1]), x) for x in dict_a])
[(1, ('c', 1)), (4, ('b', 2)), (9, ('a', 3))]
>>>

CodePudding user response：

Just do the transform beforehand and sort that:

dict_a = {('a', 3), ('b', 2), ('c', 1)}

def temp(x):
    return x**2

expected = [(1, ('c', 1)), (4, ('b', 2)), (9, ('a', 3))]
transformed = [(temp(item[1]), item) for item in dict_a]
assert sorted(transformed) == expected

Note that you typoed ('c', 3) in your question and that dict_a is not a dict, but a set of tuples..