Sort then reverse vs Sort(reverse=True)

python-sort-stability.py

# We want to sort a list by its second element in descending order.
# The example illustrates the difference in the results of different
# process of sorting in descending order.

# Sort in ascending order, then use list reverse
>>> a = [('A', 1), ('C', 5), ('A', 2), ('B', 3), ('B', 5)]
>>> a.sort(key=lambda x: x[1])
>>> print(a)
[('A', 1), ('A', 2), ('B', 3), ('C', 5), ('B', 5)]
>>> a.reverse()
>>> print(a)
[('B', 5), ('C', 5), ('B', 3), ('A', 2), ('A', 1)]

# Sort in descending order, using reverse=True
>>> b = [('A', 1), ('C', 5), ('A', 2), ('B', 3), ('B', 5)]
>>> b.sort(key=lambda x: x[1], reverse=True)
>>> print(b)
[('C', 5), ('B', 5), ('B', 3), ('A', 2), ('A', 1)]

# Note the difference in positions of the first two elements of the sorted list.
# Python sorting is stable. Hence ('C', 5) will appear before ('B', 5) after a sort() operation.

Fascinating! My intuition was that a reverse sort implied a reversal of the stability contract, but I am now of the opinion that it would be wrong to implement it that way.

showing key error in python 3.7