sort dict or list by second value of the tuple and then by the first one

You could do something as follows:

output = sorted(l, key=lambda x: (x[1], [-ord(letter) for letter in x[0]]), reverse=True)

output will contain the following:

[('oo', 11), ('aa', 10), ('ee', 10)]

We are keeping the reverse order (descending), but we are passing a list as the second option (which are compared lexicographically by Python too) for sorting which will have the negative ASCII values for each letter. This way, -ord('a') (-97) is greater than -ord('e') (-101)

I guess you should remove the reverse=True option, so that values would be sorted in the "right" order.

>>> l = [('ee',10), ('oo',11), ('aa', 10)]
>>> sorted(l, key=lambda k: (k[1], k[0]), reverse=False)
[('aa', 10), ('ee', 10), ('oo', 11)]