I have problem about sorting my dict. My code is:
x = {('S', 'A'): (5, 8), ('S', 'B'): (11, 17), ('S', 'C'): (8, 14)}
sort_x = sorted(x.items(), key=lambda kv: kv[1])
print sort_x
sort_x_dict = dict(sort_x)
print sort_x_dict
[(('S', 'A'): (5, 8)), (('S', 'C'): (8, 14)), (('S', 'B'): (11, 17))]
{('S', 'A'): (5, 8), ('S', 'B'): (11, 17), ('S', 'C'): (8, 14)}
It's apparent from your print statements that you're using Python 2.7, and yet dicts are only guaranteed to be ordered since Python 3.7. You can either upgrade to Python 3.7 for your exact code to work, or switch to collections.OrderedDict in place of dict:
from collections import OrderedDict
sort_x = sorted(x.items(), key=lambda kv: kv[1])
print(sort_x)
sort_x_dict = OrderedDict(sort_x)
print(sort_x_dict)
{('S', 'A'): (5, 8), ('S', 'C'): (8, 14), ('S', 'B'): (11, 17)} not ([(('S', 'A'), (5, 8)), (('S', 'C'), (8, 14)), (('S', 'B'), (11, 17))])
OrderedDict subclasses dict so you can use it like a dict because it really a dict. It's only when you print it that its __repr__ method would return a representation string that looks more like a list of tuples. But for all intents and purposes you can use an OrderedDict just like a regular dict. If you still want to use a regular dict to achieve what you want, again, you would have to upgrade to Python 3.7+.
dictaren't ordered, so sorting them makes no sense. Maybe you want to useOrederedDict?dicts do retain order sodict(sort_x)loses the order ofsort_x.collections.OrderedDictwill keep order.sort_x:{('S', 'A'): (5, 8), ('S', 'C'): (8, 14), ('S', 'B'): (11, 17)}.