Why is 'key' function appended to the end of my OrderedDict when sorting?

Question

I am using python 3.6.5 and I am sorting an OrderedDict, e.g. tmp.py:

from collections import OrderedDict
d = OrderedDict()
d[6] = 'a'
d[5] = 'b'
d[3] = 'c'
d[4] = 'd'
print(d)
print("keys : {}".format(d.keys()))
d = OrderedDict(sorted(d.items()), key=lambda t: t[1])
print(d)
print("keys : {}".format(d.keys()))

When I run tmp.py, I get :

OrderedDict([(6, 'a'), (5, 'b'), (3, 'c'), (4, 'd')])
keys : odict_keys([6, 5, 3, 4])
OrderedDict([(3, 'c'), (4, 'd'), (5, 'b'), (6, 'a'), ('key', <function <lambda> at 0x2ab444506bf8>)])
keys : odict_keys([3, 4, 5, 6, 'key'])

Clearly the process of sorting has appended the key() function to my new OrderedDict. I believe that I am sorting this in the same fashion prescribed in this post.

QUESTION : Why does this happen and how to I properly sort an OrderedDict?

Because you're passing `key=...` to `OrderedDict`, not to `sorted`. — deceze, Oct 11 '18 at 14:55
This question has already been answered here: https://stackoverflow.com/questions/8031418/how-to-sort-ordereddict-of-ordereddict-python — Jamie Winspear, Oct 11 '18 at 14:57

score 2 · Accepted Answer · answered Oct 11 '18 at 14:58

2

Your brackets are in the wrong place.

d = OrderedDict(sorted(d.items()), key=lambda t: t[1])

should be:

d = OrderedDict(sorted(d.items(), key=lambda t: t[1]))

answered Oct 11 '18 at 14:58

Rob Bricheno

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very good. I think you are right. – MEdwin Oct 11 '18 at 15:00

Why is 'key' function appended to the end of my OrderedDict when sorting?

1 Answers1