Sort python dictionary with value as list

Question

If we want to compare on the basis of all indices of the list and not just the 1st element. If lists are identical, then sort by key. Also length of the list is not known in advance. In that case how to sort the keys. Below is the example:

{'A': [5, 0, 0], 'B': [0, 2, 3], 'C': [0, 3, 2]}

output:

[A, C, B]

Explanation: A is at 1st position because at 0th index 5 is highest and rest is 0. C is at 2nd position because 2nd 1st index of C is 3 compared to 1st index of B. As you can see we need to compare all positions to sort it and we don't know the array size before hand.

I tried below code:

countPos = {'A': [5, 0, 0], 'B': [0, 2, 3], 'C': [0, 3, 2]}
res = sorted(countPos.items(), key=lambda x: ((-x[1][i]) for i in range(3)))

Getting an error for above code. Could someone help me on this?

Answer 1

I think got a solution, which worked. This might be naive. I encourage gurus to rectify me.

r = sorted(countPos.items(), key=lambda x: x[0])
r = dict(r)
res = sorted(r.items(), key=lambda x: x[1], reverse=True)

So, first sorted based on keys and then I sorted based on values in reverse order.