This is the format of nested list:
[['A', 0],['B', 0], ['V', 0], ['D', 0], ['E', 0], ['F', 0], ['G', 1], ['H', 1], ['I', 1], ['J', 3], ['K', 0]]
I have nested list in such format, I need to get top 3 nested lists which have max nested list integer value such as my result should be ['J', 3] , ['I', 1] , ['H', 1] I have tried to use the nested loop where i can get the max integer but struck at getting the string of those max 3 integers
Can someone please help me out here
I would do this:
First order list by number in nested list. In lambda function I return the value that has to be used for ordenation (2nd element of nested list):
a = [['A', 0],['B', 0], ['V', 0], ['D', 0], ['E', 0], ['F', 0], ['G', 1], ['H', 1], ['I', 1], ['J', 3], ['K', 0]]
sorted_list = sorted(a, key=lambda n: n[1], reverse=True)
# [['J', 3], ['G', 1], ['H', 1], ['I', 1], ['A', 0], ['B', 0], ['V', 0], ['D', 0], ['E', 0], ['F', 0], ['K', 0]]
With this I could slice 3 first elements:
sorted_list = sorted_list[:3]
# [['J', 3], ['G', 1], ['H', 1]]
And finally get a list with only letter (string):
[l[0] for l in sorted_list]
# ['J', 'G', 'H']
All at once:
[n[0] for n in sorted(a, key=lambda n: n[1], reverse=True)[:3]]
Adding to @salpreh's answer, you could also use operator.itemgetter with key for a slight speed increase:
from operator import itemgetter
a = [['A', 0],['B', 0], ['V', 0], ['D', 0], ['E', 0], ['F', 0], ['G', 1], ['H', 1], ['I', 1], ['J', 3], ['K', 0]]
print(sorted(a, key=itemgetter(0), reverse=True))
# [['V', 0], ['K', 0], ['J', 3], ['I', 1], ['H', 1], ['G', 1], ['F', 0], ['E', 0], ['D', 0], ['B', 0], ['A', 0]]
This question offers some performance stats between the two.