Get the n-smallest values from a nested python list

Question

I have the following list:

l = [(('01001', '01003'), 4.15),
 (('01001', '01005'), 2.83),
 (('01001', '01007'), 3.32),
 (('01001', '01008'), 2.32),
 (('01001', '01009'), 9.32),
 (('01001', '01007'), 0.32),
 (('01002', '01009'), 6.83),
 (('01002', '01011'), 2.53),
 (('01002', '01009'), 6.83),
 (('01002', '01011'), 2.53),
 (('01002', '01009'), 6.83),
 (('01002', '01011'), 2.53),
 (('01003', '01013'), 20.50),
 (('01003', '01013'), 10.50),
 (('01003', '01013'), 0.50),
 (('01003', '01013'), 2.50),
 (('01003', '01013'), 20.30),
 (('01003', '01013'), 12.50),
 (('01003', '01013'), 1.50),
 (('01003', '01013'), 2.40)]

I would like to select the n-smallest values for the first element of this list ('01001', '01002', and '01003').

I was able to calcualte the min value with this code:

from itertools import groupby
from statistics import mean

{k:min(v for *_, v in v) for k,v in groupby(result_map, lambda x: x[0][0])}

but would like to get the 3 smallest values and the second column to be printed:

Expected outcome would be a dictionary like this:

{'01001': ['01007', '01008', '01005'], '01002': ['01011', '01009', '01013']  , '01003': ['01013', '01013', ''01013']}

Any help would be much appreciated!

Answer 1

{k:[e[0][1] for e in sorted(v, key = lambda x: x[1])][:n] for k,v in groupby(result_map, lambda x: x[0][0])}

this above is your provided code with groupby but modified a bit to compute n-smallest list instead of min.

From your question's example it wasn't clear if you want repeated elements in n-smallest list or not (second entry '01002': ['01011', '01009', '01013'] has no repetitions, but third '01003': ['01013', '01013', ''01013'] has repetitions in your example), so I provide second one-liner to solve task without repetitions:

{k:[e[0][1] for e in sorted({f[0][1] : f for f in v}.values(), key = lambda x: x[1])][:n] for k,v in groupby(result_map, lambda x: x[0][0])}

Full version of code can be found and tried online here!

Answer 2

The following should work:

d={i:sorted([k[0][1] for k in l if k[0][0]==i])[:3] for i in set([i[0][0] for i in l])}

print(d)

{'01001': ['01003', '01005', '01007'], '01002': ['01009', '01009', '01009'], '01003': ['01013', '01013', '01013']}