根据嵌套字典 python 的值获取键

Question

I've a nested-nested dictionary like:

d = {1: {490: {'angle': 0.9439585203011613,
       'distance': 0.10334710770757652,
       'gid': 19675.0,
       'index': 1.3627196218429294,
       'length': 0.3154139938341916},
 491: {'angle': 0.6009780397464144,
       'distance': 0.1914804717173812,
       'gid': 19679.0,
       'index': 0.8333609923300924,
       'length': 0.04090248086629692},
 492: {'angle': 0.9299210633144231,
       'distance': 0.1392632853868988,
       'gid': 19680.0,
       'index': 1.1225744054577835,
       'length': 0.053390056756461614},
 493: {'angle': 0.9499097709829176,
       'distance': 0.16460473640157167,
       'gid': 19681.0,
       'index': 1.6564270961214378,
       'length': 0.5419125887369485}}}

what I would like to get is the nested key (490, 491, 492, 493) where the nested-nested value of the index key is the lowest. Basically:

for k, v in d.items():
    for kk, vv in v.items():
        print(kk, vv['index'])

490 1.3627196218429294
491 0.8333609923300924
492 1.1225744054577835
493 1.6564270961214378

the value I want is 491 .

I'm stuck with the usage of min . I know the trick should be in the min function and the usage of the key parameter, but I'm getting confused how to use it correctly within this nested situation.

Answer 1

If d is your dictionary from the question:

mn = min(
    [(i, k) for i in d.values() for k in i],
    key=lambda k: k[0][k[1]]["index"],
)
print(mn[1])

Prints:

491

Answer 2

use pandas -

import pandas as pd
print(pd.DataFrame(d[1]).transpose().sort_values('index').head(1).index) #491

Answer 3

use the key as the index value of each sub dictionary.

min(d[1],key=lambda x:d[1][x]['index'])

note that d[1] is used here, the d dictionary only has one key.

generate the list of minimums on a full dictionary like this:

[min(subd,key=lambda x:subd[x]['index']) for subd in d.values()]

and as a dictionary:

{k:min(subd,key=lambda x:subd[x]['index']) for k,subd in d.items()}