Сравнение нескольких ключей во вложенном словаре

сначала вы хотите сгруппировать по «категории» (например, «ноги», «хвост», «мясо»), которая является ключом внутреннего словаря.

затем вы можете повторить новый словарь и вывести список результатов в желаемом формате:

from collections import defaultdict

in_dic = {"Cow": {"legs": "thick", "tail": "long", "milk": "good", "meat": "alright"},
          "Goat": {"legs": "slender", "tail": "short", "milk": "ok", "meat": "excellent"},
          "Fish": {"fins": "yes", "meat": "superb", "bones": "plenty"}}


result_by_category = defaultdict(list)

for animal, categories in in_dic.items():
    for category, value in categories.items():
        result_by_category[category].append((animal, value))

for category, values in result_by_category.items():
    for animal, value in values:
        print('{} {}: {}'.format(animal, category, value))
    print('')

вывод точно:

Cow legs: thick
Goat legs: slender

Cow tail: long
Goat tail: short

Cow milk: good
Goat milk: ok

Cow meat: alright
Goat meat: excellent
Fish meat: superb

Fish fins: yes

Fish bones: plenty

Вы можете сделать это, создав список с реорганизованными ключами/значениями, используя понимание:

animals = {"Cow": {"legs": "thick", "tail": "long", "milk": "good", "meat": "alright"},
          "Goat": {"legs": "slender", "tail": "short", "milk": "ok", "meat": "excellent"},
          "Fish": {"fins": "yes", "meat": "superb", "bones": "plenty"}}

triples   = [(trait,animal,value) for animal,traits in animals.items() for trait,value in traits.items()]

lineBreak = {min(triples)[0]}
for trait,animal,value in sorted(triples):
    if not(trait in lineBreak or lineBreak.add(trait)): print("")
    print(f"{animal} {trait}: {value}")

...

Fish bones: plenty

Fish fins: yes

Cow legs: thick
Goat legs: slender

Cow meat: alright
Fish meat: superb
Goat meat: excellent

Cow milk: good
Goat milk: ok

Cow tail: long
Goat tail: short

Вы также можете попробовать с grouby:

from itertools import groupby
from operator import itemgetter

nested_dict = {
    "Cow": {"legs": "thick", "tail": "long", "milk": "good", "meat": "alright"},
    "Goat": {"legs": "slender", "tail": "short", "milk": "ok", "meat": "excellent"},
    "Fish": {"fins": "yes", "meat": "superb", "bones": "plenty"},
}

refactored_items = (
    (k1, k2, v2) for k1, v1 in nested_dict.items() for k2, v2 in v1.items()
)
sorted_refactored_items = sorted(refactored_items, key=itemgetter(1))
for _, g in groupby(sorted_refactored_items, key=itemgetter(1)):
    print("\n".join(f"{a} {b}: {c}" for a, b, c in g))
    print("")

выход:

Fish bones: plenty

Fish fins: yes

Cow legs: thick
Goat legs: slender

Cow meat: alright
Goat meat: excellent
Fish meat: superb

Cow milk: good
Goat milk: ok

Cow tail: long
Goat tail: short