比较嵌套字典中的多个键

Question

假设我有以下嵌套字典：

{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}}

我的目标是比较外键（牛，山羊和鱼），并检查它们的内键是否匹配。 理想情况下，我应该得到：

Cow legs: thick 
Goat legs: slender

Cow tail: long
Goat tail: short

Cow milk: good 
Goat milk: ok

Cow meat: alright
Goat meat: excelent
Fish meat: superb

Fish fins: yes
Fish bones: plenty

对我来说，问题是我还无法弄清楚如何比较多个词典中的内键。

我可以按照传统方式打开嵌套字典的包装：

for outerkeys, innerdicts in nestedDict:
      #but here I'm stuck on how to process multiple inner dictionaries
      #to extract matching (and unmatching) keys.

Answer 1

首先，您要按“类别”（例如“腿”，“尾巴”，“肉”）分组，这是内部词典的关键。

然后您可以迭代新词典并以所需格式列出输出：

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('')

输出是确切的：

牛腿：粗
山羊腿：纤细

牛尾巴：长
山羊尾巴：短

牛奶：好
山羊奶：可以

牛肉：好
山羊肉：极好
鱼肉：精湛

鱼鳍：是的

鱼骨头：充足

Answer 2

您可以通过使用理解力构建带有重组键/值的列表来做到这一点：

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

Answer 3

您也可以尝试使用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("")

输出：

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