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字典中的内键。
我可以按照传统方式打开嵌套字典的包:
for outerkeys, innerdicts in nestedDict:
#but here I'm stuck on how to process multiple inner dictionaries
#to extract matching (and unmatching) keys.
答案 0 :(得分:2)
首先要按“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('')
输出准确无误:
牛腿:粗
山羊腿:细长
牛尾巴:长
山羊尾巴:短
奶牛:好
山羊奶:好的
牛肉:好吧
山羊肉:极佳
鱼肉:很棒
鱼鳍:是的
鱼骨头:很多
答案 1 :(得分:0)
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
答案 2 :(得分:0)
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