Question

我想看看是否有人可以提出解决我遇到的问题的想法。

我想尝试以某种方式翻转字典的子列表和列表。以下内容应介绍我的目标。

olddict = {
  'Cat1': {
    'a': 0.9628883469409395,
    'b': 0.2215033887861984,
    'c': 0.011300919842312747,
    'd': 0.006507874528179097,
    'e': 0.0038535645472061657},
  'Cat2': {'a': 0.9984095005831831,
    'b': 0.7831176833025262,
    'c': 0.2980289093298292,
    'd': 0.04152023948978264,
    'e': 0.9772639691714836},
  'Cat3': {'a': 0.467005076142132,
    'b': 1.0,
    'c': 0.9810315430520035,
    'd': 0.37869041728517233,
    'e': 0.5812393632996296}}

变成这样：

newdict = {
  'a': {
    'Cat1': 0.9628883469409395,
    'Cat2': 0.9984095005831831,
    'Cat3': 0.467005076142132},
  'b': {
    'Cat1': 0.2215033887861984,
    'Cat2': 0.7831176833025262,
    'Cat3': 1.0},
  'c': {
    'Cat1': 0.011300919842312747,
    'Cat2': 0.2980289093298292,
    'Cat3': 0.9810315430520035},
  'd': {
    'Cat1': 0.9628883469409395,
    'Cat2': 0.04152023948978264,
    'Cat3': 0.37869041728517233},
  'e': {
    'Cat1': 0.0038535645472061657,
    'Cat2': 0.9772639691714836,
    'Cat3': 0.5812393632996296}}

以此，我可以创建新字典的总体布局。但是，我也未能添加值。

newdict = {}
for key, value in olddict.items():
    for string,v in value.items():
        newdict.setdefault(string, []).append(key)

我将不胜感激。提前非常感谢！

Answer 1

newdict = defaultdict(dict)
for cat, letters in olddict.items():
  for letter, value in letters.items():
    newdict[letter][cat] = value

Answer 2

尝试使用defaultdict。他们太神奇了。

https://docs.python.org/3/library/collections.html#collections.defaultdict

>>> from collections import defaultdict
>>> from pprint import pprint
>>>
>>> newdict = defaultdict(dict)
>>> for key, value in olddict.items():
...     for string, v in value.items():
...             newdict[string][key] = v
>>>
>>> pprint(newdict)
defaultdict(<class 'dict'>,
            {'a': {'Cat1': 0.9628883469409395,
                   'Cat2': 0.9984095005831831,
                   'Cat3': 0.467005076142132},
             'b': {'Cat1': 0.2215033887861984,
                   'Cat2': 0.7831176833025262,
                   'Cat3': 1.0},
             'c': {'Cat1': 0.011300919842312747,
                   'Cat2': 0.2980289093298292,
                   'Cat3': 0.9810315430520035},
             'd': {'Cat1': 0.006507874528179097,
                   'Cat2': 0.04152023948978264,
                   'Cat3': 0.37869041728517233},
             'e': {'Cat1': 0.0038535645472061657,
                   'Cat2': 0.9772639691714836,
                   'Cat3': 0.5812393632996296}})

Answer 3

import pandas as pd 
olddict = {'Cat1': {'a': 0.9628883469409395,
  'b': 0.2215033887861984,
  'c': 0.011300919842312747,
  'd': 0.006507874528179097,
  'e': 0.0038535645472061657},
'Cat2': {'a': 0.9984095005831831,
  'b': 0.7831176833025262,
  'c': 0.2980289093298292,
  'd': 0.04152023948978264,
  'e': 0.9772639691714836},
'Cat3': {'a': 0.467005076142132,
  'b': 1.0,
  'c': 0.9810315430520035,
  'd': 0.37869041728517233,
  'e': 0.5812393632996296}} 
df=pd.DataFrame(olddict)
df = df.T 
df.to_dict()

输出：

{'a': {'Cat1': 0.9628883469409395,
  'Cat2': 0.9984095005831831,
  'Cat3': 0.467005076142132},
 'b': {'Cat1': 0.2215033887861984, 'Cat2': 0.7831176833025262, 'Cat3': 1.0},
 'c': {'Cat1': 0.011300919842312747,
  'Cat2': 0.2980289093298292,
  'Cat3': 0.9810315430520035},
 'd': {'Cat1': 0.006507874528179097,
  'Cat2': 0.04152023948978264,
  'Cat3': 0.37869041728517233},
 'e': {'Cat1': 0.0038535645472061657,
  'Cat2': 0.9772639691714836,
  'Cat3': 0.5812393632996296}}

Answer 4

您使用setdefault的解决方案几乎是正确的，但是您使用的是字典，而不是列表，因此您应该这样做：

olddict = {'Cat1': {'a': 0.9628883469409395,
  'b': 0.2215033887861984,
  'c': 0.011300919842312747,
  'd': 0.006507874528179097,
  'e': 0.0038535645472061657},
'Cat2': {'a': 0.9984095005831831,
  'b': 0.7831176833025262,
  'c': 0.2980289093298292,
  'd': 0.04152023948978264,
  'e': 0.9772639691714836},
'Cat3': {'a': 0.467005076142132,
  'b': 1.0,
  'c': 0.9810315430520035,
  'd': 0.37869041728517233,
  'e': 0.5812393632996296}}

expected = {'a': {'Cat1': 0.9628883469409395,
  'Cat2': 0.9984095005831831,
  'Cat3': 0.467005076142132},
'b': {'Cat1': 0.2215033887861984,
  'Cat2': 0.7831176833025262,
  'Cat3': 1.0},
'c': {'Cat1': 0.011300919842312747,
  'Cat2': 0.2980289093298292,
  'Cat3': 0.9810315430520035},
'd': {'Cat1': 0.006507874528179097,
  'Cat2': 0.04152023948978264,
  'Cat3': 0.37869041728517233},
'e': {'Cat1': 0.0038535645472061657,
  'Cat2': 0.9772639691714836,
  'Cat3': 0.5812393632996296}}

newdict = {}
for key, value in olddict.items():
    for s, v in value.items():
        newdict.setdefault(s, {})[key] = v
print(newdict == expected)
# True

反向字典，键和值作为子列表

4 个答案: