我有一个 dict ：
{'Logistic Regression': u' precision recall f1-score support\n\n APAR Information 0.74 1.00 0.85 844\nAffected Products and Versions 0.00 0.00 0.00 18\n Answer 0.00 0.00 0.00 30\n Applicable component levels 0.96 0.85 0.90 241\n Error description 0.48 0.56 0.52 754\n Local fix 0.89 0.03 0.06 266\n Modules/Macros 0.96 0.87 0.91 326\n Problem 0.00 0.00 0.00 63\n Problem summary 0.51 0.73 0.60 721\n Related information 0.00 0.00 0.00 22\n Resolving The Problem 0.00 0.00 0.00 60\n Temporary fix 0.00 0.00 0.00 32\n circumvenion 0.00 0.00 0.00 124\n component 0.00 0.00 0.00 49\n temporary_fix 0.00 0.00 0.00 2\n\n micro avg 0.64 0.64 0.64 3552\n macro avg 0.30 0.27 0.26 3552\n weighted avg 0.60 0.64 0.58 3552\n'}
或
precision recall f1-score support
APAR Information 0.74 1.00 0.85 844
Affected Products and Versions 0.00 0.00 0.00 18
Answer 0.00 0.00 0.00 30
Applicable component levels 0.96 0.85 0.90 241
Error description 0.48 0.56 0.52 754
Local fix 0.89 0.03 0.06 266
Modules/Macros 0.96 0.87 0.91 326
Problem 0.00 0.00 0.00 63
Problem summary 0.51 0.73 0.60 721
Related information 0.00 0.00 0.00 22
Resolving The Problem 0.00 0.00 0.00 60
Temporary fix 0.00 0.00 0.00 32
circumvenion 0.00 0.00 0.00 124
component 0.00 0.00 0.00 49
temporary_fix 0.00 0.00 0.00 2
micro avg 0.64 0.64 0.64 3552
macro avg 0.30 0.27 0.26 3552
weighted avg 0.60 0.64 0.58 3552
我想将此字典转换为嵌套字典，例如，
{'Logistic Regression':
{'APAR Information':'0.74','1.00','0.85','844'},
{'Affected Products and Versions':'0.00','0.00','0.00','18'}
.
.
.}
如何实现？可以通过dict内置函数完成吗？
答案 0 :(得分：2)
这是一种方法。
演示：
d = {'Logistic Regression': u' precision recall f1-score support\n\n APAR Information 0.74 1.00 0.85 844\nAffected Products and Versions 0.00 0.00 0.00 18\n Answer 0.00 0.00 0.00 30\n Applicable component levels 0.96 0.85 0.90 241\n Error description 0.48 0.56 0.52 754\n Local fix 0.89 0.03 0.06 266\n Modules/Macros 0.96 0.87 0.91 326\n Problem 0.00 0.00 0.00 63\n Problem summary 0.51 0.73 0.60 721\n Related information 0.00 0.00 0.00 22\n Resolving The Problem 0.00 0.00 0.00 60\n Temporary fix 0.00 0.00 0.00 32\n circumvenion 0.00 0.00 0.00 124\n component 0.00 0.00 0.00 49\n temporary_fix 0.00 0.00 0.00 2\n\n micro avg 0.64 0.64 0.64 3552\n macro avg 0.30 0.27 0.26 3552\n weighted avg 0.60 0.64 0.58 3552\n'}
result = {}
for i, v in enumerate(d["Logistic Regression"].splitlines()):
if i == 0:
continue
val = v.strip().split(" ")
if val[0]:
result[val[0]] = " ".join(val[1:]).split()
for k, v in result.items():
print(k)
print(v)
输出：
weighted avg
[u'0.60', u'0.64', u'0.58', u'3552']
Local fix
[u'0.89', u'0.03', u'0.06', u'266']
Affected Products and Versions
[u'0.00', u'0.00', u'0.00', u'18']
component
[u'0.00', u'0.00', u'0.00', u'49']
Resolving The Problem
[u'0.00', u'0.00', u'0.00', u'60']
Error description
[u'0.48', u'0.56', u'0.52', u'754']
Problem summary
[u'0.51', u'0.73', u'0.60', u'721']
macro avg
[u'0.30', u'0.27', u'0.26', u'3552']
Related information
[u'0.00', u'0.00', u'0.00', u'22']
Applicable component levels
[u'0.96', u'0.85', u'0.90', u'241']
micro avg
[u'0.64', u'0.64', u'0.64', u'3552']
Answer
[u'0.00', u'0.00', u'0.00', u'30']
APAR Information
[u'0.74', u'1.00', u'0.85', u'844']
Problem
[u'0.00', u'0.00', u'0.00', u'63']
Modules/Macros
[u'0.96', u'0.87', u'0.91', u'326']
temporary_fix
[u'0.00', u'0.00', u'0.00', u'2']
circumvenion
[u'0.00', u'0.00', u'0.00', u'124']
Temporary fix
[u'0.00', u'0.00', u'0.00', u'32']
答案 1 :(得分：0)
您可以使用第三方熊猫通过pd.read_fwf（“固定宽度格式”）转换为数据框。您的数据混乱，您可能需要编写逻辑来计算列宽或手动添加它们。给定输入字典d：
from io import StringIO
import pandas as pd
df = pd.read_fwf(StringIO(d['Logistic Regression']), widths=[30, 11, 10, 10, 10])\
.dropna().rename(columns={'Unnamed: 0': 'index'}).set_index('index')
print(df)
precision recall f1-score support
index
APAR Information 0.74 1.00 0.85 844.0
Affected Products and Versions 0.00 0.00 0.00 18.0
Answer 0.00 0.00 0.00 30.0
Applicable component levels 0.96 0.85 0.90 241.0
Error description 0.48 0.56 0.52 754.0
Local fix 0.89 0.03 0.06 266.0
Modules/Macros 0.96 0.87 0.91 326.0
Problem 0.00 0.00 0.00 63.0
Problem summary 0.51 0.73 0.60 721.0
Related information 0.00 0.00 0.00 22.0
Resolving The Problem 0.00 0.00 0.00 60.0
Temporary fix 0.00 0.00 0.00 32.0
circumvenion 0.00 0.00 0.00 124.0
component 0.00 0.00 0.00 49.0
temporary_fix 0.00 0.00 0.00 2.0
micro avg 0.64 0.64 0.64 3552.0
macro avg 0.30 0.27 0.26 3552.0
weighted avg 0.60 0.64 0.58 3552.0
然后使用字典理解：
res = {'Logistic Regression': {idx: df.loc[idx].tolist() for idx in df.index}}
print(res)
{'Logistic Regression':
{'APAR Information': [0.74, 1.0, 0.85, 844.0],
'Affected Products and Versions': [0.0, 0.0, 0.0, 18.0],
'Answer': [0.0, 0.0, 0.0, 30.0],
'Applicable component levels': [0.96, 0.85, 0.9, 241.0],
'Error description': [0.48, 0.56, 0.52, 754.0],
'Local fix': [0.89, 0.03, 0.06, 266.0],
'Modules/Macros': [0.96, 0.87, 0.91, 326.0],
'Problem': [0.0, 0.0, 0.0, 63.0],
'Problem summary': [0.51, 0.73, 0.6, 721.0],
'Related information': [0.0, 0.0, 0.0, 22.0],
'Resolving The Problem': [0.0, 0.0, 0.0, 60.0],
'Temporary fix': [0.0, 0.0, 0.0, 32.0],
'circumvenion': [0.0, 0.0, 0.0, 124.0],
'component': [0.0, 0.0, 0.0, 49.0],
'macro avg': [0.3, 0.27, 0.26, 3552.0],
'micro avg': [0.64, 0.64, 0.64, 3552.0],
'temporary_fix': [0.0, 0.0, 0.0, 2.0],
'weighted avg': [0.6, 0.64, 0.58, 3552.0]}}