Python 3.x 将df转换为特定的熊猫词典

我有一个df，如下所示

Params                       Value
teachers                        49
students                       289
R                              3.7
holidays                       165
OS                              18
Em_from   2020-02-29T20:00:00.000Z
Em_to     2020-03-20T20:00:00.000Z
Em_F                             3
Em_C                             2
sC_from   2020-03-31T20:00:00.000Z
sC_to     2020-05-29T20:00:00.000Z
sC_F                            25
sC_C                            31

根据上面的df，我想将其转换为字典的字典，如下所示

格言：

---

panda的数据帧有一个to_json方法

这里有多个示例，但一般流程是这样的，假设您有一个名为

df

的数据帧：

import json
import pandas as pd

parsed = df.to_json()
df_json = json.loads(json_df)

阅读文档以查看更多示例和可能需要修改的不同参数。

使用：

s = df['Params'].str.split('_')
m = s.str.len().eq(1)

d1 = df[m].set_index('Params')['Value'].to_dict()
d2 = df[~m].assign(Params=s.str[-1]).agg(tuple, axis=1)\
           .groupby(s.str[0]).agg(lambda s: dict(s.tolist())).to_dict()

dct = {**d1, **d2}

结果:

{'Em': {'C': '2',
        'F': '3',
        'from': '2020-02-29T20:00:00.000Z',
        'to': '2020-03-20T20:00:00.000Z'},
 'OS': '18',
 'R': '3.7',
 'holidays': '165',
 'sC': {'C': '31',
        'F': '25',
        'from': '2020-03-31T20:00:00.000Z',
        'to': '2020-05-29T20:00:00.000Z'},
 'students': '289',
 'teachers': '49'}

---

请始终尝试以可复制的方式提供数据，更多的人将能够尝试这个问题

数据集

Params = ['teachers','students','R','holidays','OS','Em_from','Em_to','Em_F','Em_C','sC_from','sC_to','sC_F','sC_C']                       

Value = ['49','289','3.7','165','18','2020-02-29T20:00:00.000Z','2020-03-20T20:00:00.000Z','3','2','2020-03-31T20:00:00.000Z','2020-05-29T20:00:00.000Z','25','31']

df = pd.DataFrame(zip(Params,Value),columns=["col1","col2"])

{'teachers': '49',
 'students': '289',
 'R': '3.7',
 'holidays': '165',
 'OS': '18',
 'Em': {'from': '2020-02-29T20:00:00.000Z',
  'to': '2020-03-20T20:00:00.000Z',
  'F': '3',
  'C': '2'},
 'sC': {'from': '2020-03-31T20:00:00.000Z',
  'to': '2020-05-29T20:00:00.000Z',
  'F': '25',
  'C': '31'}}

你可以这样做

d = {}
for lst in df.values:
    for k,v in zip(lst[0:],lst[1:]):
        if any(name in k for name in ('Em_from', 'sC_from')):d[k.split('_')[0]] = {k.split('_')[1]:v}
        elif any(name in k for name in ('Em_to', 'Em_F','Em_C','sC_to','sC_F','sC_C')):d[k.split('_')[0]][k.split('_')[1]] = v
        else:d[k] = v

输出

Params = ['teachers','students','R','holidays','OS','Em_from','Em_to','Em_F','Em_C','sC_from','sC_to','sC_F','sC_C']                       

Value = ['49','289','3.7','165','18','2020-02-29T20:00:00.000Z','2020-03-20T20:00:00.000Z','3','2','2020-03-31T20:00:00.000Z','2020-05-29T20:00:00.000Z','25','31']

df = pd.DataFrame(zip(Params,Value),columns=["col1","col2"])

{'teachers': '49',
 'students': '289',
 'R': '3.7',
 'holidays': '165',
 'OS': '18',
 'Em': {'from': '2020-02-29T20:00:00.000Z',
  'to': '2020-03-20T20:00:00.000Z',
  'F': '3',
  'C': '2'},
 'sC': {'from': '2020-03-31T20:00:00.000Z',
  'to': '2020-05-29T20:00:00.000Z',
  'F': '25',
  'C': '31'}}