Question

我下面有一个daframe。我想将其转换为字典。但是我不希望列名作为键。

data = {'0':[0.039169993,0.023344912], '1':[0.17865846,0.01093025],'2':[0.039170124,0.023344917], '3':[0.17865846,0.01093025],'4':[0.039170124,0.023344917]}
df= pd.DataFrame(data)



         0.0            1.0          2.0        3.0          4.0
0  0.039169993   0.17865846  0.039170124   0.17865846  0.039170124
1  0.023344912   0.01093025  0.023344917   0.01093025  0.023344917

**Desired Result**:
    {{0: 0.039169993, 1:0.023344912},
     {0: 0.17865846, 1:0.01093025},
     {0: 0.039170124, 1:0.023344917},
     {0: 0.17865846, 1:0.01093025},
     {0:0.039170124, 1:0.023344917}}

MyAttempt：

df.to_dict()
{'0': {0: 0.039169993, 1: 0.023344912},
 '1': {0: 0.17865846, 1: 0.01093025},
 '2': {0: 0.039170124, 1: 0.023344917},
 '3': {0: 0.17865846, 1: 0.01093025},
 '4': {0: 0.039170124, 1: 0.023344917}}

我不希望列名作为键。有可能吗？

Answer 1

期望的结果具有一组词典的 format ，但是您不能拥有一组词典，因为词典不可散列，但是您可以有一个列表。

import pandas as pd

data = {'0': [0.039169993, 0.023344912], '1': [0.17865846, 0.01093025], '2': [0.039170124, 0.023344917],
        '3': [0.17865846, 0.01093025], '4': [0.039170124, 0.023344917]}
df = pd.DataFrame(data)

result = list(df.to_dict().values())

print(result)

输出

[{0: 0.039170124, 1: 0.023344917}, {0: 0.039169993, 1: 0.023344912}, {0: 0.17865846, 1: 0.01093025}, {0: 0.17865846, 1: 0.01093025}, {0: 0.039170124, 1: 0.023344917}]

Answer 2

您可以使用此：

df.T.to_dict(orient='records')

[{0: 0.039169993, 1: 0.023344911999999999},
 {0: 0.17865845999999999, 1: 0.010930250000000001},
 {0: 0.039170124000000001, 1: 0.023344917},
 {0: 0.17865845999999999, 1: 0.010930250000000001},
 {0: 0.039170124000000001, 1: 0.023344917}]

Answer 3

您可以使用transpose或T和.to_dict(orient='records')获得所需的输出，例如：

df.T.to_dict(orient='records')

将数据框转换为字典，但不将列名称作为python中的键

3 个答案: