Question

我有以下代码：

import numpy as np
import pandas as pd

class my_class:
    def __init__(self, vector, par1, par2, par3):
        self.vector = vector  # this is a 1D numpy object.
        self.label = '({}, {}, {})'.format(par1, par2, par3)

obj1 = my_class(np.array([0, 1, 2, 3]), 4, -8, 7)
obj2 = my_class(np.array([6, 7, 8, 9]), 10, 4, 15)
obj3 = my_class(np.array([-1, -2, 3, -4]), 9, 3, 6)
obj4 = my_class(np.array([-10, -15, -20, 3]), 1, -2, 6)

my_list_of_objects = [obj1, obj2, obj3, obj4]

df = pd.DataFrame([o.__dict__ for o in my_list_of_objects])
print(df)

这是我得到的当前结果：

               vector        label
0        [0, 1, 2, 3]   (4, -8, 7)
1        [6, 7, 8, 9]  (10, 4, 15)
2     [-1, -2, 3, -4]    (9, 3, 6)
3  [-10, -15, -20, 3]   (1, -2, 6)

我想得到下面的熊猫数据框：

   (4, -8, 7)   (10, 4, 15)   (9, 3, 6)   (1, -2, 6)
0       0            6           -1          -10
1       1            7           -2          -15
2       2            8            3          -20
3       3            9           -4            3

您将在类中看到，属性标签具有需要使用的相应列的名称。

Answer 1

尝试一下：

df.set_index('label').T.apply(pd.Series.explode).reset_index(drop=True)

输出：

label (4, -8, 7) (10, 4, 15) (9, 3, 6) (1, -2, 6)
0              0           6        -1        -10
1              1           7        -2        -15
2              2           8         3        -20
3              3           9        -4          3

Answer 2

使用unstack并从中构造一个新的数据框

df_final = pd.DataFrame.from_dict(df.set_index('label').unstack().droplevel(0)
                                                                 .to_dict())

或从label和vector构造一个字典，并将其传递给数据框构造器

df_final = pd.DataFrame.from_dict(dict(df[['label', 'vector']].to_numpy()))

Out[267]:
   (4, -8, 7)  (10, 4, 15)  (9, 3, 6)  (1, -2, 6)
0           0            6         -1         -10
1           1            7         -2         -15
2           2            8          3         -20
3           3            9         -4           3

从对象列表创建熊猫数据框

2 个答案: