计算单元格为列表的熊猫数据框的平均值

Question

假设我有以下熊猫数据框

import pandas as pd
import numpy as np
df= pd.DataFrame(np.nan, columns =["A","B","C"], index =np.arange(5))
df=df.astype(object)
for c in list(df):
    for i in df.index.values:
        df.at[i, c]=np.arange(5).tolist()

这会导致 df 的单元格是 numpy 数组

df
Out[16]: 
                 A                B                C
0  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
1  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
2  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
3  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
4  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]

我想计算数据框的平均值，但它不起作用，因为每个单元格都被视为一个字符串。例如，

type(df.loc[0][0])
Out[19]: list

因此，如果我计算它的平均值，它会返回 nan

df["Average"]= df.mean(axis=1)

df
Out[21]: 
                 A                B                C  Average
0  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]      NaN
1  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]      NaN
2  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]      NaN
3  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]      NaN
4  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]      NaN

我的问题是，如何将此 df 转换回我可以使用的数值？

Answer 1

您可能希望按照上述方式重组您的数据框。但是要使用您所拥有的，假设您想要数据框中每个元素的平均值，您可以尝试applymap 方法。

df.applymap(np.mean)

Answer 2

我认为将值转换为列的想法非常好，因为这样可以使用 pandas 矢量化函数：

df1 = pd.concat([pd.DataFrame(df[c].values.tolist()) for c in df.columns], 
                 axis=1, 
                 keys=df.columns)
df1.columns = ['{}{}'.format(i, j) for i, j in df1.columns]
print (df1)
   A0  A1  A2  A3  A4  B0  B1  B2  B3  B4  C0  C1  C2  C3  C4
0   0   1   2   3   4   0   1   2   3   4   0   1   2   3   4
1   0   1   2   3   4   0   1   2   3   4   0   1   2   3   4
2   0   1   2   3   4   0   1   2   3   4   0   1   2   3   4
3   0   1   2   3   4   0   1   2   3   4   0   1   2   3   4
4   0   1   2   3   4   0   1   2   3   4   0   1   2   3   4

但如果需要mean的所有列表一起：

df= pd.DataFrame(np.nan, columns =["A","B","C"], index =np.arange(5))
df=df.astype(object)
for c in list(df):
    for i in df.index.values:
        df.at[i, c]=np.arange(i+1).tolist()
print (df)
                 A                B                C
0              [0]              [0]              [0]
1           [0, 1]           [0, 1]           [0, 1]
2        [0, 1, 2]        [0, 1, 2]        [0, 1, 2]
3     [0, 1, 2, 3]     [0, 1, 2, 3]     [0, 1, 2, 3]
4  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]

from itertools import chain
from statistics import mean
df['Average'] = [mean(list(chain.from_iterable(x))) for x in df.values.tolist()]
print (df)
                 A                B                C  Average
0              [0]              [0]              [0]      0.0
1           [0, 1]           [0, 1]           [0, 1]      0.5
2        [0, 1, 2]        [0, 1, 2]        [0, 1, 2]      1.0
3     [0, 1, 2, 3]     [0, 1, 2, 3]     [0, 1, 2, 3]      1.5
4  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]      2.0

编辑：

如果值是字符串：

df= pd.DataFrame(np.nan, columns =["A","B","C"], index =np.arange(5))
df=df.astype(object)
for c in list(df):
    for i in df.index.values:
        df.at[i, c]=np.arange(5).tolist()

df=df.astype(str)
print (df)
                 A                B                C
0  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
1  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
2  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
3  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]
4  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]  [0, 1, 2, 3, 4]

df1 = pd.concat([df[c].str.strip('[]').str.split(', ', expand=True) for c in df.columns], 
                 axis=1, 
                 keys=df.columns).astype(float)
df1.columns = ['{}{}'.format(i, j) for i, j in df1.columns]
df1["Average"]= df1.mean(axis=1)
print (df1)
    A0   A1   A2   A3   A4   B0   B1   B2   B3   B4   C0   C1   C2   C3   C4  \
0  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0   
1  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0   
2  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0   
3  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0   
4  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0  0.0  1.0  2.0  3.0  4.0   

   Average  
0      2.0  
1      2.0  
2      2.0  
3      2.0  
4      2.0