Question

我有以下数据框

import pandas as pd
import numpy as np
d = {
    'ID':[1,2,3],
    'W1':[5,6,7],
    'W2':[9, np.nan,10],
    'w3':[11,np.nan,np.nan]
}
df = pd.DataFrame(data = d)
df


  ID    W1  W2   w3
0   1   5   9.0    11.0
1   2   6   NaN     NaN
2   3   7   10.0    NaN

我正在执行以下操作

df['Sum1'] = (df[['W1','W2']]).sum(axis = 1)/2
df['Sum2'] = (df[['W2','w3']]).sum(axis = 1)/2


    ID  W1  W2      w3  Sum1    Sum2
0   1   5   9.0    11.0 7.0     10.0
1   2   6   NaN     NaN 3.0     0.0
2   3   7   10.0    NaN 8.5     5.0

完成上述操作后如何将ID为“ 2”的Sum2设为 NaN 而不是 0 ？

Answer 1

将参数min_count=1添加到DataFrame.sum：

最低计数：整数，默认为 0
  执行操作所需的有效值数量。如果存在少于min_count个非NA值，则结果将为NA。

0.22.0版中的新增功能：添加了默认值0。这表示全NA或空系列的总和为0，全NA或空系列的乘积为1。

df['Sum1'] = (df[['W1','W2']]).sum(axis = 1, min_count=1)/2
df['Sum2'] = (df[['W2','w3']]).sum(axis = 1, min_count=1)/2

print (df)
   ID  W1    W2    w3  Sum1  Sum2
0   1   5   9.0  11.0   7.0  10.0
1   2   6   NaN   NaN   3.0   NaN
2   3   7  10.0   NaN   8.5   5.0

但是似乎您需要mean s-然后它的工作原理就像需要：

df['Sum1'] = (df[['W1','W2']]).mean(axis = 1)
df['Sum2'] = (df[['W2','w3']]).mean(axis = 1)

print (df)
   ID  W1    W2    w3  Sum1  Sum2
0   1   5   9.0  11.0   7.0  10.0
1   2   6   NaN   NaN   6.0   NaN
2   3   7  10.0   NaN   8.5  10.0

运算后如何将两个NaN设为NaN而不是使其变为零？

1 个答案: