解压缩函数返回的元组时的 Python "too many to unpack"

Question

我试图将函数应用于数据帧中的一列，该函数返回三个项目。在某些情况下，它有效，但在其他情况下它不起作用。然后我意识到这可能是因为存在 NULL 值。所以这是我代码的简化版本：

import pandas as pd
def proc(x):
    return ([x,1,2], [x+1,3,4], [x+2,5,6])
## This works fine.
df = pd.DataFrame({'a':[1,2,3]})
df['new1'],df['new2'], df['new3'] = df.a.apply(lambda x:proc(x))
## But this throws the 'too many values to unpack' error.
df2 = pd.DataFrame({'a':[1,2,3, float('nan')]})
df2['new1'],df2['new2'], df2['new3'] = df2.a.apply(lambda x:proc(x))

为什么将 float（'nan'） 添加到 df['a'] 列会导致此错误？

Answer 1

1. 使用 zip 打包值：

   def proc(x):
       return ([x,1,2], [x+1,3,4], [x+2,5,6])
   df2 = pd.DataFrame({'a':[1,2,4, float('nan')]})
   df2['new1'], df2['new2'], df2['new3'] = zip(*df2['a'].apply(proc))
        a      new1         new2        new3
     0 1.0 [1.0, 1, 2] [2.0, 3, 4] [3.0, 5, 6]
     1 2.0 [2.0, 1, 2] [3.0, 3, 4] [4.0, 5, 6]
     2 4.0 [4.0, 1, 2] [5.0, 3, 4] [6.0, 5, 6]
     3 NaN [nan, 1, 2] [nan, 3, 4] [nan, 5, 6]
   

2. 使用正确数量的列表元素进行解压缩，并在 proc 中使用相同数量的元素返回：

   def proc(x):
       return ([x,1,2], [x+1,3,4], [x+2,5,6])
   df2 = pd.DataFrame({'a':[1,2,4, float('nan')]})
   df2['new1'], df2['new2'], df2['new3'] = zip(*df2['a'].apply(proc))
        a  new1         new2        new3        new4
   0   1.0 [1.0, 1, 2] [2.0, 1, 2] [4.0, 1, 2] [nan, 1, 2]
   1   2.0 [2.0, 3, 4] [3.0, 3, 4] [5.0, 3, 4] [nan, 3, 4]
   2   4.0 [3.0, 5, 6] [4.0, 5, 6] [6.0, 5, 6] [nan, 5, 6]
   3   NaN [4.0, 7, 8] [5.0, 7, 8] [7.0, 7, 8] [nan, 7, 8]