Python 在数据帧中连接包含NAN的列表列

我有一个带有两列的数组，两列要么有列表，要么有NaN值两列中都没有包含NaN的行。我想创建第三列，以以下方式合并其他两列的值：-

if row df.a is NaN -> df.c = df.b

if row df.b is Nan -> df.c = df.a

else df.c = df.a + df.b

输入：-

df
                                 a                                    b
0   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
1   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
2   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
3   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
4   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
5   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
6   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
7   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                  NaN   
8   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]   
9   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]   
10                             NaN  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]   
11                             NaN  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14] 

输出：

df.c

0   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
1   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
2   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
3   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
4   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
5   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
6   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
7   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]                                     
8   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]   
9   [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]   
10  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
11  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

我尝试将此嵌套条件用于apply

df['c'] = df.apply(lambda x: x.a if x.b is float else (x.b if x.a is float else (x['a'] + x['b'])), axis = 1)

但是他给了我这个错误：

TypeError:（“只能将列表（而不是“浮点”）连接到列表”，u“出现在索引0处”）

我正在使用（而且它确实有效）

因为这是我发现的唯一将列表与NaN值分开的方法。

您可以先将

NaN

s替换为

空列表：

df = pd.DataFrame({'a': [[0, 1, 2], np.nan, [0, 1, 2]],
                   'b':[np.nan,[0, 1, 2],[ 5, 6, 7, 8, 9]]})  
print (df)

s = pd.Series([[]], index=df.index)
df['c'] = df['a'].fillna(s) + df['b'].fillna(s)
print (df)
           a                b                         c
0  [0, 1, 2]              NaN                 [0, 1, 2]
1        NaN        [0, 1, 2]                 [0, 1, 2]
2  [0, 1, 2]  [5, 6, 7, 8, 9]  [0, 1, 2, 5, 6, 7, 8, 9]

您可以将
NaN
s转换为list，然后应用
np.sum
：

In [718]: df['c'] = df[['a', 'b']].applymap(lambda x: [] if x != x else x).apply(np.sum, axis=1); df['c']
Out[718]: 
0                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
1                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
2                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
3                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
4                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
5                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
6                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
7     [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, ...
8     [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, ...
9                   [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
10                  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
Name: c, dtype: object

这适用于具有列表/NaN内容的任意数量的列。
当您使用pd.DataFrame.stack时，默认情况下会删除空值。然后，我们可以按索引的第一级进行分组，并将列表与
sum

df.stack().groupby(level=0).sum()

0                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
1                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
2                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
3                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
4                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
5                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
6                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
7                                        [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
8     [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
9     [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
10                                  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
11                                  [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
dtype: object

然后，我们可以使用
assign

df.assign(c=df.stack().groupby(level=0).sum())

或者将其添加到新的列中

df['c'] = df.stack().groupby(level=0).sum()

df['c'] = df.stack().groupby(level=0).sum()