在 Pandas DataFrame 中连接带有“NaN”值的列值

Question

我正在尝试将带有NaN值的Pandas DataFrame列连接起来。

In [96]:df = pd.DataFrame({'col1' : ["1","1","2","2","3","3"],
                'col2'  : ["p1","p2","p1",np.nan,"p2",np.nan], 'col3' : ["A","B","C","D","E","F"]})

In [97]: df
Out[97]: 
  col1 col2 col3
0    1   p1    A
1    1   p2    B
2    2   p1    C
3    2  NaN    D
4    3   p2    E
5    3  NaN    F

In [98]: df['concatenated'] = df['col2'] +','+ df['col3']
In [99]: df
Out[99]: 
  col1 col2 col3 concatenated
0    1   p1    A         p1,A
1    1   p2    B         p2,B
2    2   p1    C         p1,C
3    2  NaN    D          NaN
4    3   p2    E         p2,E
5    3  NaN    F          NaN

我希望在“连接”列中，代替'NaN'值，对于这个例子分别得到“D”和“F”？

- Nilani Algiriyage

Answer 1

我不认为您的问题是琐碎的。但是，这里有一个使用NumPy矢量化的解决方法：

In [49]: def concat(*args):
    ...:     strs = [str(arg) for arg in args if not pd.isnull(arg)]
    ...:     return ','.join(strs) if strs else np.nan
    ...: np_concat = np.vectorize(concat)
    ...: 

In [50]: np_concat(df['col2'], df['col3'])
Out[50]: 
array(['p1,A', 'p2,B', 'p1,C', 'D', 'p2,E', 'F'], 
      dtype='|S64')

In [51]: df['concatenated'] = np_concat(df['col2'], df['col3'])

In [52]: df
Out[52]: 
  col1 col2 col3 concatenated
0    1   p1    A         p1,A
1    1   p2    B         p2,B
2    2   p1    C         p1,C
3    2  NaN    D            D
4    3   p2    E         p2,E
5    3  NaN    F            F

[6 rows x 4 columns]

- Kiwi

---

嗨，谢谢Kiwi，看起来这是最简单的方法。 :) - Nilani Algiriyage

不知道为什么，但我必须进行一些更改，比如 strs = [str(arg) for arg in args if not arg == 'nan'] 和 return ','.join(filter(None, strs)) if strs else ''。 - n8-da-gr8

Answer 2

您可以先将整个数据框或您想要的列的NaN替换为空字符串。

In [6]: df = df.fillna('')

In [7]: df['concatenated'] = df['col2'] +','+ df['col3']

In [8]: df
Out[8]:
  col1 col2 col3 concatenated
0    1   p1    A         p1,A
1    1   p2    B         p2,B
2    2   p1    C         p1,C
3    2         D           ,D
4    3   p2    E         p2,E
5    3         F           ,F

- sl1129

Answer 3

我们可以使用 stack 函数来去掉 NaN，然后使用 groupby.agg 函数和 ','.join 函数来拼接字符串：

df['concatenated'] = df[['col2', 'col3']].stack().groupby(level=0).agg(','.join)

  col1 col2 col3 concatenated
0    1   p1    A         p1,A
1    1   p2    B         p2,B
2    2   p1    C         p1,C
3    2  NaN    D            D
4    3   p2    E         p2,E
5    3  NaN    F            F

- Erfan