Python-连接两个相同的列，而其他保持不变

Question

我有一些数据，例如

>>> df1
  Col1     Col2
0   Hi    Hello
1  How     Fine
2  Why  Because
3  You       Me
4   We     They
>>> df2
      Col1   Col2
0       Mr    Mrs
1  Welcome    Bye
2      Wow    Ohh
3      Sit  Stand

我只希望连接这两个数据帧的第一列Col1，并更新df1，而使Col2保持不变，但我无法做到

我尝试了很多方法，但没有成功

我想要的输出

>>> df2
  Col1     Col2
0   Hi    Hello
1  How     Fine
2  Why  Because
3  You       Me
4  We     They
5  Mr
6  Welcome
7  Wow
8  Sit

索引不同还是出现NaN而不是空单元格都没关系

我尝试过的事情

1

>>> r=pd.concat([df1["Col1"],df2["Col1"]])
>>> df1["Col1"] = r.reset_index(drop=True)
>>> df1
  Col1     Col2
0   Hi    Hello
1  How     Fine
2  Why  Because
3  You       Me
4   We     They

2

>>> pd.concat([df1["Col1"],df2])
     0     Col1   Col2
0   Hi      NaN    NaN
1  How      NaN    NaN
2  Why      NaN    NaN
3  You      NaN    NaN
4   We      NaN    NaN
0  NaN       Mr    Mrs
1  NaN  Welcome    Bye
2  NaN      Wow    Ohh
3  NaN      Sit  Stand

3

I Tried append(), merge() and some more techniques but failed

Answer 1

append

df1.append(df2[['Col1']])

      Col1     Col2
0       Hi    Hello
1      How     Fine
2      Why  Because
3      You       Me
4       We     They
0       Mr      NaN
1  Welcome      NaN
2      Wow      NaN
3      Sit      NaN

---

concat

pd.concat([df1, df2[['Col1']]])

      Col1     Col2
0       Hi    Hello
1      How     Fine
2      Why  Because
3      You       Me
4       We     They
0       Mr      NaN
1  Welcome      NaN
2      Wow      NaN
3      Sit      NaN

---

无论哪种情况，您都可以通过ignore_index=True选项忽略索引。

df1.append(df2[['Col1']], ignore_index=True)

      Col1     Col2
0       Hi    Hello
1      How     Fine
2      Why  Because
3      You       Me
4       We     They
5       Mr      NaN
6  Welcome      NaN
7      Wow      NaN
8      Sit      NaN