Python 如何将数据帧中具有不同信息或NAN的两个重复行合并为一行？

我有一个数据框，如下所示：

 Name           Email Assessment 1 Assessment 2 Assessment 3 Assessment 4    Assessment 5
0   abc   abc@email.com         Good          NaN          NaN          NaN          NaN
1   abc   abc@email.com          NaN        Good          Good          NaN          NaN   
2   abc   abc@email.com          NaN          NaN          NaN         Good          NaN   
3   abc   abc@email.com          NaN          NaN          NaN          NaN         Good 
4  john  john@email.com         Good        Good          Fail          NaN          NaN   
5  john  john@email.com          NaN          NaN         Good          NaN          NaN
6  john  john@email.com          NaN          NaN          NaN         Good         Good   
7   joe   joe@email.com         Good        Good          Fail         Fail          NaN
8   joe   joe@email.com          NaN          NaN         Fail         Good         Good   
9   joe   joe@email.com          NaN          NaN         Fail          NaN          NaN

在这里，我尝试合并重复的记录，使用电子邮件作为键，并保留以下行中未丢失的信息作为最终信息。在上面的示例中，以下是我的预期输出：

Name           Email Assessment 1 Assessment 2 Assessment 3 Assessment 4      Assessment 5
0   abc   abc@email.com         Good        Good          Good         Good         Good   
1  john  john@email.com         Good        Good          Good         Good         Good   
2   joe   joe@email.com         Good        Good          Fail         Good         Good   

我在这里看到了许多关于行组合的解决方案，但它们大多涉及内容的串联，即，它们为电子邮件创建一个行值，如

Good Good

或

Good Good Fail

，但不是以示例输出中所示的方式。请帮忙

---

样本数据

data_dict = pd.DataFrame({'Name': ['abc','abc','abc','abc','john','john','john','joe','joe','joe'],
             'Email': ['abc@email.com','abc@email.com','abc@email.com','abc@email.com','john@email.com','john@email.com','john@email.com','joe@email.com','joe@email.com','joe@email.com'],
             'Assessment 1': ['Good', np.nan, np.nan, np.nan, 'Good', np.nan, np.nan, 'Good', np.nan, np.nan],
             'Assessment 2': [np.nan,'Good',np.nan,np.nan,'Good',np.nan,np.nan,'Good ',np.nan,np.nan],
             'Assessment 3': [np.nan,'Good',np.nan,np.nan,'Fail','Good',np.nan,'Fail','Fail','Fail'],
             'Assessment 4': [np.nan,np.nan,'Good',np.nan,np.nan,np.nan,'Good','Fail','Good',np.nan],
             'Assessment 5': [np.nan,np.nan,np.nan,'Good',np.nan,np.nan,'Good','Fail','Good',np.nan]} )

---

如果需要最后一次唯一排序，且每组不缺少值，请使用：

df = pd.DataFrame(data_dict)

def f(x):
    try:
        return np.sort(x.dropna().unique())[-1]
    except:
        return np.nan

df = df.groupby(['Name','Email'], as_index=False, sort=False).agg(f)
print (df)
   Name           Email Assessment 1 Assessment 2 Assessment 3 Assessment 4  \
0   abc   abc@email.com         Good         Good         Good         Good   
1  john  john@email.com         Good         Good         Good         Good   
2   joe   joe@email.com         Good        Good          Fail         Good   

  Assessment 5  
0         Good  
1         Good  
2         Good  

编辑：

如果需要最后一个非缺失值，请使用：

---

如果每组的值为

故障良好

或

故障良好

，则逻辑是什么？在任何情况下，应以最新一行的结果为准

Fail-Good

，将

Good

保持在后一行中，按照相同的逻辑，

Good-Fail

，

Fail

将是组合行中的值。确定，然后编辑了答案。

df = df.groupby(['Name','Email'], as_index=False, sort=False).last()
print (df)
   Name           Email Assessment 1 Assessment 2 Assessment 3 Assessment 4  \
0   abc   abc@email.com         Good         Good         Good         Good   
1  john  john@email.com         Good         Good         Good         Good   
2   joe   joe@email.com         Good        Good          Fail         Good   

  Assessment 5  
0         Good  
1         Good  
2         Good