合并行基于索引

Question

我有一个熊猫据帧像这样，

Timestamp Meter1 Meter2 
0   234 NaN 
1   235 NaN 
2   236 NaN 
0   NaN 100 
1   NaN 101 
2   NaN 102 

，我有麻烦了基于索引时间戳是这样的合并行，

Timestamp Meter1 Meter2 
0   234 100 
1   235 101 
2   236 102 

Answer 1

使用groupby：

df.groupby(level=0).max() 

OR

df.groupby('Timestamp').max() 

输出

  Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0 

Answer 2

使用groupby和聚集sum：

df = df.groupby(level=0).sum() 
print (df) 
      Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0 

如果只ints：

df = df.groupby(level=0).sum().astype(int) 
print (df) 
      Meter1 Meter2 
Timestamp     
0    234  100 
1    235  101 
2    236  102 

但也许问题是您在concat忘记axis=1：

print (df1) 
      Meter1 
Timestamp   
0    234 
1    235 
2    236 

print (df2) 
      Meter2 
Timestamp   
0    100 
1    101 
2    102 

print (pd.concat([df1, df2])) 
      Meter1 Meter2 
Timestamp     
0   234.0  NaN 
1   235.0  NaN 
2   236.0  NaN 
0    NaN 100.0 
1    NaN 101.0 
2    NaN 102.0 

print (pd.concat([df1, df2], axis=1)) 
      Meter1 Meter2 
Timestamp     
0    234  100 
1    235  101 
2    236  102 

Answer 3

选项0

df.max(level=0) 

      Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0 

---

选项1

df.sum(level=0) 

      Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0 

---

的Opti 2
令人不安的回答

df.stack().unstack() 

      Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0 

---

正如@jezrael长大并链接到发行here

然而，正如我理解groupby.first和groupby.last是，它会返回第一个（或最后一个）组中的每个列的有效值。换句话说，我相信这是按预期工作的。

选项3

df.groupby(level=0).first() 

      Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0 

---

选项4

df.groupby(level=0).last() 

      Meter1 Meter2 
Timestamp     
0   234.0 100.0 
1   235.0 101.0 
2   236.0 102.0