尝试将数据框列表组合为一个数据框。数据如下：
Date station_id Hour Temp
0 2004-01-01 1 1 46.0
1 2004-01-01 1 2 46.0
2 2004-01-01 1 3 45.0
3 2004-01-01 1 4 41.0
...
433730 2008-06-30 11 3 64.0
433731 2008-06-30 11 4 64.0
433732 2008-06-30 11 5 64.0
433733 2008-06-30 11 6 64.0
这给了我一个数据帧列表：
stations = [x for _,x in df.groupby('station_id')]
当我重置“ stations”和concat的索引时，我可以得到一个数据框，但它看起来不像我想要的：
for i in range(0,11):
stations[i].reset_index(drop=True,inplace=True)
pd.concat(stations,axis=1)
Date station_id Hour Temp Date station_id Hour Temp
0 2004-01-01 1 1 46.0 2004-01-01 2 1 38.0
1 2004-01-01 1 2 46.0 2004-01-01 2 2 36.0
2 2004-01-01 1 3 45.0 2004-01-01 2 3 35.0
3 2004-01-01 1 4 41.0 2004-01-01 2 4 30.0
我更愿意成为这样的df：
Date Hour Stn1 Stn2
0 2004-01-01 1 46.0 38.0
1 2004-01-01 2 46.0 6.0
2 2004-01-01 3 45.0 35.0
3 2004-01-01 4 41.0 30.0
我该怎么做？
答案 0 :(得分：0)
根据您的预期输出，您正在寻找带有.Close
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</div>答案 1 :(得分：0)
对于它的价值，这到达了我想要去的地方。
stations = [x for _,x in df.groupby('station_id')] #,as_index=True)]
for i in range(0,11):
stations[i].reset_index(drop=True,inplace=True)
stations[i].rename(columns={'Temp':'Stn'+str(i+1)},inplace=True)
stations[i].drop(columns='station_id',inplace=True)
if i>0:
stations[i].drop(columns=['Date','Hour'],inplace=True)
stations = pd.concat(stations,axis=1)
不过，对我来说有点蛮力。欢迎其他pythonic建议。