Python 将具有公共列值的行堆叠到熊猫中

将具有相同时间请求的行放在一起而不是根据错误代码分组的方法是什么。目前我正在获取所有0错误代码报告，然后是所有错误代码1报告，如下所示

>>> data.groupby([data['ErrorCode'], pd.Grouper(freq='15T')])['latency'].describe().unstack().reset_index()
    ErrorCode            Time_req  count           mean             std  \
0           0 2017-03-08 04:30:00      1  603034.000000             NaN   
1           0 2017-03-08 04:45:00      2  174720.000000    38101.741797   
2           0 2017-03-08 05:00:00      2  674942.500000   786118.185810   
3           0 2017-03-08 07:45:00     10  266653.200000   165867.496817   
4           0 2017-03-08 08:00:00     23  208949.304348   124902.942685   
5           0 2017-03-08 08:15:00     31  247282.064516   181780.519320   
6           0 2017-03-08 08:30:00     35  249332.857143   340084.918015   
7           0 2017-03-08 08:45:00      7  250066.000000   195051.871617   
8           1 2017-03-08 04:45:00      4  227747.500000   148185.181566   
9           1 2017-03-08 05:00:00      2  126633.000000     1337.846030   
10          1 2017-03-08 07:45:00     10  421781.900000   464249.118555   
11          1 2017-03-08 08:00:00     22  188122.272727    82110.336132   
12          1 2017-03-08 08:15:00     32  294896.968750   229498.560222   
13          1 2017-03-08 08:30:00     35  501679.628571  1353873.878385   
14          1 2017-03-08 08:45:00      6  531606.000000   582290.903396

ErrorCode Time_req count
0 2017-03-08 04:30:00      1
1 NaN        NaN           NaN
0 2017-03-08 04:45:00      2
1 2017-03-08 04:45:00      4
AND SO ON

但我需要下面这样的替代品

>>> data.groupby([data['ErrorCode'], pd.Grouper(freq='15T')])['latency'].describe().unstack().reset_index()
    ErrorCode            Time_req  count           mean             std  \
0           0 2017-03-08 04:30:00      1  603034.000000             NaN   
1           0 2017-03-08 04:45:00      2  174720.000000    38101.741797   
2           0 2017-03-08 05:00:00      2  674942.500000   786118.185810   
3           0 2017-03-08 07:45:00     10  266653.200000   165867.496817   
4           0 2017-03-08 08:00:00     23  208949.304348   124902.942685   
5           0 2017-03-08 08:15:00     31  247282.064516   181780.519320   
6           0 2017-03-08 08:30:00     35  249332.857143   340084.918015   
7           0 2017-03-08 08:45:00      7  250066.000000   195051.871617   
8           1 2017-03-08 04:45:00      4  227747.500000   148185.181566   
9           1 2017-03-08 05:00:00      2  126633.000000     1337.846030   
10          1 2017-03-08 07:45:00     10  421781.900000   464249.118555   
11          1 2017-03-08 08:00:00     22  188122.272727    82110.336132   
12          1 2017-03-08 08:15:00     32  294896.968750   229498.560222   
13          1 2017-03-08 08:30:00     35  501679.628571  1353873.878385   
14          1 2017-03-08 08:45:00      6  531606.000000   582290.903396

ErrorCode Time_req count
0 2017-03-08 04:30:00      1
1 NaN        NaN           NaN
0 2017-03-08 04:45:00      2
1 2017-03-08 04:45:00      4
AND SO ON

我认为您需要添加缺少的值：

df = data.groupby([data['ErrorCode'], pd.Grouper(freq='15T')])['latency'].describe()
df = df.unstack(0).stack(dropna=False).unstack(1).reset_index()
print (df)
               Time_req  ErrorCode  count           mean           std
0   2017-03-08 04:30:00          0    1.0  603034.000000           NaN
1   2017-03-08 04:30:00          1    NaN            NaN           NaN
2   2017-03-08 04:45:00          0    2.0  174720.000000  3.810174e+04
3   2017-03-08 04:45:00          1    4.0  227747.500000  1.481852e+05
4   2017-03-08 05:00:00          0    2.0  674942.500000  7.861182e+05
5   2017-03-08 05:00:00          1    2.0  126633.000000  1.337846e+03
6   2017-03-08 07:45:00          0   10.0  266653.200000  1.658675e+05
7   2017-03-08 07:45:00          1   10.0  421781.900000  4.642491e+05
8   2017-03-08 08:00:00          0   23.0  208949.304348  1.249029e+05
9   2017-03-08 08:00:00          1   22.0  188122.272727  8.211034e+04
10  2017-03-08 08:15:00          0   31.0  247282.064516  1.817805e+05
11  2017-03-08 08:15:00          1   32.0  294896.968750  2.294986e+05
12  2017-03-08 08:30:00          0   35.0  249332.857143  3.400849e+05
13  2017-03-08 08:30:00          1   35.0  501679.628571  1.353874e+06
14  2017-03-08 08:45:00          0    7.0  250066.000000  1.950519e+05
15  2017-03-08 08:45:00          1    6.0  531606.000000  5.822909e+05

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我认为df.to_csv'my_file.csv'，index=False可以完美地工作。不是csv，而是我在python终端中看到的内容。谢谢，我想你需要pd。设置选项“扩展框架报告”，错误检查。我的个人资料中有电子邮件。但请接受我的解决方案-要将答案标记为已接受，请单击答案旁边的复选标记，将其从灰色变为已填写。非常感谢。