Question

我一直在使用pandas对时间序列数据进行分析，并且已经陷入将它们集成到数据透视表中。我在csv中有一个数据：

gov start   end
a   2015-12-08T16:05:00.980+03  2015-12-08T16:14:31.765+03
a   2015-12-08T16:07:53.356+03  2015-12-08T16:34:43.413+03
b   2015-12-08T16:08:43.371+03  2015-12-08T16:54:32.257+03
b   2015-12-08T15:56:12.006+03  2015-12-08T17:35:04.499+03

我有一个简单的数据集start和end时间，从中找出两者之间的时差：

piv_t_subset = pd.read_csv('time_test.csv', parse_dates=['start','end'])

piv_t_subset['time_diff'] = piv_t_subset['end'] - piv_t_subset['start']

我可以将时间的独立平均值计算为：

t = piv_t_subset['time_diff'].mean()
print t

0 days 00:18:53.703286

我想创建一个包含此时间信息的数据透视表，但是当我尝试时：

pd.pivot_table(piv_t_subset,index=["gov"],values=['time_diff'],aggfunc=[np.mean])

我收到错误：

DataError：无聚合的数字类型

我是否需要进行更多预处理才能将其从timeseries转换为float？

Answer 1

现在不支持link。

但您可以Series将timedelta64的{{1}}转换为Series float的{{1}}：

piv_t_subset['time_diff1'] = [td.total_seconds() for td in piv_t_subset['time_diff']]

print piv_t_subset
  gov                   start                     end
0   a 2015-12-08 13:05:00.980 2015-12-08 13:14:31.765
1   a 2015-12-08 13:07:53.356 2015-12-08 13:34:43.413
2   b 2015-12-08 13:08:43.371 2015-12-08 13:54:32.257
3   b 2015-12-08 12:56:12.006 2015-12-08 14:35:04.499

piv_t_subset['time_diff'] = piv_t_subset['end'] - piv_t_subset['start']

piv_t_subset['time_diff1'] = [td.total_seconds() for td in piv_t_subset['time_diff']]
print piv_t_subset
  gov                   start                     end       time_diff  \
0   a 2015-12-08 13:05:00.980 2015-12-08 13:14:31.765 00:09:30.785000   
1   a 2015-12-08 13:07:53.356 2015-12-08 13:34:43.413 00:26:50.057000   
2   b 2015-12-08 13:08:43.371 2015-12-08 13:54:32.257 00:45:48.886000   
3   b 2015-12-08 12:56:12.006 2015-12-08 14:35:04.499 01:38:52.493000   

   time_diff1  
0     570.785  
1    1610.057  
2    2748.886  
3    5932.493  

print piv_t_subset.groupby('gov').agg({'time_diff1':np.mean})
     time_diff1
gov            
a     1090.4210
b     4340.6895

#omit aggfunc, in pivot_table is default numpy.mean
print pd.pivot_table(piv_t_subset,index=["gov"],values=['time_diff1'])
     time_diff1
gov            
a     1090.4210
b     4340.6895

pandas数据透视表与平均时间

1 个答案: