从一天开始的n天将Pandas DataFrame分组

Question

我刚刚发现了熊猫的力量而我喜欢它,但我无法弄清楚这个问题:

我有一个DataFrame df.head():

   lon   lat  h  filename                  time
0  19.961216  80.617627    -0.077165     60048 2002-05-15 12:59:31.717467
1  19.923916  80.614847    -0.018689     60048 2002-05-15 12:59:31.831467
2  19.849396  80.609257    -0.089205     60048 2002-05-15 12:59:32.059467
3  19.830776  80.607857     0.076485     60048 2002-05-15 12:59:32.116467
4  19.570708  80.588183     0.162943     60048 2002-05-15 12:59:32.888467

我想将我的数据分组为九天

gb = df.groupby(pd.TimeGrouper(key='time', freq='9D'))

第一组:

2002-05-15 12:59:31.717467       lon   lat  h filename                  time
0    19.961216  80.617627    -0.077165     60048 2002-05-15 12:59:31.717467
1    19.923916  80.614847    -0.018689     60048 2002-05-15 12:59:31.831467
2    19.849396  80.609257    -0.089205     60048 2002-05-15 12:59:32.059467
3    19.830776  80.607857     0.076485     60048 2002-05-15 12:59:32.116467
...

下一组:

2002-05-24 12:59:31.717467        lon   lat  height  filename                  time
815   18.309498  80.457024     0.187387     60309 2002-05-24 16:35:39.553563
816   18.291458  80.458514     0.061446     60309 2002-05-24 16:35:39.610563
817   18.273408  80.460014     0.129255     60309 2002-05-24 16:35:39.667563
818   18.255358  80.461504     0.046761     60309 2002-05-24 16:35:39.724563
...

因此,数据从第一次(12:59:31.717467)开始计算,分为9天,而不是按照我的意愿从一天开始.

分组一天时:

gb = df.groupby(pd.TimeGrouper(key='time', freq='D'))

给我:

2002-05-15 00:00:00       lon   lat  h  filename                  time
0    19.961216  80.617627    -0.077165     60048 2002-05-15 12:59:31.717467
1    19.923916  80.614847    -0.018689     60048 2002-05-15 12:59:31.831467
2    19.849396  80.609257    -0.089205     60048 2002-05-15 12:59:32.059467
3    19.830776  80.607857     0.076485     60048 2002-05-15 12:59:32.116467
...

我可以循环过去,直到我得到一个九天的间隔,但我认为它可以更聪明地完成,我正在寻找一个freq相当于YS(年初)的Grouper 选项只是几天,一种设定开始时间的方法(也许是通过Grouper选项convention : {‘start’, ‘end’, ‘e’, ‘s’}),或???

我正在运行Python 3.5.2并且Pandas的版本是:0.19.0

Answer 1

删除第一个时间行：

您最好的选择是normalize该列的第一行datetime，以便时间重置为00:00:00（午夜）并根据9D间隔进行分组：

df.loc[0, 'time'] = df['time'].iloc[0].normalize()
for _, grp in df.groupby(pd.TimeGrouper(key='time', freq='9D')):
    print (grp)

#          lon        lat         h  filename                       time
# 0  19.961216  80.617627 -0.077165     60048 2002-05-15 00:00:00.000000
# 1  19.923916  80.614847 -0.018689     60048 2002-05-15 12:59:31.831467
# 2  19.849396  80.609257 -0.089205     60048 2002-05-15 12:59:32.059467
# 3  19.830776  80.607857  0.076485     60048 2002-05-15 12:59:32.116467
# 4  19.570708  80.588183  0.162943     60048 2002-05-15 12:59:32.888467
# ......................................................................

这会恢复其他行中的时间，因此您不会丢失该信息。

保留第一行：

如果您想保留第一个时间行而不对其进行任何更改，但只想从午夜开始分组，您可以这样做：

df_t_shift = df.shift()    # Shift one level down
df_t_shift.loc[0, 'time'] = df_t_shift['time'].iloc[1].normalize()
# Concat last row of df with the shifted one to account for the loss of row
df_t_shift = df_t_shift.append(df.iloc[-1], ignore_index=True)  

for _, grp in df_t_shift.groupby(pd.TimeGrouper(key='time', freq='9D')):
    print (grp)

#          lon        lat         h  filename                       time
# 0        NaN        NaN       NaN       NaN 2002-05-15 00:00:00.000000
# 1  19.961216  80.617627 -0.077165   60048.0 2002-05-15 12:59:31.717467
# 2  19.923916  80.614847 -0.018689   60048.0 2002-05-15 12:59:31.831467
# 3  19.849396  80.609257 -0.089205   60048.0 2002-05-15 12:59:32.059467
# 4  19.830776  80.607857  0.076485   60048.0 2002-05-15 12:59:32.116467
# 5  19.570708  80.588183  0.162943   60048.0 2002-05-15 12:59:32.888467