用于 Python 的 Pandas，分组

Question

我有一个数据集，每个时间戳由多个元组组成 - 每个元组都有一个计数。每个时间戳可能存在不同的元组。我想在 5 分钟内将它们组合在一起，并为每个唯一的元组添加计数。使用 Pandas group-by 有没有一种很好的干净的方法来做到这一点？

它们具有以下形式:
((u'67.163.47.231', u'8.27.82.254', 50186, 80, 6, 1377565195000), 2)

这是当前的一个列表，有一个 6 元组(最后一个条目是时间戳)，然后计数。

每个时间戳都会有一个 5 元组的集合:

(5-tuple), t-time-stamp, count, example (for only one time stamp)

[((u'71.57.43.240', u'8.27.82.254', 33108, 80, 6, 1377565195000), 1),
 ((u'67.163.47.231', u'8.27.82.254', 50186, 80, 6, 1377565195000), 2),
 ((u'8.27.82.254', u'98.206.29.242', 25159, 80, 6, 1377565195000), 1),
 ((u'71.179.102.253', u'8.27.82.254', 50958, 80, 6, 1377565195000), 1)]

In [220]: df = DataFrame ( { 'key1' : [ (u'71.57.43.240', u'8.27.82.254', 33108, 80, 6), (u'67.163.47.231', u'8.27.82.254', 50186, 80, 6) ], 'data1' : np.array((1,2)), 'data2': np.array((1377565195000,1377565195000))})

In [226]: df
Out[226]: 
   data1          data2                                        key1
0      1  1377565195000   (71.57.43.240, 8.27.82.254, 33108, 80, 6)
1      2  1377565195000  (67.163.47.231, 8.27.82.254, 50186, 80, 6)

或转换:

In [231]: df = DataFrame ( { 'key1' : [ (u'71.57.43.240', u'8.27.82.254', 33108, 80, 6), (u'67.163.47.231', u'8.27.82.254', 50186, 80, 6) ], 'data1' : np.array((1,2)), 
   .....: 'data2': np.array(( datetime.utcfromtimestamp(1377565195),datetime.utcfromtimestamp(1377565195) )) })

In [232]: df
Out[232]: 
   data1               data2                                        key1
0      1 2013-08-27 00:59:55   (71.57.43.240, 8.27.82.254, 33108, 80, 6)
1      2 2013-08-27 00:59:55  (67.163.47.231, 8.27.82.254, 50186, 80, 6)


Here's a simpler example:

time         count       city
00:00:00       1         Montreal
00:00:00       2         New York
00:00:00       1         Chicago
00:01:00       2         Montreal
00:01:00       3         New York

after bin-ing

time         count       city
00:05:00       3         Montreal
00:05:00       5         New York
00:05:00       1         Chicago

以下是似乎运作良好的内容:

times = [ parse('00:00:00'), parse('00:00:00'), parse('00:00:00'), parse('00:01:00'), parse('00:01:00'),
parse('00:02:00'), parse('00:02:00'), parse('00:03:00'), parse('00:04:00'), parse('00:05:00'),
parse('00:05:00'), parse('00:06:00'), parse('00:06:00') ]
cities = [ 'Montreal', 'New York', 'Chicago', 'Montreal', 'New York', 
'New York', 'Chicago', 'Montreal', 'Montreal', 'New York', 'Chicago', 'Montreal', 'Chicago']
counts = [ 1, 2, 1, 2, 3, 1, 1, 1, 2, 2, 2, 1, 1]
frame = DataFrame( { 'city': cities, 'time': times, 'count': counts } )

In [150]: frame
Out[150]: 
        city  count                time
0   Montreal      1 2013-09-07 00:00:00
1   New York      2 2013-09-07 00:00:00
2    Chicago      1 2013-09-07 00:00:00
3   Montreal      2 2013-09-07 00:01:00
4   New York      3 2013-09-07 00:01:00
5   New York      1 2013-09-07 00:02:00
6    Chicago      1 2013-09-07 00:02:00
7   Montreal      1 2013-09-07 00:03:00
8   Montreal      2 2013-09-07 00:04:00
9   New York      2 2013-09-07 00:05:00
10   Chicago      2 2013-09-07 00:05:00
11  Montreal      1 2013-09-07 00:06:00
12   Chicago      1 2013-09-07 00:06:00

frame['time_5min'] = frame['time'].map(lambda x: pd.DataFrame([0],index=pd.DatetimeIndex([x])).resample('5min').index[0])

In [152]: frame
Out[152]: 
        city  count                time           time_5min
0   Montreal      1 2013-09-07 00:00:00 2013-09-07 00:00:00
1   New York      2 2013-09-07 00:00:00 2013-09-07 00:00:00
2    Chicago      1 2013-09-07 00:00:00 2013-09-07 00:00:00
3   Montreal      2 2013-09-07 00:01:00 2013-09-07 00:00:00
4   New York      3 2013-09-07 00:01:00 2013-09-07 00:00:00
5   New York      1 2013-09-07 00:02:00 2013-09-07 00:00:00
6    Chicago      1 2013-09-07 00:02:00 2013-09-07 00:00:00
7   Montreal      1 2013-09-07 00:03:00 2013-09-07 00:00:00
8   Montreal      2 2013-09-07 00:04:00 2013-09-07 00:00:00
9   New York      2 2013-09-07 00:05:00 2013-09-07 00:05:00
10   Chicago      2 2013-09-07 00:05:00 2013-09-07 00:05:00
11  Montreal      1 2013-09-07 00:06:00 2013-09-07 00:05:00
12   Chicago      1 2013-09-07 00:06:00 2013-09-07 00:05:00

In [153]: df = frame.groupby(['time_5min', 'city']).aggregate('sum')

In [154]: df
Out[154]: 
                              count
time_5min           city           
2013-09-07 00:00:00 Chicago       2
                    Montreal      6
                    New York      6
2013-09-07 00:05:00 Chicago       3
                    Montreal      1
                    New York      2

In [155]: df.reset_index(1)
Out[155]: 
                         city  count
time_5min                           
2013-09-07 00:00:00   Chicago      2
2013-09-07 00:00:00  Montreal      6
2013-09-07 00:00:00  New York      6
2013-09-07 00:05:00   Chicago      3
2013-09-07 00:05:00  Montreal      1
2013-09-07 00:05:00  New York      2

Answer 1

如果将日期设置为索引，则可以使用 TimeGrouper(例如，它允许您按 5 分钟间隔进行分组):

In [11]: from pandas.tseries.resample import TimeGrouper

In [12]: df.set_index('data2', inplace=True)

In [13]: g = df.groupby(TimeGrouper('5Min'))

然后，您可以使用 nunique 计算每 5 分钟间隔内唯一项目的数量:

In [14]: g['key1'].nunique()
Out[14]: 
2013-08-27 00:55:00    2
dtype: int64

如果您正在寻找每个元组的计数，您可以使用 value_counts:

In [15]: g['key1'].apply(pd.value_counts)
Out[15]: 
2013-08-27 00:55:00  (71.57.43.240, 8.27.82.254, 33108, 80, 6)     1
                     (67.163.47.231, 8.27.82.254, 50186, 80, 6)    1
dtype: int64

注意:这是一个带有 MultiIndex 的系列(使用 reset_index 使其成为 DataFrame)。

In [16]: g['key1'].apply(pd.value_counts).reset_index(1)
Out[16]: 
                                                        level_1  0
2013-08-27 00:55:00   (71.57.43.240, 8.27.82.254, 33108, 80, 6)  1
2013-08-27 00:55:00  (67.163.47.231, 8.27.82.254, 50186, 80, 6)  1

您可能希望提供这些信息更丰富的列名称 :)。

更新:以前我黑了得到 get_dummies ，见编辑历史。