Pandas DataFrame由两列组成,并获得第一个和最后一个

Question

我有一个DataFrame喜欢的追随者.

df = pd.DataFrame({'id' : [1,1,2,3,2],
                'value'  : ["a","b","a","a","c"], 'Time' : ['6/Nov/2012 23:59:59 -0600','6/Nov/2012 00:00:05 -0600','7/Nov/2012 00:00:09 -0600','27/Nov/2012 00:00:13 -0600','27/Nov/2012 00:00:17 -0600']})

我需要获得如下输出.

combined_id | enter time | exit time | time difference

combined_id应该通过分组'id'和'value'来创建

g = df.groupby(['id', 'value'])

以下不适用于按两列分组.(如何使用first(),last()这里作为进入和退出时间？)

df['enter'] = g.apply(lambda x: x.first())

为了获得差异会跟着工作吗？

df['delta'] = (df['exit']-df['enter'].shift()).fillna(0)

Answer 1

首先确保您的列是正确的日期时间列:

In [11]: df['Time'] = pd.to_datetime(df['Time'])

现在,您可以执行groupby并使用agg first和lastgroupby方法:

In [12]: g = df.groupby(['id', 'value'])

In [13]: res = g['Time'].agg({'first': 'first', 'last': 'last'})

In [14]: res = g['Time'].agg({'enter': 'first', 'exit': 'last'})

In [15]: res['time_diff'] = res['exit'] - res['enter']

In [16]: res
Out[16]:
                        exit               enter  time_diff
id value
1  a     2012-11-06 23:59:59 2012-11-06 23:59:59     0 days
   b     2012-11-06 00:00:05 2012-11-06 00:00:05     0 days
2  a     2012-11-07 00:00:09 2012-11-07 00:00:09     0 days
   c     2012-11-27 00:00:17 2012-11-27 00:00:17     0 days
3  a     2012-11-27 00:00:13 2012-11-27 00:00:13     0 days

注意:这是一个无聊的例子,因为每个组中只有一个项目......