我有一些日志数据,我想先按 user_id 分组,然后选择第二个条目。下面就是这样做的。缺少的步骤是分组后每个条目相对于第一个条目的年龄。
dd = pd.DataFrame({'item_id': {0: 0, 1: 4, 2: 6, 3: 8, 4: 9, 5: 1}, 'date': {0: '2013-12-29T17:56:01Z', 1: '2013-12-29T19:44:09Z', 2: '2013-12-29T19:58:05Z', 3: '2013-12-29T20:00:09Z', 4: '2013-12-29T20:13:35Z', 5: '2013-12-29T20:19:56Z'}, 'user_id': {0: 6, 1: 8, 2: 3, 3: 3, 4: 6, 5: 6}})
print "Step 1: Original DataFrame, sorted by date:\n", dd
g = dd.groupby(by='user_id', sort=False)
print "\nStep 2: Grouped by User ID:\n", g.head()
# Print the 2nd entey (if it exists)
print "\nStep 3: The 2nd user for each entry:\n", g.nth(1).dropna(how='all')
# age?
返回:
Step 1: Original DataFrame, sorted by date:
date item_id user_id
0 2013-12-29T17:56:01Z 0 6
1 2013-12-29T19:44:09Z 4 8
2 2013-12-29T19:58:05Z 6 3
3 2013-12-29T20:00:09Z 8 3
4 2013-12-29T20:13:35Z 9 6
5 2013-12-29T20:19:56Z 1 6
Step 2: Grouped by User ID:
date item_id user_id
user_id
6 0 2013-12-29T17:56:01Z 0 6
4 2013-12-29T20:13:35Z 9 6
5 2013-12-29T20:19:56Z 1 6
8 1 2013-12-29T19:44:09Z 4 8
3 2 2013-12-29T19:58:05Z 6 3
3 2013-12-29T20:00:09Z 8 3
Step 3: The 2nd user for each entry:
date item_id
user_id
6 2013-12-29T20:13:35Z 9
3 2013-12-29T20:00:09Z 8
但我想在第 2 步打印相对于该用户消费的第一个 item_id 的年龄(例如,十进制天数),这样我就可以判断第 3 步中日志条目的年龄。是否有一种无需迭代即可执行此操作的 pythonic 方法?
期望的输出是:
user_id date item_id age
0 3 2013-12-29 20:00:09 8 0:02:04
1 6 2013-12-29 20:13:35 9 2:17:34
最佳答案
首先将日期从字符串列转换为 datetime64[ns] dtype
In [21]: dd['date'] = pd.to_datetime(dd['date'])
In [22]: dd
Out[22]:
date item_id user_id
0 2013-12-29 17:56:01 0 6
1 2013-12-29 19:44:09 4 8
2 2013-12-29 19:58:05 6 3
3 2013-12-29 20:00:09 8 3
4 2013-12-29 20:13:35 9 6
5 2013-12-29 20:19:56 1 6
[6 rows x 3 columns]
按日期排序
In [23]: dd.sort_index(by='date')
Out[23]:
date item_id user_id
0 2013-12-29 17:56:01 0 6
1 2013-12-29 19:44:09 4 8
2 2013-12-29 19:58:05 6 3
3 2013-12-29 20:00:09 8 3
4 2013-12-29 20:13:35 9 6
5 2013-12-29 20:19:56 1 6
[6 rows x 3 columns]
定义一个函数来区分该列(并只返回组的其余部分)
In [4]: def f(x):
...: x['diff'] = x['date']-x['date'].iloc[0]
...: return x
...:
In [5]: dd.sort_index(by='date').groupby('user_id').apply(f)
Out[5]:
date item_id user_id diff
0 2013-12-29 17:56:01 0 6 00:00:00
1 2013-12-29 19:44:09 4 8 00:00:00
2 2013-12-29 19:58:05 6 3 00:00:00
3 2013-12-29 20:00:09 8 3 00:02:04
4 2013-12-29 20:13:35 9 6 02:17:34
5 2013-12-29 20:19:56 1 6 02:23:55
[6 rows x 4 columns]
差异现在是 timedelta64[ns],参见 here了解如何转换/舍入到特定频率(例如天)。
这是 pandas 0.13(隔天或第二天发布)。其中大部分也适用于 0.12。
关于python - Dataframe groupby - 返回日志条目的增量时间,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20889337/