使用 pandas groupby 根据时间间隔获取频率计数答案

【问题标题】：Use pandas groupby to fetch frequency count based on time intervals使用 pandas groupby 根据时间间隔获取频率计数
【发布时间】：2021-04-01 21:05:13
【问题描述】：

我有一个如下所示的数据框

df = pd.DataFrame({'subject_id':[1,1,1,2,2,2],
              'start_time':['2130-03-25 18:51:47','2130-04-23 18:51:47','2130-04-23 18:51:47','2120-01-11 18:51:47','2120-01-11 18:51:47','2120-04-28 18:51:47'],
              'test_time':['2130-03-26 14:51:47','2130-04-24 18:51:47','2130-04-25 18:51:47','2121-02-26 18:51:47','2121-02-26 18:51:47','2120-04-28 19:51:47'],
              'test':['test1','test2','test2','test2','test3','test3']})
df['start_time'] = pd.to_datetime(df['start_time'])
df['test_time'] = pd.to_datetime(df['test_time'])

我想做的是

a) 从start_time 获取每 24 小时为每个主题完成的测试次数。测试时间可以从test_time栏目中查到

示例 - 24 小时，我的意思是 0-24hours、24-48hours、48-72hours 等。

我尝试了以下

df['time_diff'] = (df.test_time - df.start_time) / pd.Timedelta(hours=1)
conditions = [
    (df['time_diff'] >= 0) & (df['time_diff'] <= 24),
    (df['time_diff'] >24 ) & (df['time_diff'] <= 48),
    (df['time_diff'] > 48) & (df['time_diff'] <= 72)]
choices = ['0-24hrs','24-48hrs','48-72hrs']
df['op'] = np.select(conditions, choices, default='Greater than 3 days')
df.groupby(['subject_id','test','op'])['test'].count()

但是上面的输出格式不正确。

我希望我的输出如下所示

【问题讨论】：

标签： python pandas dataframe numpy pandas-groupby

【解决方案1】：

你可以添加unstack

out = df.groupby(['subject_id','test','op'])['test'].count().unstack(fill_value=0).reset_index()
out
op  subject_id   test  0-24hrs  24-48hrs  Greater than 3 days
0            1  test1        1         0                    0
1            1  test2        1         1                    0
2            2  test2        0         0                    1
3            2  test3        1         0                    1

【讨论】：