如何计算每个月或一个月范围内的活动日期

Question

我有一个DataFrame 喜欢：

Student_id	actvity_timestamp
1001	2019-09-05 08:26:12
1001	2019-09-06 09:26:12
1001	2019-09-21 10:11:01
1001	2019-10-24 11:44:01
1001	2019-10-25 11:31:01
1001	2019-10-26 12:13:01
1002	2019-09-11 12:21:01
1002	2019-09-12 13:11:01
1002	2019-11-23 16:22:01

我想要输出类似：

Student_id	total_active_days_in_Sept	total_active_days_in_Oct	total_active_days_in_Nov
1001	3	3	0
1002	2	0	1

如何做到这一点（actvity_timestamp 的输出列必须用几个月）？

Answer 1

你可以尝试做类似这样的事情：

df = pd.DataFrame.from_dict({
    "Student_id": [1001,1001,1001,1001,1001,1001,1002,1002,1002],
    "actvity_timestamp": ["2019-09-05 08:26:12", "2019-09-06 09:26:12", "2019-09-21 10:11:01", "2019-10-24 11:44:01", "2019-10-25 11:31:01", "2019-10-26 12:13:01", "2019-09-11 12:21:01", "2019-09-12 13:11:01", "2019-11-23 16:22:01"]
})

months = pd.to_datetime(df.actvity_timestamp).dt.strftime("%B")

result = pd.crosstab(
    df.Student_id,
    months,
    values=df.activity_timestamp.dt.date,
    aggfunc=pd.Series.nunique # These last two parameters make it so that if a Student_id has been active more than once in a single day, to count it only once. (Thanks to @tlentali)
).fillna(0)

Series.dt.strftime 适用于日期时间系列，%B 格式化日期时间以仅显示月份名称。

result 将产生

actvity_timestamp  November  October  September
Student_id                                     
1001                      0        3          3
1002                      1        0          2

Answer 2

您可以达到所需的布局（列名按正确的月份顺序排序：'Sep' -> 'Oct' -> 'Nov' 而不是 'Nov' -> 'Oct' -> 'Sep ') 在以下步骤中：

1) 创建一个带有月份短名称的列。然后使用.pivot_table() 转换数据框 （每个Student_id 下每个月的活动日期计数汇总）：

df['actvity_timestamp'] = pd.to_datetime(df['actvity_timestamp']) # to datetime format 
df['activity_month'] = df['actvity_timestamp'].dt.strftime('%b')  # get month short name
df['activity_date'] = df['actvity_timestamp'].dt.date     # get activity dates

df_out = (df.pivot_table(index='Student_id',   # group under each student id
                         columns='activity_month',  # month short name as new columns
                         values='activity_date',    # aggregate on dates
                         aggfunc='nunique',    #activities on the same date counted once
                         fill_value=0)
            .rename_axis(columns=None)                          
         )


            Nov  Oct  Sep
Student_id               
1001          0    3    3
1002          1    0    2

2) 将月份短名的列名按.sort_index与排序键参数排序回日历序列，如下：

df_out = df_out.sort_index(axis=1, key=lambda x: pd.to_datetime(x, format='%b').month)


            Sep  Oct  Nov
Student_id               
1001          3    3    0
1002          2    0    1

3) 通过.add_prefix()进一步转换成想要的布局：

df_out = df_out.add_prefix('total_active_days_in_').reset_index()

结果：

print(df_out)

   Student_id  total_active_days_in_Sep  total_active_days_in_Oct  total_active_days_in_Nov
0        1001                         3                         3                         0
1        1002                         2                         0                         1

Answer 3

从您的Dataframe 开始：

>>> import pandas as pd

>>> df = pd.DataFrame({'Student_id': [1001, 1001, 1001, 1001, 1001, 1001, 1002, 1002, 1002],
...                    'activity_timestamp': ['2019-09-05 08:26:12', '2019-09-06 09:26:12', '2019-09-21 10:11:01', '2019-10-24 11:44:01', '2019-10-25 11:31:01', '2019-10-26 12:13:01', '2019-09-11 12:21:01', '2019-09-12 13:11:01', '2019-11-23 16:22:01']}, 
...                   index = [0, 1, 2, 3, 4, 5, 6, 7, 8])
>>> df
    Student_id  activity_timestamp
0   1001        2019-09-05 08:26:12
1   1001        2019-09-06 09:26:12
2   1001        2019-09-21 10:11:01
3   1001        2019-10-24 11:44:01
4   1001        2019-10-25 11:31:01
5   1001        2019-10-26 12:13:01
6   1002        2019-09-11 12:21:01
7   1002        2019-09-12 13:11:01
8   1002        2019-11-23 16:22:01

我们将activity_timestamp 转换为datetime，然后像这样提取日期和月份：

>>> df['activity_timestamp'] = pd.to_datetime(df['activity_timestamp'], format='%Y-%m-%d %H:%M:%S.%f')
>>> df['date'] = df['activity_timestamp'].dt.date
>>> df['month'] = df['activity_timestamp'].dt.month_name()
>>> df
    Student_id  activity_timestamp  date        month
0   1001        2019-09-05 08:26:12 2019-09-05  September
1   1001        2019-09-05 08:26:13 2019-09-05  September
2   1001        2019-09-06 09:26:12 2019-09-06  September
3   1001        2019-09-21 10:11:01 2019-09-21  September
4   1001        2019-10-24 11:44:01 2019-10-24  October
5   1001        2019-10-25 11:31:01 2019-10-25  October
6   1001        2019-10-26 12:13:01 2019-10-26  October
7   1002        2019-09-11 12:21:01 2019-09-11  September
8   1002        2019-09-12 13:11:01 2019-09-12  September
9   1002        2019-11-23 16:22:01 2019-11-23  November

然后，我们使用pivot_table() 方法和nunique 函数代替count 来获取唯一日期的数量：

>>> df_result = (df.pivot_table(index='Student_id', 
...                             columns='month', 
...                             values='date', 
...                             aggfunc=pd.Series.nunique, 
...                             fill_value=0).rename_axis(columns=None)).add_prefix('total_active_days_in_').reset_index(drop=False)
>>> df_result
    Student_id  total_active_days_in_November   total_active_days_in_October    total_active_days_in_September
0   1001        0                               3                               3
1   1002        1                               0                               2

感谢@SeaBean 提供add_prefix 方法。