我有一個數據框，其中包含有關誰在工作、執行哪個任務以及他/她開始工作的時間的“日志”信息： index | Entrance time | Name | Last name | Employee_ID | Task -------------------------------------------------------------------- 0 |2000-01-01 00:00:00 | John | Fischer | 001 | Maintenance 1 |2000-01-01 00:04:30 | John | Fischer | 001 | Development 2 |2000-01-01 00:04:30 | Bob | Conrad | 002 | Maintenance 3 |2000-01-01 00:10:00 | Mary | Smith | 003 | Multitasking 4 |2000-01-01 00:09:30 | John | Fischer | 001 | Maintenance 5 |2000-01-01 00:15:30 | John | Fischer | 001 | Maintenance 6 |2000-01-02 00:04:30 | Bob | Conrad | 002 | Maintenance 7 |2000-01-02 00:10:00 | Mary | Smith | 003 | Multitasking然后，如果我們正在查找的任務和其他任務之間的進入時間差小于 10 分鐘，并且任務和名稱相同，我想消除重復項。所以生成的數據框應該是： index | Entrance time | Name | Last name | Employee_ID | Task -------------------------------------------------------------------- 0 |2000-01-01 00:00:00 | John | Fischer | 001 | Maintenance 1 |2000-01-01 00:04:30 | John | Fischer | 001 | Development 2 |2000-01-01 00:04:30 | Bob | Conrad | 002 | Maintenance 3 |2000-01-01 00:10:00 | Mary | Smith | 003 | Multitasking 5 |2000-01-01 00:15:30 | John | Fischer | 001 | Maintenance 6 |2000-01-02 00:04:30 | Bob | Conrad | 002 | Maintenance 7 |2000-01-02 00:10:00 | Mary | Smith | 003 | Multitasking我使用過drop_duplicates(subset=["Name", "Last name", "Task"])，但我不知道如何應用時間條件來將每一行與其余行進行比較。希望你能幫助我，提前謝謝你
1 回答
蕪湖不蕪
TA貢獻1796條經驗 獲得超7個贊
對于時差的計算，這可能會對您有所幫助。但是，您還需要根據重復案例應用您的條件。
# Make df sequential in ["Name", "Last name", "Task"]
df.sort_values(["Name", "Last name", "Task"], inplace=True)
# Compute time difference
temp = df['Entrance time'] - df['Entrance time'].shift()
# converts the difference in terms of minutes (taking into account absolute values)
df['diff_mins'] = temp.abs() /np.timedelta64(1,'m')
輸出：
2 2 2000-01-01 00:04:30 Bob Conrad 2 Maintenance nan
6 6 2000-01-02 00:04:30 Bob Conrad 2 Maintenance 1440
1 1 2000-01-01 00:04:30 John Fischer 1 Development 1440
0 0 2000-01-01 00:00:00 John Fischer 1 Maintenance 4.5
4 4 2000-01-01 00:09:30 John Fischer 1 Maintenance 9.5
5 5 2000-01-01 00:15:30 John Fischer 1 Maintenance 6
3 3 2000-01-01 00:10:00 Mary Smith 3 Multitasking 5.5
7 7 2000-01-02 00:10:00 Mary Smith 3 Multitasking 1440
添加回答
舉報
0/150
提交
取消