熊猫测量自条件以来经过的时间

Question

我有以下数据帧:

               Time   Work
2018-12-01 10:00:00     Off
2018-12-01 10:00:02     On
2018-12-01 10:00:05     On
2018-12-01 10:00:06     On
2018-12-01 10:00:07     On
2018-12-01 10:00:09    Off
2018-12-01 10:00:11    Off
2018-12-01 10:00:14     On
2018-12-01 10:00:16     On
2018-12-01 10:00:18     On
2018-12-01 10:00:20    Off

我想创建一个新列,其中包含自设备开始工作以来经过的时间.

               Time   Work    Elapsed Time
2018-12-01 10:00:00    Off               0
2018-12-01 10:00:02     On               2
2018-12-01 10:00:05     On               5
2018-12-01 10:00:06     On               6
2018-12-01 10:00:07     On               7
2018-12-01 10:00:09    Off               0
2018-12-01 10:00:11    Off               0
2018-12-01 10:00:14     On               3
2018-12-01 10:00:16     On               5
2018-12-01 10:00:18     On               7
2018-12-01 10:00:20    Off               0

我该怎么做？

Answer 1

你可以使用groupby:

# df['Time'] = pd.to_datetime(df['Time'], errors='coerce') # Uncomment if needed.
sec = df['Time'].dt.second
df['Elapsed Time'] = (
    sec - sec.groupby(df.Work.eq('Off').cumsum()).transform('first'))

df
                  Time Work  Elapsed Time
0  2018-12-01 10:00:00  Off             0
1  2018-12-01 10:00:02   On             2
2  2018-12-01 10:00:05   On             5
3  2018-12-01 10:00:06   On             6
4  2018-12-01 10:00:07   On             7
5  2018-12-01 10:00:09  Off             0
6  2018-12-01 10:00:11  Off             0
7  2018-12-01 10:00:14   On             3
8  2018-12-01 10:00:16   On             5
9  2018-12-01 10:00:18   On             7
10 2018-12-01 10:00:20  Off             0

想法是提取秒部分并从状态从"关"变为"开"的第一时刻减去经过的时间.这是使用transform和完成的first.

cumsum 用于识别组:

df.Work.eq('Off').cumsum()

0     1
1     1
2     1
3     1
4     1
5     2
6     3
7     3
8     3
9     3
10    4
Name: Work, dtype: int64

如果您的设备有可能在"开启"时跨越多分钟,则初始化sec为:

sec = df['Time'].values.astype(np.int64) // 10e8

df['Elapsed Time'] = (
    sec - sec.groupby(df.Work.eq('Off').cumsum()).transform('first'))

df
                  Time Work  Elapsed Time
0  2018-12-01 10:00:00  Off           0.0
1  2018-12-01 10:00:02   On           2.0
2  2018-12-01 10:00:05   On           5.0
3  2018-12-01 10:00:06   On           6.0
4  2018-12-01 10:00:07   On           7.0
5  2018-12-01 10:00:09  Off           0.0
6  2018-12-01 10:00:11  Off           0.0
7  2018-12-01 10:00:14   On           3.0
8  2018-12-01 10:00:16   On           5.0
9  2018-12-01 10:00:18   On           7.0
10 2018-12-01 10:00:20  Off           0.0

Answer 2

IIUC first与transform

(df.Time-df.Time.groupby(df.Work.eq('Off').cumsum()).transform('first')).dt.seconds
Out[1090]: 
0     0
1     2
2     5
3     6
4     7
5     0
6     0
7     3
8     5
9     7
10    0
Name: Time, dtype: int64

Answer 3

你可以用两个groupbys.第一个计算每个组内的时差.然后第二个对每个组内的那些进行求和.

s = (df.Work=='Off').cumsum()
df['Elapsed Time'] = df.groupby(s).Time.diff().dt.total_seconds().fillna(0).groupby(s).cumsum()

产量

                  Time Work  Elapsed Time
0  2018-12-01 10:00:00  Off           0.0
1  2018-12-01 10:00:02   On           2.0
2  2018-12-01 10:00:05   On           5.0
3  2018-12-01 10:00:06   On           6.0
4  2018-12-01 10:00:07   On           7.0
5  2018-12-01 10:00:09  Off           0.0
6  2018-12-01 10:00:11  Off           0.0
7  2018-12-01 10:00:14   On           3.0
8  2018-12-01 10:00:16   On           5.0
9  2018-12-01 10:00:18   On           7.0
10 2018-12-01 10:00:20  Off           0.0