Question

最近我问到如何根据Count number of registers in interval中的答案按间隔计算寄存器的数量。

该解决方案效果很好，但我必须对其进行调整以考虑一些本地化密钥。

我是通过以下代码完成的：

def time_features(df, time_key, T, location_key, output_key):
    """
    Create features based on time such as: how many BDs are open in the same GRA at this moment (hour)?
    """ 
    from datetime import date
    assert np.issubdtype(df[time_key], np.datetime64)
    output = pd.DataFrame()

    grouped = df.groupby(location_key)
    for name, group in grouped:
        # initialize times registers open as 1, close as -1
        start_times = group.copy()
        start_times[time_key] = group[time_key]-pd.Timedelta(hours=T)
        start_times[output_key] =  1

        aux = group.copy()
        all_times = start_times.copy()
        aux[output_key] = -1  
        all_times = all_times.append(aux, ignore_index=True)

        # sort by time and perform a cumulative sum to get opened registers
        # (subtract 1 since you don't want to include the current time as opened)
        all_times = all_times.sort_values(by=time_key)
        all_times[output_key] = all_times[output_key].cumsum() - 1

        # revert the index back to original order, and truncate closed times
        all_times = all_times.sort_index().iloc[:len(all_times)//2]
        output = output.append(all_times, ignore_index=True)
    return output

输出：

time    loc1    loc2
0   2013-01-01 12:56:00 1   "a"
1   2013-01-01 12:00:12 1   "b"
2   2013-01-01 10:34:28 2   "c"
3   2013-01-01 09:34:54 2   "c"
4   2013-01-01 08:34:55 3   "d"
5   2013-01-01 08:34:55 5   "d"
6   2013-01-01 16:35:19 4   "e"
7   2013-01-01 16:35:30 4   "e"

time_features(df, time_key='time', T=2, location_key='loc1', output_key='count')

这适用于小数据，但对于较长的数据（我使用带有100万行的文件），它需要＆＃34;永远＆＃34;跑步。我想知道我是否能以某种方式优化这个计算。

计数间隔和寄存器中的寄存器数量地点

0 个答案: