python - 统计间隔内的寄存器数量

Question

我想我最好通过一个例子来解释我想要实现的目标。假设我有这个数据框:

     time
0     2013-01-01 12:56:00
1     2013-01-01 12:00:12
2     2013-01-01 10:34:28
3     2013-01-01 09:34:54
4     2013-01-01 08:34:55
5     2013-01-01 16:35:19
6     2013-01-01 16:35:30

我想，给定一个时间间隔 T，计算每行在该时间间隔内“打开”了多少个寄存器。例如，考虑到 T = 2 小时，这将是输出:

     time                  count
0     2013-01-01 12:56:00  1     # 12:56-2 = 10:56 -> 1 register between [10:56, 12:56)
1     2013-01-01 12:00:12  1 
2     2013-01-01 10:34:28  2     # 10:34:28-2 = 8:34:28 -> 2 registers between [8:34:28, 10:34:28) 
3     2013-01-01 09:34:54  1
4     2013-01-01 08:34:55  0
5     2013-01-01 16:35:19  0
6     2013-01-01 16:35:30  1

我想知道如何使用 pandas 获得这个结果。例如，如果我只考虑 dt.hour 访问器(accessor)，对于 T 等于 1，我可以创建每小时的列计数，然后将其移动 1，对 count[i] + count[i-1] 的结果求和。 。但我不知道是否可以将其概括为所需的输出。

Answer 1

这里的想法是将所有寄存器开放时间标记为+1，将所有寄存器关闭时间标记为-1。然后按时间排序并对 +/- 1 值执行累积求和，以获得给定时间打开的计数。

# initialize interval start times as 1, end times as -1
start_times= df.assign(time=df['time'] - pd.Timedelta(hours=2), count=1)
all_times = start_times.append(df.assign(count=-1), ignore_index=True)

# sort by time and perform a cumulative sum get the count of overlaps at a given time
# (subtract 1 since you don't want to include the current value in the overlap)
all_times = all_times.sort_values(by='time')
all_times['count'] = all_times['count'].cumsum() - 1

# reassign to the original dataframe, keeping only the original times
df['count'] = all_times['count']

结果输出:

                 time  count
0 2013-01-01 12:56:00      1
1 2013-01-01 12:00:12      1
2 2013-01-01 10:34:28      2
3 2013-01-01 09:34:54      1
4 2013-01-01 08:34:55      0
5 2013-01-01 16:35:19      0
6 2013-01-01 16:35:30      1