Question

我想我可以通过一个例子更好地解释我想要实现的目标。假设我有这个数据帧：

     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，给定一个间隔T，计算每行，有多少寄存器被打开＆＃34;在那段时间里。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获得此结果。如果我只考虑dt.hour acessor，例如，对于T等于1，我可以创建每小时的列数，然后将它移动1，将count[i] + count[i-1]的结果相加。但是，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打开时间标记为+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值执行累积和以使计数在给定时间打开。

let a, b, c;
a = b = c = 2+3;

结果输出：

# 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']

计算间隔中的寄存器数

1 个答案: