对于列中的每个新值，请删除以下两行

Question

我有以下数据帧：

time   alarm
0       0
1       1
2       0
3       1
4       1
5       1
6       1
7       0
8       0
9       1
10      0

列alarm表示报警。如果它响起，则取值为1
每次闹钟响起，我都想"静音"；接下来的两行。然后，如果它在静音期后再次响起，我想静音接下来的两行，以此类推

换句话说，我想获得以下数据帧：

time   alarm    silenced
0       0       no
1       1       no
2       0       yes
3       1       yes
4       1       no
5       1       yes
6       1       yes
7       0       no
8       0       no
9       1       no
10      0       yes

我设法使用for循环或lambda函数来实现，但我必须加快计算速度
有人能帮我吗？提前谢谢！

---

p.S。由于我最终将删除"；静音"；行，也将接受直接删除此类行的解决方案。在这种情况下，结果应该是：

time   alarm
0       0
1       1
4       1
7       0
8       0
9       1

---

在辅助功能中使用for循环的MY ATTEMPT

import numpy as np
import pandas as pd
df = pd.DataFrame({"time":[0,1,2,3,4,5,6,7,8,9,10], "alarm":[0,1,0,1,1,1,1,0,0,1,0]})
df
def fun_silence(df):

# bool: if True,  we are in a "silent" period 
#       if False, we can consider the current time as a possible alarm
flag_silent = False

# time of the *last* alarm
alarm_time = np.nan

# loop over rows
for index, row in df.iterrows():

# if we are in a silent period
if flag_silent:

# if 2 time steps passed from the last alarm, we end the silent period
if row['time'] - alarm_time > 2:
flag_silent = False

# otherwise, we mark this row as silenced
else:
df.at[index, 'silenced'] = 1

# if there is an alarm and we are not in a silent period
if row['alarm'] == 1 and not flag_silent:
# save the alarm time
alarm_time = row['time']
# enter in a silent period
flag_silent = True

return df

df['silenced'] = 0
df_silenced = fun_silence(df)
df_silenced

Answer 1

我认为你无法避免这个问题中的for循环，但你当然可以优化函数，然后使用numba编译它，以在大型数据集上实现类似C的速度

from numba import njit
@njit
def silence(alarm):
count = 0
for a in alarm:
if count > 0:
yield True
count -= 1
elif count == 0 and a == 1:
count = 2
yield False
else:
yield False

df['silenced'] = [*silence(df['alarm'].to_numpy())]

---

time  alarm  silenced
0      0      0     False
1      1      1     False
2      2      0      True
3      3      1      True
4      4      1     False
5      5      1      True
6      6      1      True
7      7      0     False
8      8      0     False
9      9      1     False
10    10      0      True