问题描述
我有一个Time列，该列不是空对象，并且无法将其转换为timedelta或datetime.
I have a column Time which is non null object and I cannot convert it to timedelta or datetime.
Time msg
12:29:36.306000 Setup
12:29:36.507000 Alerting
12:29:38.207000 Service
12:29:39.194000 Setup
12:30:05.773000 Alerting
12:30:06.205000 Service
12:32:07.315000 Setup
12:32:17.194000 Service
12:32:26.889000 Setup
12:36:06.274000 Alerting
12:36:08.523000 Service
12:37:59.200000 Setup
12:47:10.652000 Alerting
12:47:43.921000 Setup
当我输入df.info()时，我发现时间"列不是非空对象，因此无法将其转换为timedelta或datetime(为此，很明显为什么我不能做到这一点).因此，找到连续的msg(时间增量)之间的差异的解决方案是什么，但是如果timedelta <比通过5秒.输出:
When I type df.info(), I got that a 'Time' column is non null object and I couldn't convert it to timedelta or datetime(for this it's obvious why I can't do it). So, what's is the solution to find the difference between consecutive msg (time delta), but if is timedelta < 5sec than pass.Output:
Time msg diff
12:29:36.306000 Setup
12:29:36.507000 Alerting
12:29:38.207000 Service
12:29:39.194000 Setup
12:30:05.773000 Alerting
12:30:06.205000 Service
12:32:07.315000 Setup
12:32:17.194000 Service
12:32:26.889000 Setup
12:36:06.274000 Alerting 6.30***
12:36:08.523000 Service
12:37:59.200000 Setup
12:47:10.652000 Alerting 11.02***
12:47:43.921000 Setup
我已经尝试过这样的事情:
I've tried with something like this:
df['diff'] = (df['Time']df['Time'].shift()).fillna(0)
但是我不知道在5秒间隔内写条件.
But I didn't know to write condition for 5sec interval.
推荐答案
我认为首先需要转换为str，然后调用 to_timedelta .
I think first need convert to str and then call to_timedelta.
然后获取 diff 并与5s进行比较.
最后一个新列使用 mask 通过遮罩:
Last for new column use mask by mask:
df['Time'] = pd.to_timedelta(df['Time'].astype(str))
df['diff'] = df['Time'].diff()
df['mask'] = df['Time'].diff() > pd.Timedelta(5, unit='s')
print (df)
Time msg diff mask
0 12:29:36.306000 Setup NaT False
1 12:29:36.507000 Alerting 00:00:00.201000 False
2 12:29:38.207000 Service 00:00:01.700000 False
3 12:29:39.194000 Setup 00:00:00.987000 False
4 12:30:05.773000 Alerting 00:00:26.579000 True
5 12:30:06.205000 Service 00:00:00.432000 False
6 12:32:07.315000 Setup 00:02:01.110000 True
7 12:32:17.194000 Service 00:00:09.879000 True
8 12:32:26.889000 Setup 00:00:09.695000 True
9 12:36:06.274000 Alerting 00:03:39.385000 True
10 12:36:08.523000 Service 00:00:02.249000 False
11 12:37:59.200000 Setup 00:01:50.677000 True
12 12:47:10.652000 Alerting 00:09:11.452000 True
13 12:47:43.921000 Setup 00:00:33.269000 True
df['Time'] = pd.to_timedelta(df['Time'])
diff = df['Time'].diff()
mask = df['Time'].diff() > pd.Timedelta(5, unit='s')
df['new'] = diff.where(mask)
print (df)
Time msg new
0 12:29:36.306000 Setup NaT
1 12:29:36.507000 Alerting NaT
2 12:29:38.207000 Service NaT
3 12:29:39.194000 Setup NaT
4 12:30:05.773000 Alerting 00:00:26.579000
5 12:30:06.205000 Service NaT
6 12:32:07.315000 Setup 00:02:01.110000
7 12:32:17.194000 Service 00:00:09.879000
8 12:32:26.889000 Setup 00:00:09.695000
9 12:36:06.274000 Alerting 00:03:39.385000
10 12:36:08.523000 Service NaT
11 12:37:59.200000 Setup 00:01:50.677000
12 12:47:10.652000 Alerting 00:09:11.452000
13 12:47:43.921000 Setup 00:00:33.269000
这篇关于如何在Python Pandas中强制转换时间列并查找有条件的timedelta的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！