Question

我有一个列时间是非空对象，我无法将其转换为timedelta或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，我得到了一个＆＃39; Time＆＃39; column是非null对象，我无法将其转换为timedelta或datetime（为此我明白为什么我不能这样做）。那么，找到连续msg（时间增量）之间的差异的解决方案是什么，但是如果是timedelta＆lt;通过5秒。输出：

     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

我尝试过这样的事情：

df['diff'] = (df['Time']df['Time'].shift()).fillna(0)

但我不知道写5秒间隔的条件。

Answer 1

我认为首先需要转换为str，然后调用to_timedelta。

然后获取diff和comapre 5s。

最后通过掩码使用mask新列：

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

如何在python pandas中建立时间列并在条件中找到timedelta

1 个答案: