我想保留最后几行,但是一旦在100ms以上有时间间隔,就切断其余的数据帧。例如:

输入:

           Time  X
0   12:30:00.00  A
1  12:30:00.100  B
2  12:30:00.202  C
3  12:30.00.300  D


输出量

           Time  X
2  12:30:00.202  C
3  12:30.00.300  D


说明:B行和C行之间的距离超过100毫秒,因此我们将C行上方的所有内容都丢弃了。

最佳答案

您可以使用diffTimedeltato_timedelta进行比较,然后将cumsum1进行比较。上次使用boolean indexing

df['Time']= pd.to_datetime(df['Time'], format='%H:%M:%S.%f')

print (df)
                     Time  X
0 1900-01-01 12:30:00.000  A
1 1900-01-01 12:30:00.100  B
2 1900-01-01 12:30:00.202  C
3 1900-01-01 12:30:00.300  D

print (df.Time.diff())
0               NaT
1   00:00:00.100000
2   00:00:00.102000
3   00:00:00.098000
Name: Time, dtype: timedelta64[ns]

mask = (((df.Time.diff() > pd.to_timedelta('00:00:00.100000')).cumsum()) >= 1)
print (mask)
0    False
1    False
2     True
3     True
Name: Time, dtype: bool

print (df[mask])
                     Time  X
2 1900-01-01 12:30:00.202  C
3 1900-01-01 12:30:00.300  D


如果需要列Time不变,则将第一个值拆分为更高的100ms

df['Time1']= pd.to_datetime(df['Time'], format='%H:%M:%S.%f')
print (df)
           Time  X                   Time1
0   12:30:00.00  A 1900-01-01 12:30:00.000
1  12:30:00.100  B 1900-01-01 12:30:00.100
2  12:30:00.202  C 1900-01-01 12:30:00.202
3  12:30:00.300  D 1900-01-01 12:30:00.300
1  12:30:00.100  E 1900-01-01 12:30:00.100
2  12:30:00.202  F 1900-01-01 12:30:00.202

print (df.Time1.diff())
0                        NaT
1            00:00:00.100000
2            00:00:00.102000
3            00:00:00.098000
1   -1 days +23:59:59.800000
2            00:00:00.102000
Name: Time1, dtype: timedelta64[ns]

mask = (((df.Time1.diff() > pd.to_timedelta('00:00:00.100000')).cumsum()) >= 1)
print (mask)
0    False
1    False
2     True
3     True
1     True
2     True
Name: Time1, dtype: bool

print (df[mask].drop('Time1',axis=1))
           Time  X
2  12:30:00.202  C
3  12:30:00.300  D
1  12:30:00.100  E
2  12:30:00.202  F


如果需要除以最后一个值:

print (df)
           Time  X
0   12:30:00.00  A
1  12:30:00.100  B
2  12:30:00.202  C
3  12:30:00.300  D
1  12:30:00.100  E
2  12:30:00.202  F

#create helper series
time_ser= pd.to_datetime(df['Time'], format='%H:%M:%S.%f')
#get differences
print (time_ser.diff())
0                        NaT
1            00:00:00.100000
2            00:00:00.102000
3            00:00:00.098000
1   -1 days +23:59:59.800000
2            00:00:00.102000
Name: Time, dtype: timedelta64[ns]




#compare with 100ms timedalta
mask = (((time_ser.diff() > pd.to_timedelta('00:00:00.100000')).cumsum()))
print (mask)
0    0
1    0
2    1
3    1
1    1
2    2
Name: Time, dtype: int32

#get last value of mask
last_val = mask.iat[-1]
print(last_val)
2

#compare mask with last value and use boolean indexing
print (df[mask == last_val])
           Time  X
2  12:30:00.202  F

10-02 22:11