背景资讯

电子表格A每33毫秒（在时间戳列中）有时间戳记值。此外，它（in）方便地跳过第二个＆＃34; S.000＆＃34;。

的每个顶部

SPREADSHEET A

时间戳|位置1 5/4/2017 10：00：00.000 | 0.005430355
5/4/2017 10：00：00.033 | 0.004475154
5/4/2017 10：00：00.066 | 0.004958829
5/4/2017 10：00：00.099 | 0.010678668
5/4/2017 10：00：00.133 | 0.014313085
5/4/2017 10：00：00.166 | 0.004182263
5/4/2017 10：00：00.199 | 0.00232128
5/4/2017 10：00：00.233 | 0.004263132
5/4/2017 10：00：00.266 | 0.007513777
。
。
5/4/2017 10：00：00.999 | 0.011229943 #always跳过第二个＆＃34; S.000＆＃34;
5/4/2017 10：00：01.033 | 0.016148495 #always跳过第二个＆＃34; S.000＆＃34;
5/4/2017 10：00：01.066 | 0.009239103
5/4/2017 10：00：01.099 | 0.015364848
5/4/2017 10：00：01.133 | 0.032139104
5/4/2017 10：00：01.166 | 0.023679454
5/4/2017 10：00：01.199 | 0.002503840

如何使用pandas来舍入每个具有＆＃34; X.999＆＃34;到最近的第二个？我想

see if a value is "X.999" in the to "Time Stamp" column if it is "X.999": change it to "X+1.000" / or round it up else leave the time stamp as it is

然后继续重复上述步骤，直到我为新数据框创建新的设置时间戳。或者甚至更好，用新的时间戳替换当前的数据帧时间戳。

Answer 1

到目前为止，这是我的解决方案......花了很长时间，但它现在有效。考虑微秒值为9700000（970毫秒）或更高的时间戳，将其差异从1秒加到自身，然后用这个新总和替换时间戳（总是在X.000秒）。

这是代码。

%matplotlib inline
import seaborn as sns
import matplotlib as mp
import pandas as pd
import numpy as np
import datetime
dfa=pd.read_excel("may2017.xlsx",sheetname=0)
dfb=pd.read_excel("logs.xlsx",sheetname=0, skip_rows=746243)
dfb = b.iloc[746244:754431] #got the index value range from trial and error
a = dfa
b = dfb
adummy = a
bdummy = b

for i in range(len(adummy.index)): #get the length of all rows
    #find the difference between each time stamp from 1 second
    dif = 1000000 - adummy.iloc[i]["Time Stamp"].microsecond
    #change microseconds to timedelta object for later
    tdelta = datetime.timedelta(microseconds=dif)
    #now check if this row timestamp is near a 1-second interval
    if adummy.iloc[i]["Time Stamp"].microsecond > 975000:
        #if near 1 second, 'round up' to the next second by adding tdelta
        tnew = tdelta + adummy.iloc[i]["Time Stamp"]
        #replace the time stamp in this row with the nearest second above
        adummy["Time Stamp"] = adummy["Time Stamp"].replace(adummy.iloc[i]["Time Stamp"], tnew)
        #print this new time to verify the dataframe got updated
        print(adummy.iloc[i]["Time Stamp"])
    else:
#else if the timestamp is not near 1 from the low side, do nothing
        pass
#Restart the loop

输出：

2017-05-04 10:00:01
2017-05-04 10:00:02
2017-05-04 10:00:03
2017-05-04 10:00:04
2017-05-04 10:00:05
2017-05-04 10:00:06
2017-05-04 10:00:07

pandas python round timestamp到最近的第二个

背景资讯

1 个答案: