PROSAGA码农传奇-matlab-时间序列的时间百分比[关闭]

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  <P>
    指出的问题不是很清楚，但也许这可能会有所帮助，使用基数R：
  </p>
   <pre>
    <code>
      # first you define your time as date
dats$time <-strptime(dats$time, " %d/%m/%Y %I:%M:%S %p")

# now a lagged vector, that points out the "lasting"
lagged <- c(tail(dats$time , -1),NA)

# here added as column
dats <- cbind(dats, lagged)

dats
                  time rpm              lagged
1  2018-07-08 20:08:16   8 2018-07-08 20:08:21
2  2018-07-08 20:08:21   9 2018-07-08 20:08:41
3  2018-07-08 20:08:41   8 2018-07-08 20:09:30
4  2018-07-08 20:09:30   9 2018-07-08 20:19:02
5  2018-07-08 20:19:02  10 2018-07-08 20:23:16
6  2018-07-08 20:23:16   9 2018-07-08 20:23:22
7  2018-07-08 20:23:22  10 2018-07-08 20:23:24
8  2018-07-08 20:23:24   9 2018-07-08 20:23:36
9  2018-07-08 20:23:36   9 2018-07-08 20:23:45
10 2018-07-08 20:23:45  10                <NA>

# calculate the full time
 overall <- max(dats$time)-min(dats$time)
overall
Time difference of 15.48333 mins

# select only the rpm you want, unluckily due your data, I choose 8, because I do not
# have 1, you can easily change it
dats_rpm <- dats[dats$rpm <= 8,]

# let's calculate each duration of the rpm <= 8
dats_rpm$duration <- dats_rpm$lagged - dats_rpm$time

# lastly the percentage of the duration of rpm<=8 / overall duration
as.double(sum(dats_rpm$duration), units='secs')/as.double(overall, units='secs')*100
[1] 5.812702

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  </pre>
  <P>
    希望能帮助到你！
  </p>
  <HR />
  <P>
    有了数据：
  </p>
   <pre>
    <code>
      dats <-  structure(list(time = c("08/07/2018 08:08:16 PM", "08/07/2018 08:08:21 PM", 
"08/07/2018 08:08:41 PM", "08/07/2018 08:09:30 PM", "08/07/2018 08:19:02 PM", 
"08/07/2018 08:23:16 PM", "08/07/2018 08:23:22 PM", "08/07/2018 08:23:24 PM", 
"08/07/2018 08:23:36 PM", "08/07/2018 08:23:45 PM"), rpm = c(8L, 
9L, 8L, 9L, 10L, 9L, 10L, 9L, 9L, 10L)), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10"))

</code>
  </pre>
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