Question

我有一个数据框，Phys。它包含一列时间和两列其他变量，如下所示： Data frame "Phys"

在某个时间点，两个变量达到某个阈值（例如，etatg> 0.5，etco2> 2.5）。我需要报告这些值在这些阈值之上的初始时间至少为以下9个元素（持续90秒）。我正在寻找最有效的方法来测试＆＃34;以下9个要素，看它们是否符合标准。

我目前有以下代码：

  #Find all instances of relevant heuristic
  tempalgEval = which(Phys$etagt > 0.5 & Phys$etco2>2.5)
  #Reduce tempalgEval by length 9 to avoid index error when searching data frame
  tempalgEval = head(tempalgEval, length(tempalgEval)-9)

  if (length(tempalgEval) < 9) {
    algEval = tempalgEval
  } else{
    for (m in tempalgEval) {
      if ((
        Phys$etagt[m + 1] > 0.5 &
        Phys$etagt[m + 2] > 0.5 &
        Phys$etagt[m + 3] > 0.5 &
        Phys$etagt[m + 4] > 0.5 &
        Phys$etagt[m + 5] > 0.5 &
        Phys$etagt[m + 6] > 0.5 &
        Phys$etagt[m + 7] > 0.5 &
        Phys$etagt[m + 8] > 0.5 &
        Phys$etagt[m + 9] > 0.5
      ) |
      (
        Phys$etco2[m + 1] > 2.5 &
        Phys$etco2[m + 2] > 2.5 &
        Phys$etco2[m + 3] > 2.5 &
        Phys$etco2[m + 4] > 2.5 &
        Phys$etco2[m + 5] > 2.5 &
        Phys$etco2[m + 6] > 2.5 &
        Phys$etco2[m + 7] > 2.5 &
        Phys$etco2[m + 8] > 2.5 & Phys$etco2[m + 9] > 2.5
      )) {
        algEval = tempalgEval
      }
    }
  }
  if(length(algEval) > 0){
    algTime = min(Phys$time[algEval], na.rm=T)
  }else{
    algTime = NA
  }

提前谢谢。

编辑：最小工作数据集

structure(
  list(
    time = c(
      1070,
      1080,
      1090,
      1100,
      1110,
      1120,
      1130,
      1160,
      1170,
      1180,
      1190,
      1200,
      1210,
      1220,
      1230,
      1240,
      1250,
      1260,
      1270,
      1280,
      1290,
      1300,
      1310,
      1320,
      1330,
      1340,
      1350,
      1360,
      1370,
      1380,
      1390
    ),
    etagt = c(
      0,
      0,
      0,
      0,
      0,
      0,
      0,
      2.92,
      2.33379310344828,
      1.74758620689655,
      1.21689655172414,
      1.18586206896552,
      1.1548275862069,
      1.11965517241379,
      1.06793103448276,
      1.01620689655172,
      0.997586206896552,
      1.05620689655172,
      1.1148275862069,
      1.16241379310345,
      1.19344827586207,
      1.22448275862069,
      1.23655172413793,
      1.22965517241379,
      1.22275862068966,
      1.74965517241379,
      2.63241379310345,
      3.5151724137931,
      3.59655172413793,
      3.33448275862069,
      3.07241379310345
    ),
    etco2 = c(
      0,
      0.871379310344828,
      2.11620689655172,
      3.36103448275862,
      2.61413793103448,
      1.36931034482759,
      0.124482758620689,
      0,
      1.5448275862069,
      3.08965517241379,
      4.49379310344828,
      4.63172413793103,
      4.76965517241379,
      4.92620689655172,
      5.15724137931034,
      5.38827586206897,
      5.53551724137931,
      5.48724137931034,
      5.43896551724138,
      5.37551724137931,
      5.28931034482759,
      5.20310344827586,
      5.16,
      5.16,
      5.16,
      4.15034482758621,
      2.46758620689655,
      0.784827586206896,
      1.56896551724138,
      3.41034482758621,
      5.25172413793103
    )
  ),
  .Names = c("time",
             "etagt", "etco2"),
  row.names = c(
    108L,
    109L,
    110L,
    111L,
    112L,
    113L,
    114L,
    117L,
    118L,
    119L,
    120L,
    121L,
    122L,
    123L,
    124L,
    125L,
    126L,
    127L,
    128L,
    129L,
    130L,
    131L,
    132L,
    133L,
    134L,
    135L,
    136L,
    137L, 138L, 139L, 140L), class = "data.frame")

Answer 1

您可以按照以下方式执行此操作：

require(data.table)
setDT(dat)
# tr := both Threshold Reached
dat[, tr:=etagt>0.5 & etco2 > 2.5] 
# Get grouping variable - in case have a look at ?rleid
dat[, run := rleid(tr)]
# Get indices where run was long enough 
# 10 means the first one and the 9 following were > threshold
ind <- dat[,.N, run][N>=10] # For >=9 you would get 2 matches
# Get the first timeing per run
dat[ind, on="run", mult="first"]

这给了你：

   time    etagt    etco2   tr run  N
1: 1180 1.747586 3.089655 TRUE   2 17

要查看最新情况，请查看dat，dat[,.N, run]和ind

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1 个答案: