在 R 中有效地重采样数据(线性外推)

Resampling data efficiently in R (linear extrapolation)

我有一组以 1 秒为间隔记录的观察结果(伽玛和时间)。我想在 0.1 秒时重新采样这些数据。数据如下所示:

38804.96    12.59222222
38805.12    12.5925
38805.38    12.59277778
38805.4     12.59305556
38805.27    12.59333333
38805.36    12.59361111
38805.33    12.59388889
38805.23    12.59416667
38805.3     12.59444444
38805.18    12.59472222
38805.21    12.595
38805.28    12.59527778

我想出了以下代码来重新采样(线性外推)伽玛,但它非常耗时,因为我的数据集有超过 30000 个观测值。

    #Resampling la diurnal drift
    j <- (0:9)
    A <- 0
    VectorT <- numeric()
    VectorG <- numeric()
    for (I in 1:nrow(R20140811)){ 
       # Calculate the increment of time
       Rate <- (R20140811[I+1,2]- R20140811[I,2])/10
       Time <- R20140811[I,2]
       # Calculate the increment of gamma
       nT <- (R20140811[I+1,1] - R20140811[I,1])/10
       Gamma <- R20140811[I,1]
       print(I)
       for (j in 0:9){ 
          A <- A + 1
          VectorT[A] <- Time + (j*Rate)
          VectorG[A] <- Gamma + (j*nT)
          R20140811[A,3] <- VectorG[A]
          R20140811[A,4] <- VectorT[A]
       }
    }

你知道更有效的方法吗?

您需要矢量化您的计算。

鉴于您的矩阵:

R20140811 <- matrix(
c(38804.96   ,12.59222222,
38805.12   ,12.5925    ,
38805.38   ,12.59277778,
38805.4    ,12.59305556,
38805.27   ,12.59333333,
38805.36   ,12.59361111,
38805.33   ,12.59388889,
38805.23   ,12.59416667,
38805.3    ,12.59444444,
38805.18   ,12.59472222,
38805.21   ,12.595     ,
38805.28   ,12.59527778),ncol=2,byrow=TRUE)

单独的时间和伽玛:

times <- R20140811[,2]
gammas <- R20140811[,1]

然后定义外推函数:

# given a vector vect, extrapole nInt points between points
Extrapol <- function(vect,nInt){
        # the starting points of your intervals
        zeros <- vect[1:(length (vect)-1)]
        # determine the increments
        increments <- (vect[2:length (vect)]-zeros)/nInt
        # get the new sample
        newSample <- rep(zeros[1: (length (times)-1)],each=10) + as.vector(outer (0:9,increments))
        return(newSample)
}

并将外推函数应用于您的时间和 gammas

newSampleGamma <- Extrapol(gammas,10)
newSampleTimes <- Extrapol(times,10)

应该快几个数量级:-)