[R] efficient sine interpolation

Marc Girondot marc_grt at yahoo.fr
Tue May 13 08:08:21 CEST 2014


You can try this:
http://max2.ese.u-psud.fr/epc/conservation/Girondot/Publications/Blog_r/Entrees/2013/6/4_GLM_with_periodic_(annual)_transformation_of_factor.html

Sincerely,

Marc

Le 13/05/2014 05:42, Ortiz-Bobea, Ariel a écrit :
> Hello,
>
> I'm trying to fit a sine curve over successive temperature readings (i.e. minimum and maximum temperature) over several days and for many locations. The code below shows a hypothetical example of 5000 locations with  7 days of temperature data. Not very efficient when you have many more locations and days.
>
> The linear interpolation takes 0.7 seconds, and the sine interpolations take 2 to 4 seconds depending on the approach.
>
> Any ideas on how to speed this up? Thanks in advance.
>
> Ariel
>
> ### R Code ######
>
> # 1- Prepare data fake data
>    days<- 7
>    n <- 5000*days
>    tmin <- matrix(rnorm(n, mean=0) , ncol=days, nrow=5000)
>    tmax <- matrix(rnorm(n, mean=10), ncol=days, nrow=5000)
>    m <- matrix(NA, ncol=days*2, nrow=5000)
>    m[,seq(1,ncol(m),2)]  <- tmin
>    m[,seq(2,ncol(m)+1,2)]<- tmax
>    # check first row
>    plot(1:ncol(m), m[1,], type="l")
>
> # 2 -linear interpolation: 0.66 seconds
>    xout <- seq(0,ncol(m),0.25/24*2)[-1] # time step = 0.25 hours or 15 minutes
>    system.time( m1 <- t(apply(m,1, function(y) approx(x=1:ncol(m), y=y, xout=xout, method="linear")$y)) )
>    # Check first row
>    plot(1:ncol(m), m[1,], type="l")
>    points(xout, m1[1,], col="red", cex=1)
>
>
> # 3- sine interpolation
>    sine.approx1 <- function(index, tmin, tmax) {
>      b <- (2*pi)/24  # period = 24 hours
>      c <- pi/2 # horizontal shift
>      xout <- seq(0,24,0.25)[-1]
>      yhat <- apply(cbind(tmin[index,],tmax[index,]), 1, function(z) diff(z)/2 * sin(b*xout-c) + mean(z))
>      #yhat <- yhat[-nrow(yhat),]
>      yhat <- c(yhat)
>      #plot(yhat, type="l")
>    }
>    sine.approx2 <- function(index, tmin, tmax) {
>      b <- (2*pi)/24  # period = 24 hours
>      c <- pi/2 # horizontal shift
>      xout1 <- seq(0 ,12,0.25)
>      xout2 <- seq(12,24,0.25)[-1]
>      xout2 <- xout2[-length(xout2)]
>      yhat1 <- apply(cbind(tmin[index,]                       ,tmax[index,]    ), 1, function(z) diff(z)/2 * sin(b*xout1-c) + mean(z))
>      yhat2 <- apply(cbind(tmax[index,][-length(tmax[index,])],tmin[index,][-1]), 1, function(z) diff(z)/2 * sin(b*xout2+c) + mean(z))
>      yhat2 <- cbind(yhat2,NA)
>      yhat3 <- rbind(yhat1,yhat2)
>      #yhat3 <- yhat3[-nrow(yhat3),]
>      yhat3 <- c(yhat3)
>      yhat <- yhat3
>      #plot(c(yhat1))
>      #plot(c(yhat2))
>      #plot(yhat, type="l")
>    }
>
>    # Single sine: 2.23 seconds
>    system.time( m2 <- t(sapply(1:nrow(m), function(i) sine.approx1(i, tmin=tmin, tmax=tmax))) )
>
>    # Double sine: 4.03 seconds
>    system.time( m3 <- t(sapply(1:nrow(m), function(i) sine.approx2(i, tmin=tmin, tmax=tmax))) )
>
>    # take a look at approach 1
>    plot(seq(-1,ncol(m)-1,1)[-1], m[1,], type="l")
>    points(xout, m2[1,], col="red", cex=1)
>
>    # take a look at approach 2
>    plot(seq(-1,ncol(m)-1,1)[-1], m[1,], type="l")
>    points(xout, m3[1,], col="blue", cex=1)
>
>
> ---
> Ariel Ortiz-Bobea
> Fellow
> Resources for the Future
> 1616 P Street, N.W.
> Washington, DC 20036
> 202-328-5173
>
>
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>
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