[R] linear interpolation of multiple random time series

[jim holtman]

This should do it for you:

> x <- read.table(textConnection("trial   time    x
+ 1       1       1
+ 1       5       4
+ 1       7       9
+ 1       12      20
+ 2       1       0
+ 2       3       5
+ 2       9       10
+ 2       13      14
+ 2       19      22
+ 2       24      32"), header=TRUE)
> # compute for each trial
> trial.list <- lapply(split(x, x$trial), function(set){
+     .xval <- seq(min(set$time), max(set$time))
+     .yval <- approx(set$time, set$x, xout=.xval)$y
+     cbind(trial=set$trial[1], time=.xval, x=.yval)
+ })
> do.call('rbind', trial.list)
      trial time         x
 [1,]     1    1  1.000000
 [2,]     1    2  1.750000
 [3,]     1    3  2.500000
 [4,]     1    4  3.250000
 [5,]     1    5  4.000000
 [6,]     1    6  6.500000
 [7,]     1    7  9.000000
 [8,]     1    8 11.200000
 [9,]     1    9 13.400000
[10,]     1   10 15.600000
[11,]     1   11 17.800000
[12,]     1   12 20.000000
[13,]     2    1  0.000000
[14,]     2    2  2.500000
[15,]     2    3  5.000000
[16,]     2    4  5.833333
[17,]     2    5  6.666667
[18,]     2    6  7.500000
[19,]     2    7  8.333333
[20,]     2    8  9.166667
[21,]     2    9 10.000000
[22,]     2   10 11.000000
[23,]     2   11 12.000000
[24,]     2   12 13.000000
[25,]     2   13 14.000000
[26,]     2   14 15.333333
[27,]     2   15 16.666667
[28,]     2   16 18.000000
[29,]     2   17 19.333333
[30,]     2   18 20.666667
[31,]     2   19 22.000000
[32,]     2   20 24.000000
[33,]     2   21 26.000000
[34,]     2   22 28.000000
[35,]     2   23 30.000000
[36,]     2   24 32.000000
>


On 7/19/07, Mike Lawrence <Mike.Lawrence at dal.ca> wrote:
> Hi all,
>
> Looking for tips on how I might more optimally solve this. I have
> time series data (samples from a force sensor) that are not
> guaranteed to be sampled at the same time values across trials. ex.
>
> trial   time    x
> 1       1       1
> 1       5       4
> 1       7       9
> 1       12      20
> 2       1       0
> 2       3       5
> 2       9       10
> 2       13      14
> 2       19      22
> 2       24      32
>
> Within each trial I'd like to use linear interpolation between each
> successive time sample to fill in intermediary timepoints and x-
> values, ex.
>
> trial   time    x
> 1       1       1
> 1       2       1.75
> 1       3       2.5
> 1       4       3.25
> 1       5       4
> 1       6       6.5
> 1       7       9
> 1       8       11.2
> 1       9       13.4
> 1       10      15.6
> 1       11      17.8
> 1       12      20
> 2       1       0
> 2       2       2.5
> 2       3       5
> 2       4       5.83333333333333
> 2       5       6.66666666666667
> 2       6       7.5
> 2       7       8.33333333333333
> 2       8       9.16666666666667
> 2       9       10
> 2       10      11
> 2       11      12
> 2       12      13
> 2       13      14
> 2       14      15.3333333333333
> 2       15      16.6666666666667
> 2       16      18
> 2       17      19.3333333333333
> 2       18      20.6666666666667
> 2       19      22
> 2       20      24
> 2       21      26
> 2       22      28
> 2       23      30
> 2       24      32
>
>
> The solution I've coded (below) involves going through the original
> data frame line by line and is thus very slow (indeed, I had to
> resort to writing to file as with a large data set I started running
> into memory issues if I tried to create the new data frame in
> memory). Any suggestions on a faster way to achieve what I'm trying
> to do?
>
> #assumes the first data frame above is stored as 'a'
> arows = (length(a$x)-1)
> write('', 'temp.txt')
> for(i in 1:arows){
>        if(a$time[i+1] > a$time[i]){
>                write.table(a[i,], 'temp.txt', row.names = F, col.names = F, append
> = T)
>                x1 = a$time[i]
>                x2 = a$time[i+1]
>                dx = x2-x1
>                if(dx != 1){
>                        y1 = a$x[i]
>                        y2 = a$x[i+1]
>                        dy = y2-y1
>                        slope = dy/dx
>                        int = -slope*x1+y1
>                        temp=a[i,]
>                        for(j in (x1+1):(x2-1)){
>                                temp$time = j
>                                temp$x = slope*j+int
>                                write.table(temp, 'temp.txt', row.names = F, col.names = F,
> append = T)
>                        }
>                }
>        }else{
>                write.table(a[i,], 'temp.txt', row.names = F, col.names = F, append
> = T)
>        }
> }
> i=i+1
> write.table(a[i,], 'temp.txt', row.names = F, col.names = F, append = T)
>
> b=read.table('temp.txt',skip=1)
> names(b)=names(a)
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?