Rolling standard deviation for every 5th day in R

Question

I have a time series dataframe which looks like

2014-02-05 2014-02-06 2014-02-07 2014-02-12 2014-02-14 2014-02-17 2014-02-18 2014-02-19 ......
    0.0379    -0.0008     0.0352     0.0379     0.0392     0.0173     0.0360     0.0371

I want to compute moving standard deviation for every 5th day data from this list in R. What I mean is that, I wish to select a sample in the form such that sample1[1] = 2014-02-05, 0.0379 , sample1[2] =2014-02-12, 0.0379.....and then find the std dev of this sample and then use a rolling standard deviation to move on to the next date ie sample2[1] =2014-02-06, -0.0008 , sample2[2] =2014-02-12, 0.0379 and find the standard deviation of this list and so on. Since day available is irregular, I cannot use seq(1:l, by = ). In rollapply, the function would take every consecutive numbers to compute the standard deviation. Is there a way to sample every 5th day data from this list in an efficient way, or modify the standard deviation function somehow, to make it select every 5th day data and then compute the standard deviation on the available data. Any suggestion in this regard will be highly appreciated.

Answer 1

Restating Question I am assuming you want to fill in missing days and then if z is the resulting series calculate the following

sd(c(z[1], z[6], z[11], z[16], z[21]))
sd(c(z[2], z[7], z[12], z[17], z[22]))
etc.

but only keeping those sd's which start at times found in sample1 .

If that is not the intent of the question please clarify with further explanation and by giving an actual example of input and output.

Answer Create a daily grid g and merge with sample1 filling in NAs from the end giving the filled in series z . (Note that if points have gaps greater than 4 days then we do not fill those gaps since that would involve including points more than once in the sd.) Then use rollapply to compute the desired sd keeping only the original times.

g <- zoo(, seq(start(sample1), end(sample1), "day"))
z <- na.locf(merge(sample1, g), fromLast = TRUE, maxgap = 4)
r <- rollapply(z, 21, function(x) sd(x[seq(1, 21, 5)]), align = "left")
r[time(sample1)]

Note The rollapply statement could alternatively be written like this:

r <- rollapply(z, list(seq(0, length = 5, by = 5)), sd)

since the width argument can be specified as a list containing a vector of offsets.

Update Revised again after re-reading question. Also provided alternate rollapply expression.

Answer 2

Following may be useful:

xx = structure(c(0.0379, -8e-04, 0.0352, 0.0379, 0.0392, 0.0173, 0.036, 
0.0371), .Names = c("2014-02-05", "2014-02-06", "2014-02-07", 
"2014-02-12", "2014-02-14", "2014-02-17", "2014-02-18", "2014-02-19"
))

xx
2014-02-05 2014-02-06 2014-02-07 2014-02-12 2014-02-14 2014-02-17 2014-02-18 2014-02-19 
    0.0379    -0.0008     0.0352     0.0379     0.0392     0.0173     0.0360     0.0371 
yy = as.numeric()
for(i in 5:length(xx)){
    yy[i]= sd(xx[(i-4):i])
}
yy
[1]          NA          NA          NA          NA 0.017212408 0.017278108 0.008982038 0.009130991

For a data frame version:

ddf = structure(list(date = structure(1:8, .Label = c("2014-02-05", 
"2014-02-06", "2014-02-07", "2014-02-12", "2014-02-14", "2014-02-17", 
"2014-02-18", "2014-02-19"), class = "factor"), value = c(0.0379, 
-8e-04, 0.0352, 0.0379, 0.0392, 0.0173, 0.036, 0.0371)), .Names = c("date", 
"value"), class = "data.frame", row.names = c(NA, -8L))


ddf
        date   value
1 2014-02-05  0.0379
2 2014-02-06 -0.0008
3 2014-02-07  0.0352
4 2014-02-12  0.0379
5 2014-02-14  0.0392
6 2014-02-17  0.0173
7 2014-02-18  0.0360
8 2014-02-19  0.0371


ddf$rolling_sd=0
for(i in 5:nrow(ddf)){
    ddf$rolling_sd[i]= sd(ddf$value[(i-4):i])
}
ddf
        date   value  rolling_sd
1 2014-02-05  0.0379 0.000000000
2 2014-02-06 -0.0008 0.000000000
3 2014-02-07  0.0352 0.000000000
4 2014-02-12  0.0379 0.000000000
5 2014-02-14  0.0392 0.017212408
6 2014-02-17  0.0173 0.017278108
7 2014-02-18  0.0360 0.008982038
8 2014-02-19  0.0371 0.009130991