search for: nfrequenc

...ggregate(x, nfreq=1/3) Error in aggregate.ts(x, nfreq = 1/3) : cannot change frequency from 1 to 0.333333333333333 In fact aggregate.ts only accepts a new frequency that is a negative power of two in this example. The problem with the current test for compatible frequencies if ((ofrequency%%nfrequency) != 0) stop(paste("cannot change frequency from", ofrequency, "to", nfrequency)) is that numerical errors may cause it to fail. I suggest if (abs(ofrequency%%nfrequency) < ts.eps) instead. Martyn --please do not edit the information below-- Ver...

..., I want the points for the seasonal plot to line up with cycle 6 of the first plot. Thanks. dat <- ts(rnorm(12*20), start = c(1980,1), frequency = 12) plot(dat) dat.sub <- dat dat.sub[cycle(dat.sub) < 6] <- NA dat.sub[cycle(dat.sub) > 8] <- NA dat.sub <- aggregate(dat.sub, nfrequency = 1, FUN = mean, na.rm = T) tsp(dat) tsp(dat.sub) par(mfrow = c(2, 1)) plot(dat) plot(dat.sub, type = "b")

...can't get the codes right to do what I want. Here are examples of unsuccessful attempts: try = aggregate(test.irts$value, by= list(time=test.ts$time), FUN= mean) ERROR MESS: Error in FUN(X[[1L]], ...) : arguments must have same length try = aggregate.ts(test.irts$value,frequency(test.ts$time),nfrequency=1, ndeltat=1) PROBLEM: gives me only one vector for my values (no time vector) but they are NOT aggregated by hour. try = list(x=test.ts$time, y=aggregate.ts(test.irts$value, nfrequency=1)) PROBLEM: Gives 2 separate vectors :hourly time step and values - but again, values are not aggregated by ho...

...1 1987 12 13 14 15 1988 16 17 18 19 1989 20 If I do this manually, the mean for the 1st quarter would be mean(c(4,8,12,16,20)), which is 12. But I am wondering if there is a R function that could do this faster. I tried aggregate.ts but it didn't work: > aggregate(ts.data,nfrequency=4,mean) Qtr1 Qtr2 Qtr3 Qtr4 1984 1 2 3 1985 4 5 6 7 1986 8 9 10 11 1987 12 13 14 15 1988 16 17 18 19 1989 20 Does anyone know what am I doing wrong? -- Tom [[alternative HTML version deleted]]

...uency = 0.5 [1] 2 4 6 8 S> x <- rts(1:10) S> aggregate(x, nf=0.5, fun = min) [1] 1 3 5 7 9 start deltat frequency 1 2 0.5 I have edited the aggregate.ts function to reproduce the S-PLUS behaviour, which I find more intuitive: old: #nstart <- ceiling(tsp(x)[1] * nfrequency)/nfrequency #x <- as.matrix(window(x, start = nstart)) new: nstart <- tsp(x)[1] # Can't use nstart <- start(x) as this causes problems if # you get a vector of length 2. x <- as.matrix(x) I have tried this on a range of examples and have not encountered any...