Re: [R] lag function row name with

Hi John, dplyr::lag expects a vector. The following should work dplyr::lag(temp[,1],2) HTH, Eric On Mon, Mar 25, 2019 at 9:45 AM John wrote: > Hi, > >I have a dataset whose row names corresponds to months. When I apply lag > function (dplyr package) on this dataset, I get NAs with warning

[R] lag function row name with

Hi, I have a dataset whose row names corresponds to months. When I apply lag function (dplyr package) on this dataset, I get NAs with warning messages. Is there any lag function that carries out the lag but keep the row names? I will have two datasets. The dates of the datasets are not exact

Re: [R] Lag function

Well, maybe. Whether it's +24 or -24 depends on what you mean by "one day lag." I suspect you mean -24, but perhaps this will help you decide: test <- ts(1:72, frequency = 24) plot(lag(test,24)) plot(lag(test,-24)) Note that the +24 moves the time base back 24 observable units (= hours) and -24 m

[R] Lag function

I have an hourly database and I defined a variable as follows: time<-ts(data$variable, frequency=24) If i need to create the variables with one day lag, the corresponding command is lag(time,24)? Thank you [[alternative HTML version deleted]]

Re: [R] Lag Function

It seems to mean that it adds a Tsp attribute but it does not change the class to "ts": > dput(lag(1:3)) structure(1:3, .Tsp = c(0, 2, 1)) Try this: > ts(1:3) - structure(lag(1:3), class = "ts") Time Series: Start = 1 End = 2 Frequency = 1 [1] -1 -1 or > ts(1:3) - lag(ts(1:3)) Time Series: Sta

Re: [R] Lag Function

Hello, Downey, Patrick wrote: Can anyone tell me what's going on here? x <- matrix(data=c(1,2,3,4,5),ncol=1) x1 <- lag(x,k=1) x x1 x - x1 That's with x specified as a column vector, but the same thing happens when it's a row vector. x <- c(1,2,3,4,5) x1 <- lag(x,k=1) x x1 x - x1 I'm not su

[R] Lag Function

Can anyone tell me what's going on here? x <- matrix(data=c(1,2,3,4,5),ncol=1) x1 <- lag(x,k=1) x x1 x - x1 That's with x specified as a column vector, but the same thing happens when it's a row vector. x <- c(1,2,3,4,5) x1 <- lag(x,k=1) x x1 x - x1 When the documentation says "Vector or matrix

Re: [R] lag function doesn't work - what am i doing wrong?

Note that there is actually a good reason why this is so. ts can only represent regularly spaced series but the result of a subscripting operation might not be a regularly spaced series so you would not be able to guarantee that it could be a ts series as well. That's the reason for window.ts too

Re: [R] lag function doesn't work - what am i doing wrong?

thanks. i'll try zoo or xts. ts is a p.i.t.a. Gabor Grothendieck wrote: > > lag1resid is a time series, try > > str(residsq) > str(lag1residsq) > > The problem is that when you subscript lag1resid you > don't get a time series out from that. See ?window.ts > or try the zoo or xts packages

Re: [R] lag function doesn't work - what am i doing wrong?

lag1resid is a time series, try str(residsq) str(lag1residsq) The problem is that when you subscript lag1resid you don't get a time series out from that. See ?window.ts or try the zoo or xts packages where subscripting of time series works. On Fri, Oct 10, 2008 at 1:27 PM, Scotty Nelson <[EMAIL

[R] lag function doesn't work - what am i doing wrong?

I am trying to lag a time series. My data is in a matrix, but I coerce it into a ts object. But when I lag it and then look at the result, nothing has changed. What am I doing wrong? residsq<-resid^2 residsq<-as.ts(residsq) lag1residsq<-lag(residsq,-1) > residsq[1:5] 1