Hi all, I have a time series which contain data collected weekly from week 26 to week 25 the following year. How do I plot this data, so that the x-axis is displaying the week numbers, ordered as in the data? Thanks in advance, Gustaf --- x<-c(26:52,1:25) y<-rnorm(52)+1:52 plot(x,y) ## How do I get the x axis to be ordered by the current ordering of x? -- Gustaf Rydevik, M.Sci. tel: +46(0)703 051 451 address:Essingetorget 40,112 66 Stockholm, SE skype:gustaf_rydevik
This should do what you want. x<-c(26:52,1:25) y<-rnorm(52)+1:52 plot(seq_along(x), y, xaxt='n') axis(1, at=seq_along(x), labels=x) On 9/14/07, Gustaf Rydevik <gustaf.rydevik at gmail.com> wrote:> Hi all, > > I have a time series which contain data collected weekly from week 26 > to week 25 the following year. How do I plot this data, so that the > x-axis is displaying the week numbers, ordered as in the data? > > Thanks in advance, > > Gustaf > --- > x<-c(26:52,1:25) > y<-rnorm(52)+1:52 > plot(x,y) ## How do I get the x axis to be ordered by the current > ordering of x? > > > > -- > Gustaf Rydevik, M.Sci. > tel: +46(0)703 051 451 > address:Essingetorget 40,112 66 Stockholm, SE > skype:gustaf_rydevik > > ______________________________________________ > R-help at r-project.org 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?
On 9/17/07, Nair, Murlidharan T <mnair at iusb.edu> wrote:> > I am calculating the median absolute deviation using mad function, and > it tends to ignore the parameter constant=1, when I am calculating it > for x=seq(1:5). Am I missing something here? > > x<-seq(1:5) > mad(x)# gives [1] 1.4826 > mad(x, constant=1)# gives [1] 1 > #Here is the long form > dev.from.median<-abs((x-median(x))) > dev.from.median # Gives [1] 2 1 0 1 2 > sum(dev.from.median) # Gives [1] 6 > sum(dev.from.median)/length(x) # Gives [1] 1.2 > # The long form does not match the output from the function > > # When x<-seq(1:10) they match > x<-seq(1:10) > dev.from.median<-abs((x-median(x))) > sum(dev.from.median)/length(x) # Gives 2.5 > mad(x, constant=1) # Gives 2.5 > #The long form matches the output from the function > > Did I miss anything here?yes; mad := Median (not mean) absolute deviation (from the median, by default). -Deepayan
-----Original Message----- From: Deepayan Sarkar [mailto:deepayan.sarkar at gmail.com] Sent: Monday, September 17, 2007 5:10 PM To: Nair, Murlidharan T Cc: r-help at stat.math.ethz.ch Subject: Re: [R] MAD On 9/17/07, Nair, Murlidharan T <mnair at iusb.edu> wrote:> > I am calculating the median absolute deviation using mad function, and > it tends to ignore the parameter constant=1, when I am calculating it > for x=seq(1:5). Am I missing something here? > > x<-seq(1:5) > mad(x)# gives [1] 1.4826 > mad(x, constant=1)# gives [1] 1 > #Here is the long form > dev.from.median<-abs((x-median(x))) > dev.from.median # Gives [1] 2 1 0 1 2 > sum(dev.from.median) # Gives [1] 6 > sum(dev.from.median)/length(x) # Gives [1] 1.2 > # The long form does not match the output from the function > > # When x<-seq(1:10) they match > x<-seq(1:10) > dev.from.median<-abs((x-median(x))) > sum(dev.from.median)/length(x) # Gives 2.5 > mad(x, constant=1) # Gives 2.5 > #The long form matches the output from the function > > Did I miss anything here?yes; mad := Median (not mean) absolute deviation (from the median, by default). -Deepayan Indeed, its median and that what I am calculating in the long form. So, what is that you found I was doing differently? May be I missed your point. Thx../M