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plot time series / dates (basic)

5 messages · bogdan romocea, Brian Ripley, Gabor Grothendieck +1 more

Original
bogdan romocea
# 
Dear R users,

I'm having a hard time with some very simple things. I have a time
series where the dates are in the format 7-Oct-04. I imported the
file with read.csv so the date column is a factor. The series is
rather long and I want to plot it piece by piece. The function below
works fine, except that the labels for date are meaningless (ie
9.47e+08 or 1098000000 - apparently the number of seconds since
whatever). I don't want to convert the data frame to a ts object
because there are missing days and I don't want any interpolation.

1. How do I replace the date labels with something like 'Mar04',
instead of 9.47e+08 / 1098000000?

2. In the PDF file, the space between the two graphs printed pair by
pair is fairly large. Can I remove/reduce the area that seems
reserved for Title and X label so that, on a page, the space between
the graph at the top and the one at the bottom is minimized?

3. Given the function below, I haven't discovered a way to have
"vara" appear as the Title or Y label in graphs.
main=as.character(vara) lists all the values of vara (which is a
column from the data frame d). So, how can I use the name of a vector
as title or label in a plot?

Thank you,
b.


d <- ('data.csv', header = T, sep = ",", quote="", dec=".", 
	fill = T, skip=0)
attach(d)
#function to plot a long time series piece by piece
pl <- function(vara, varb, points)
	{
	date <- as.POSIXct(strptime(as.character(Date), "%d-%b-%y"), tz =
"GMT")
	pr1 <- vector(mode="numeric")
	pr2 <- vector(mode="numeric")
	dat <- vector()
	for (j in 1:(round(length(Vol)/points)+1)) #number of plots
		{
		for (i in ((j-1)*points+1):(j*points)) 
			{
			pr1[i-points*(j-1)] <- vara[i]
			pr2[i-points*(j-1)] <- varb[i]
			dat[i-points*(j-1)] <- date[i]
			}
		par(mfrow=c(2,1)) 
		plot(dat, pr1, type="b")
		plot(dat, pr2, type="b")
		}
	}

pdf("Rplots.pdf")
pl(Vol, atr, 50)
dev.off()





		
__________________________________
Brian Ripley
#
On Mon, 1 Nov 2004, bogdan romocea wrote:

            

      
      
      
    

        
So why use as.POSIXct for a date, rather than as.Date?

            

      
      
      
    

        
Just don't convert them to that format.  You set up

            

      
      
      
    

        
which is not a dates object.  If you use standard R indexing, it will
work. If you throw the class away, it will not.  Try

       dat <- date[(j-1)*points+1):(j*points)]

etc (no for loop required).

If you want a different format, see ?axis.Date

            

      
      
      
    

        
There's a whole chapter on this in `An Introduction to R': have you read 
it?

            

      
      
      
    

        
That's almost an FAQ.  Use deparse(substitute(vara))

            

      
      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      bogdan romocea
    

    

      
      #
    

  


  

      
Thank you for the suggestions. I managed to fix everything except the
first part. 
	dat <- date[(j-1)*points+1):(j*points)]
causes a syntax error. If I do 
	dat <- vector() 
I end up with numbers (which is fine by me - just like SAS dates).
However, after checking a couple of sources I still have no idea how
to format numbers as dates (for plotting/printing). Does anyone have
an example for formatting 12710 (# of days since 1 Jan 1970) as
19-Oct-04 (in the x axis of a plot)?

Regards,
b.


#function to plot a long time series piece by piece
pl <- function(vara, varb, points)
	{
	date <- as.Date(as.character(Date), "%d-%b-%y")
	pr1 <- vector(mode="numeric")
	pr2 <- vector(mode="numeric")
	#dat <- vector()
	dat <- date[(j-1)*points+1):(j*points)]
	for (j in 1:(round(length(Vol)/points)+1)) #number of plots
		{
		for (i in ((j-1)*points+1):(j*points)) 
			{
			pr1[i-points*(j-1)] <- vara[i]
			pr2[i-points*(j-1)] <- varb[i]
			#dat[i-points*(j-1)] <- date[i]
			#dat <- date[i]
			}
		par(mfrow=c(2,1), mai=c(0.4, 0.5, 0.3, 0.1), omi=c(0.2, 0, 0, 0), 
			cex.axis=0.7, cex=1.2, cex.main=0.7, pch="*") 
		plot(dat, pr1, main=deparse(substitute(vara)), type="o")
		#axis.Date(1,dat,format="%b%y") 
		plot(dat, pr2, main=deparse(substitute(varb)), type="o")
		}
	}

        
--- Prof Brian Ripley <ripley at stats.ox.ac.uk> wrote:

        

            

      
      
      
    

        
__________________________________

  
  


  


      

    

    

      
      

        


  

        

      Gabor Grothendieck
    

    

      
      #
    

  


  

          

      
      
      
    

        
: first part. 
: 	dat <- date[(j-1)*points+1):(j*points)]
: causes a syntax error. If I do 

You have unbalanced parentheses.

: 	dat <- vector() 
: I end up with numbers (which is fine by me - just like SAS dates).
: However, after checking a couple of sources I still have no idea how
: to format numbers as dates (for plotting/printing). Does anyone have
: an example for formatting 12710 (# of days since 1 Jan 1970) as
: 19-Oct-04 (in the x axis of a plot)?

You can use chron or Date classes:

	library(chron)
	dd <- chron(12710:12721, out.format = "dd-mmm-yy")
	plot(dd, 1:12)
or
	as.Date.integer <- function(x) structure(x, class = "Date")
	dd <- as.Date(12710:12721)
	plot(dd, 1:12)

If you don't like that labelling you can use axis to set up
your own.  Continuing the last example:

	plot(dd, 1:12, xaxt = "n")
	axis.Date(1, dd, format = "%d-%b-%y", cex.axis = .5)

see ?axis, ?strptime and the Help Desk article in R News 4/1.

  
  


  


      

    

    

      
      

        


  

        

      Peter Dalgaard
    

    

      
      #
    

  


  

      
bogdan romocea <br44114 at yahoo.com> writes:

            

      
      
      
    

        
Why not just fix the syntax error? Can't take that long to spot that
there are more ")" than "(" in that line, so presumably what was meant
was

  dat <- date[((j-1)*points+1):(j*points)]