Thanks David, your way of constructing df is much more compact than what I was using, so I've incorporated it. I also had my rows and columns transposed relative to how xyplot wanted them (though I had tested for that, other problems interfered).
In my case, I may have varying numbers of y columns, from y.1 to y.n let's say. Is there an easy way of creating the phrase y.1+y.2+...y.n to pass to xyplot, or even better, some sort of syntax that says "take all y.n" and plot them against x? Thanks, Bryan On 9/6/09 12:51 AM, "David Winsemius" <dwinsem...@comcast.net> wrote: > I'm not exactly sure what structure df has. Here's my effort to > duplicate it: > > df <- data.frame(y=matrix(rnorm(24), nrow=6), x=1:6) >> df > y.1 y.2 y.3 y.4 x > 1 0.1734636 0.2348417 -1.2375648 -1.3246439 1 > 2 1.9551669 -1.1027262 -0.7307332 0.3953752 2 > 3 -0.7645778 1.6297861 0.4743805 -0.4476145 3 > 4 -0.5308756 -0.5246534 -0.3854609 -1.6097777 4 > 5 0.7406525 -0.8691720 -0.8194084 1.6122059 5 > 6 -0.9625619 -1.0774165 1.0760829 0.3659436 6 > > And this seems to accomplish the desired task. Presumably you have > assigned off-stage the value of title to a meaningful character string? > >> p <- xyplot(y.1+y.2+y.3+y.4 ~ x |1:4, data = df, main = > "title" ,layout=c(1,4) ) >> p > > > > > On Sep 5, 2009, at 11:52 PM, Bryan Hanson wrote: > >> Hello R Folks... >> >> I have a list with the following structure: >> >>> str(df) >> List of 3 >> $ y : num [1:4, 1:1242] -0.005379 0.029874 -0.023274 0.000655 >> -0.004537 >> .. >> $ x : num [1:1242] 501 503 505 507 509 ... >> $ names: Factor w/ 4 levels "PC Loading 1",..: 1 2 3 4 >> >> I want to plot each row of df$y against df$x, and have each plot in >> it¹s own >> panel according to the levels of df$names. The following works in >> the sense >> that the layout is right, but the y values have clearly been >> recycled or >> skipped in some fashion (and an error is thrown for each panel that >> the >> length of x and y aren¹t the same): >> >> p <- xyplot(y ~ x | names, data = df, main = title, >> layout = c(1, dim(y)[1]) >> >> In reviewing the extended formula interface in the Lattice Book, >> what I want >> to happen is y1 + y2 + y3 + y4 ~ x | names, outer = TRUE >> >> I see two options: figure out a way to create the extended formula >> on the >> fly (and the actual number of rows in y may vary), which seems >> potentially >> tricky, or create a data frame by stacking each row of y and >> repeating x and >> names to match. This seems like a waste of memory. >> >> I¹ve looked through the archives and haven¹t come across something >> quite >> like this, or at least I don¹t recognize it if I have! Is there a >> more >> elegant way to tell xyplot I want to use each row of y repeatedly >> with the >> same x, in a loop-like fashion? >> >> TIA. Bryan >> ************* >> Bryan Hanson >> Professor of Chemistry & Biochemistry >> DePauw University, Greencastle IN USA >> >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> R-help@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. > > David Winsemius, MD > Heritage Laboratories > West Hartford, CT > ______________________________________________ R-help@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.
