search for: gray80

...ke to be able to convert that to a data.frame and vice versa, but can't quite figure out how. pats <- list(structure(list(shape = 0, shape.col = "black", shape.lty = 1, cell.fill = "white", back.fill = "white", label = 1, label.size = 1, ref.col = "gray80", ref.grid = "yes", scale.max = 100), .Names = c("shape", "shape.col", "shape.lty", "cell.fill", "back.fill", "label", "label.size", "ref.col", "ref.grid", "scale.max")), structure(li...

...36, 1.7828406,1.8732236,1.9645632,2.0566696,2.1493940,2.2426199, 2.3362558,2.4302300,2.5244855,2.6189767,2.7136672,2.8085272, 2.9035325,2.9986630,3.0939022,3.1892363,3.2846536,3.3801443, 3.4757000,3.5713134,3.6669785,3.7626899,3.8584430) plot(c(),c(),ylim=c(0,4),xlim=c(0,4)) lines(x1,y1,col="gray80",lwd=1.5) lines(x2,y2,col="gray80",lwd=1.5) [[alternative HTML version deleted]]

...0),xlab="",ylab="",type="p", data=ordinate) abline(v=0,h=0) par(new=TRUE) # Define cars vector with 8 values cars <- c(1,1,1,1,1,1,1,1) # Define colors in each sector color1 <- c("grey80","grey80","grey80","grey80","gray80","grey30","grey20","gray10" ) ## gray10=<0.0001, gray20<0.001, gray30<0.01, gray40=<0.05, gray 80=>0.05 # Putting sector levels car_labels <- c(1,2,3,4,5,6,7,8) car_labels <- paste(car_labels,sep="") # Create a pie chart with define...

...n on a plot of a time series. I'm going to add lines from a smooth.spline interpolation and other annotation to it. But here's the general idea: # Start code n <- 121 dat <- rnorm(n) plot(dat, type="n") polygon(c(1:n,n:1), c(dat,c(rep(0,n))), col = "gray80", border = NA) # End code What if dat is a time series? # Start code that won't parse. dat.ts <- ts(rnorm(n), start = 1980, frequency = 12) plot(dat.ts, type="n") polygon(...) # How do I specify x and y for the time series? # Is it easier to...

...any results. Any help in this matter will be appreciated. The code is posted below: par(mfrow=c(1,1)) fit <- survfit(Surv(revreg$Tcuprev , revreg$Rcup_n_j_n)~revreg$RAceGrp_1Maz) fit survplot (fit, col=c("gray1","gray40"), lty=1, lwd = 1, col.fill=c("gray50","gray80"), mark.time= T, conf="bands" , conf.int=.95, add=F , xlim=c(0, 20) , time.inc = 4, ylim=c(0.5, 1), xlab="Years Postoperative" , ylab="Cumulative Survival Probability" , label.curves=F , abbrev.label=F ,...

...ear R users, I cannot display ylab with the following code while trying to stack 7 rows of 3 histograms each. Thanks in advance! par(mfrow=c(7,3)) par(mar=c(2,2,2,2)) par(oma=c(5,5,0,0)) with (newdata <- subset(table, Month1 == "1"), hist(newdata$CarapWid, breaks=5*(0:80), col="gray80", main="All individuals", xlab="", xlim =c(15,85), ylab="April 2007")) - Kumar -- Section of Integrative Biology University of Texas at Austin Austin, Texas 78712, USA [[alternative HTML version deleted]]

...This `fails' (on a virgin device) clab <- paste(lab,collapse=",") plot(1,1, xlim = c(0,15), ylim = c(-.8,1), type ="n", axes = FALSE, main=paste("plot(*,axes=F); par(lab= c(",clab,")); axis(*) x 2",sep="")) box(col="gray80") op <- par(lab = lab) ; on.exit(par(op)) axis(1, pos = 0) axis(2, pos = 0, las = 1) par(op) op } p2 <- function(lab = c(8,12,7)) { ## This works [par() *BEFORE* plot() ] clab <- paste(lab,collapse=",") op <- par(lab = lab) ; on.exit(pa...

...irst of all a more simpler formula (x^2+y^2), where I adjusted color and things like that: farbe<-rgb(85, 141, 85, maxColorValue=255) x <- seq(-40, 40, length= 10) y <- c(1:22) f <- function(x,y) { r <- x^2+y^2; r } z <- outer(x, y, f) z[is.na(z)] <- 1 op <- par(bg = "gray80") persp(x, y, z, theta = -60, phi = 30, expand = 0.5, col = farbe, ticktype = "detailed", xlab = "X", ylab = "Y", zlab = "Z" ) now I tried to put in the log-normal density formula with an increasing mean and variance. That means the mean and...

Perhaps this is not a bug but rather my misunderstanding of lty and col. I would like to generate a legend for plots as follows: par(mfcol = c(2, 1), mar = c(5.1, 6.1, 4.1, 2.1)) matplot(t(matrix(1:8,8,20)), type="l") par(mfg = c(2, 1, 2, 1)) box(col = 0) # Rbug workaround. Thanks Martin legend((par()$usr)[1:2], (par()$usr)[3:4], as.character(1:8), lty=1:8, col=1:8,

Perhaps this is not a bug but rather my misunderstanding of lty and col. I would like to generate a legend for plots as follows: par(mfcol = c(2, 1), mar = c(5.1, 6.1, 4.1, 2.1)) matplot(t(matrix(1:8,8,20)), type="l") par(mfg = c(2, 1, 2, 1)) box(col = 0) # Rbug workaround. Thanks Martin legend((par()$usr)[1:2], (par()$usr)[3:4], as.character(1:8), lty=1:8, col=1:8,

...panel.abline(v = c(2.5, 7.5), col = trellis.par.get('reference.line')$col) if(current.row() == 2) panel.abline(h = 0, col = 'gray50') # Add placebo quantile reference with(medians, panel.polygon(c(x, rev(x)), c(h1, rev(h2)), col = 'gray80', border = NA, alpha = 0.75)) # Basic bar and whisker plot panel.bwplot(..., subscripts = subscripts) # Add placebo median reference with(medians, panel.lines(x, median, col = 'gray40')) }, as.table = TRUE, ylab = '',...

...quot;", se1.mult = 1, se2.mult = 2, partial.resids = FALSE, rug = TRUE, se = TRUE, scale = -1, n = 100, n2 = 40, pers = FALSE, theta = 30, phi = 30, jit = FALSE, xlab = NULL, ylab = NULL, main = NULL, ylim = NULL, xlim = NULL, too.far = 0.1, shade = FALSE, shade.col = "gray80", shift = 0, trans = I, by.resids = FALSE, scheme = NULL, ...) $ where : chr "namespace:mgcv" $ visible: logi FALSE $ dups : logi FALSE - attr(*, "class")= chr "getAnywhere" Both `methods()` and `getS3method()` don't list/find this method, but t...

...re so dark. You can't write on top ### of them. You can grab shades of gray like "gray30" or such. ### I'll just specify 4 colors I know will work. I prefer ### the light gray colors because they can be written over. mycols <- c("lightgray","gray70","gray80","gray90","gray75") barplot(x, names=mynames, las=2, col=mycols) legend(2,20,legend=c("A","B","C","D"),fill=mycols) ### Still, I don't like the solid shades so much. ### fill up the bars with angled lines. ### density determine...

...ver-fitting. I'd like to eliminate that. require('e1071') colors <- c(2, 3, 4, 5, 6) set.seed(23) # set random seed for repeatability # log(x) + cos(x) + noise f <- function(x) log(x) + cos(x) x <- seq(0.1, 5, by = 0.05) y <- f(x) + rnorm(x, sd = 0.2) plot(x, y, col="gray80") legend("topleft", c("log(x) + cos(x)", "SVM, untuned", "SVM, tuned"), bty="n", col=colors, pch=20) lines(x, f(x), col = colors[1]) # overlay noiseless data # SVM, untuned svmmodel1 <- svm(x, y) print(summary(...

I would like to organize my data as follows: I have a table that contains various data, and the numbers represent a level of similarity between these data, eg RF00013 has 100% similarity with the data RF00014. I would leave my table as a heatmap where darker colors represent higher similarity, and the lighter colors represent less level of similarity. I'm using version 2.11 of R. these

Dear R users, is there a way to define the color of the title for the legend in lattice? The help page on xyplot has a lot of details on key options just as the new book, but no mentioning of a color attribute for the title. Should I use ltext or is there any other way? Best wishes, Erik [[alternative HTML version deleted]]

Hi everyone, I have a certain number of samples and I want to visualize the groups those samples belong to. For example, suppose to have three variables, age, sex, and smoker/nonsmoker, and three samples, S1, S2, S3. S1 is 35, male, nonsmoker S2 is 24, female, nonsmoker S3 is 24, female, smoker at the end I have the following data frame: S1 S2 S3 age 35 24 30 sex M F F smk N N S What I

Dear useRs, I have the following data "z" of two variables "x"(z[,1]) and "y"(z[,2]). > dput(z) structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66,

Hello, I've a question concerning the display of interval data. A sample dataset where X is an interval between Xa and Xb which should be displayed: Y=c(15,14,23,18,19,9,19,13) Xa=c(17,22,21,18,19,25,8,19) Xb=c(22,22,29,34,19,26,17,22) X = (Xa+Xb)/2 It's easily possible to plot the mean of the interval like: plot(X,Y) afterwards I can create lines for the interval with:

>library(RODBC) >ch<- odbcConnect("durga") > sqlQuery(ch, paste("SELECT * from emp")) >sqlQuery(ch, ("SELECT * from xyz")) store revenue year_y 1 bigbazar 787875 2008 2 more 87876 2008 plot(revenue ~ year_y, data=xyz, pch=16) can i get any diffrent types of graphs like pie charts, bar