I want to create a plot.
x values are matt 's colnames : count.4, count.5, ...
y values are range of matt elements
the scatter POINTS on plot are corresponding matt[i,j] elemnts, such that points for each row have the same color.
for (rr in 2:8) {
rownames(matt)[rr-1] <- paste('class', rr, sep='.')
for(cc in 4:20) {
matt[rr-1, cc-3] <- rr * cc
colnames(matt)[cc-3] <- paste('count', cc, sep='.')
}
}
I am really stuck at how to get such plot. Any help or hint is very appreciated. I found
matplot (c(4:20), cbind(matt[1,]:mat[7,]), pch = 19, ylim = range(c(matt[1,]:mat[7,]))
I see cbind(matt[1,]:matt[7,]) is not of correct format. but I do not know how to write this such than for a matrix with many more rows I do not have to write all matt[j,]
how can I get to this?
Edit. this is the plot I get when I use cbind(matt[1,],matt[2,],...,matt[7,])
Other than the efficient way coding this, I do not know why there are two sets of black points. can I label colors to corresponding "class" so that the plot gets easier to be read.
Normally series are stored in columns, not in rows, but if they are in rows just use transpose. There are other color palettes available at ?rainbow and a long vector of colors available via colors() if you don't like the colors we used here. (We used the input in the Note at the end.)
col <- rainbow(length(rr))
matplot(cc, t(mat), pch = 19, col = col, type = "o")
legend("topleft", legend = rownames(mat), pch = 19, col = col, lty = 1)
Note
rr <- 2:8
cc <- 4:20
mat <- outer(rr, cc)
dimnames(mat) <- list(paste0("class.", rr), paste0("count.", cc))
Related
Consider this data, where we have several groups with 10 observations each, and we conduct a pairwise.t.test():
set.seed(123)
data <- data.frame(group = rep(letters[1:18], each = 10),
var = rnorm(180, mean = 2, sd = 5))
ttres <- pairwise.t.test(x=data$var, g=data$group, p.adjust.method = "none")#just to make sure i get some sigs for the example
Now lets get the matrix of p values, convert them to a binary matrix showing significant and non-significant values, and plot them with corrplot(), so that we can visualize which groups are different:
library(corrplot)
pmat <- as.matrix(ttres$p.value)
pmat<-round(pmat,2)
pmat <- +(pmat <= 0.1)
pmat
corrplot(pmat, insig = "blank", type = "lower")
Does anyone know a way to color the background of each square according to a grouping label? For instance, say we want the squares for groups a:g to be yellow, the squares for groups h:n to be blue, and the squares for groups o:r to be red. Or is there an alternative way to do this with ggplot?
You can pass a vector of background colors via the bg= parameter. The trick is just making sure they are in the right order. Here's on way to do that
bgcolors <- matrix("white", nrow(pmat), ncol(pmat),dimnames = dimnames(pmat))
bgcolors[1:6, ] <- "yellow"
bgcolors[7:15, ] <- "blue"
bgcolors[14:17, ] <- "red"
bgcolors <- bgcolors[lower.tri(bgcolors, diag=TRUE)]
corrplot(pmat, insig = "blank", type = "lower", bg=bgcolors)
Basically we just make a matrix the same shape as our input, then we set the colors we want for the different rows, and then we just pass the lower triangle of that matrix to the function.
I am trying to plot 18 individual plots on a 3x6 multiplot in R. To be more efficient I have created these plots as a loop, however I would like the plots in each column to have their own color (i.e. the all the plots in column 1 would be red, all the plots in column 2 would be blue etc.). Is there a way I can do this while still retaining loop format?
par(mfcol = c(3,6))
for(i in 1:6)
{
plot(sigma_trace[,i], type ='l', main = paste("Sigma Traceplot Chain", i))
plot(theta_1_trace[,i], type = 'l', main = paste("Theta[1] Traceplot Chain", i))
plot(theta_2_trace[,i], type = 'l', main = paste("Theta[2] Traceplot Chain", i))
}
So basically, I think I want each loop statement to follow the same pattern of colours. Is this possible?
Thanks.
You can make a colour palette using RColorBrewer and then call each colour in your loop. For example.
library(RColorBrewer)
# set the colour palette
cols <- brewer.pal(4,'Set2')
# variables to plot
x = (1:250)/10
y = cos(x)
# plot in the loop
op <- par(mfrow = c(2, 2))
for (i in 1:4){
plot(x, y, col=cols[i], type='l', lwd=3)
}
par(op)
Here's an overview of the package.
In Base R you can use colorRampPalette() to create gradient, or you can even just make an object with the colours that you wan to reference:
plotcolors <- colorRampPalette(c("gold","blue"))(6)
par(mfrow = c(2, 3))
for(i in 1:6){
plot(1:10,1:10,type='l',col=plotcolors[i])
}
If you want to specify all 6 of your colours its as easy as modifying the above code
plotcolors <- c("red","blue","green","black","yellow","purple")
I'm trying to plot lines and color the lines based on the probability of that connection. Given a vector of probabilities, I use:
colfunc <- colorRamp(c("white", "red"))
colors <- colfunc(probs)
colors is then an nx3 matrix of rgb values. However, colfunc quite often returns a 0 value, so when i attempt to plot using these colors, R complains
Error in col2rgb(colors) : numerical color values must be positive
Is there an error in the way I am defining my color function?
Your function works fine, I think, but it doesn't return colors you can use with plot, because plot wants a color, not RGB values in a matrix.
There's probably a better way, but you can simply covert the matrix:
probs <- runif(10)
colors <- colfunc(probs)
my_col = apply(colors, MARGIN = 1, function(x) rgb(x[1]/255, x[2]/255, x[3]/255))
plot(1:10, 1:10, col = my_col) # should work fine
or you could just wrap your function
better_colfunc <- function(x, ramp = colorRamp(c("white", "red"))) {
colors <- ramp(x)
colors = apply(colors, MARGIN = 1, function(x) rgb(x[1]/255, x[2]/255, x[3]/255))
return(colors)
}
plot(1:10, 1:10, col = better_colfunc(probs, ramp = colfunc))
As for "colfunc quite often returns a 0 value", and other issues, you'll need to share both some data (what do your probs look like?) as well as perhaps the actual plotting code. See here for tips on making reproducible questions.
I am a bit confused what you are trying to do...the col2rgb function returns rgb values, so if you already have those then what do you want?
Or if you want rgb, why not use:
col2rgb(c("white", "red"))
Refer to the above plot. I have drawn the equations in excel and then shaded by hand. You can see it is not very neat. You can see there are six zones, each bounded by two or more equations. What is the easiest way to draw inequalities and shade the regions using hatched patterns ?
To build up on #agstudy's answer, here's a quick-and-dirty way to represent inequalities in R:
plot(NA,xlim=c(0,1),ylim=c(0,1), xaxs="i",yaxs="i") # Empty plot
a <- curve(x^2, add = TRUE) # First curve
b <- curve(2*x^2-0.2, add = TRUE) # Second curve
names(a) <- c('xA','yA')
names(b) <- c('xB','yB')
with(as.list(c(b,a)),{
id <- yB<=yA
# b<a area
polygon(x = c(xB[id], rev(xA[id])),
y = c(yB[id], rev(yA[id])),
density=10, angle=0, border=NULL)
# a>b area
polygon(x = c(xB[!id], rev(xA[!id])),
y = c(yB[!id], rev(yA[!id])),
density=10, angle=90, border=NULL)
})
If the area in question is surrounded by more than 2 equations, just add more conditions:
plot(NA,xlim=c(0,1),ylim=c(0,1), xaxs="i",yaxs="i") # Empty plot
a <- curve(x^2, add = TRUE) # First curve
b <- curve(2*x^2-0.2, add = TRUE) # Second curve
d <- curve(0.5*x^2+0.2, add = TRUE) # Third curve
names(a) <- c('xA','yA')
names(b) <- c('xB','yB')
names(d) <- c('xD','yD')
with(as.list(c(a,b,d)),{
# Basically you have three conditions:
# curve a is below curve b, curve b is below curve d and curve d is above curve a
# assign to each curve coordinates the two conditions that concerns it.
idA <- yA<=yD & yA<=yB
idB <- yB>=yA & yB<=yD
idD <- yD<=yB & yD>=yA
polygon(x = c(xB[idB], xD[idD], rev(xA[idA])),
y = c(yB[idB], yD[idD], rev(yA[idA])),
density=10, angle=0, border=NULL)
})
In R, there is only limited support for fill patterns and they can only be
applied to rectangles and polygons.This is and only within the traditional graphics, no ggplot2 or lattice.
It is possible to fill a rectangle or polygon with a set of lines drawn
at a certain angle, with a specific separation between the lines. A density
argument controls the separation between the lines (in terms of lines per inch)
and an angle argument controls the angle of the lines.
here an example from the help:
plot(c(1, 9), 1:2, type = "n")
polygon(1:9, c(2,1,2,1,NA,2,1,2,1),
density = c(10, 20), angle = c(-45, 45))
EDIT
Another option is to use alpha blending to differentiate between regions. Here using #plannapus example and gridBase package to superpose polygons, you can do something like this :
library(gridBase)
vps <- baseViewports()
pushViewport(vps$figure,vps$plot)
with(as.list(c(a,b,d)),{
grid.polygon(x = xA, y = yA,gp =gpar(fill='red',lty=1,alpha=0.2))
grid.polygon(x = xB, y = yB,gp =gpar(fill='green',lty=2,alpha=0.2))
grid.polygon(x = xD, y = yD,gp =gpar(fill='blue',lty=3,alpha=0.2))
}
)
upViewport(2)
There are several submissions on the MATLAB Central File Exchange that will produce hatched plots in various ways for you.
I think a tool that will come handy for you here is gnuplot.
Take a look at the following demos:
feelbetween
statistics
some tricks
I am trying to create a simple heatmap in R, using a diverging colour palette. I want to use a gradient so that all numbers below a threshold N are designated a color (say purple), and all numbers above the threshold are designated another color (say orange). The further away the number is from the threshold, the darker the color should be.
Here is a sample dataset:
Division,COL1,COL2,COL3,COL4,COL5,COL6,COL7
Division 1,31.9221884012222,75.8181694429368,97.0480443444103,96.295954938978,70.5677134916186,63.0451830103993,93.0396212730557
Division 2,85.7012346852571,29.0621076244861,16.9130333233625,94.6443660184741,19.9103083927184,61.9562198873609,72.3791105207056
Division 3,47.1665125340223,99.4153356179595,8.51091076619923,79.1276383213699,41.915355855599,7.45079894550145,24.6946100145578
Division 4,66.0743870772421,24.6163331903517,78.694460215047,42.04714265652,50.2694897353649,73.0409651994705,87.3745442833751
Division 5,29.6664374880493,35.4036891367286,19.2967326845974,5.48460693098605,32.4517334811389,15.5926876701415,76.0523204226047
Division 6,95.4969164915383,8.63230894319713,61.7535551078618,24.5590241160244,25.5453423131257,56.397921172902,44.4693325087428
Division 7,87.5015622004867,28.7770316936076,56.5095080062747,34.6680747810751,28.1923673115671,65.0204187724739,13.795713102445
Division 8,70.1077231671661,72.4712177179754,38.4903231170028,36.1821102909744,97.0875509083271,17.184783378616,78.2292529474944
Division 9,47.3570406902581,90.2257485780865,65.6037972308695,77.0234781783074,25.6294377148151,84.900529962033,82.5080851092935
Division 10,58.0811711959541,0.493217632174492,58.5604055318981,53.5780876874924,9.12552657537162,20.313960686326,78.1371118500829
Division 11,34.6708688884974,76.711881859228,22.6064443588257,22.1724311355501,5.48891355283558,79.1159523651004,56.8405059166253
Division 12,33.6812808644027,44.1363711375743,70.6362190190703,3.78900407813489,16.6075889021158,9.12654218263924,39.9711143691093
Here is a simple snippet to produce a heatmap from the above data
data <- read.csv("dataset.csv", sep=",")
row.names(data) <- data$Division
data <- data[,2:7]
data_matrix <- data.matrix(data)
heatmap(data_matrix, Rowv=NA, Colv=NA, col = heat.colors(256), scale="column", margins=c(5,10))
How can I modify the above code to produce:
a color gradient (orange) for all numbers ABOVE 50 (darker the further the number is from 50)
a color gradient (purple) for all numbers BELOW 50 (darker the further the number is from 50)
Nice to have (but optional) write the number value in the grid cell
Nice to have (but optional), use a different color for grid cell that is EXACTLY the threshold number (50 in this case)
[[Edit]]
I have just seen this question on SO, which seems to be very similar. The answer uses ggplot (which I have no experience of), and I have so far, been unable to adapt the ggplot solution to my slightly more complicated data.
This should get you most of the way. (Note that you'll need to set scale="none" if you want the plotted colors to correspond to the actual (rather than the rescaled) values of the cells).
ncol <- 100
## Make a vector with n colors
cols <- RColorBrewer:::brewer.pal(11,"PuOr") # OR c("purple","white","orange")
rampcols <- colorRampPalette(colors = cols, space="Lab")(ncol)
rampcols[(n/2) + 1] <- rgb(t(col2rgb("green")), maxColorValue=256)
## Make a vector with n+1 breaks
rampbreaks <- seq(0, 100, length.out = ncol+1)
## Try it out
heatmap(data_matrix, Rowv = NA, Colv = NA, scale="none",
col = rampcols, breaks = rampbreaks)
EDIT
For finer control over the placement of the threshold, I'd suggest creating two separate palettes -- one for values less than the threshold and one for values above the threshold -- and then "suturing" them together. Try something like this, playing around with different values for Min, Max, Thresh, etc.:
nHalf <- 50
Min <- 0
Max <- 100
Thresh <- 50
## Make vector of colors for values below threshold
rc1 <- colorRampPalette(colors = c("purple", "white"), space="Lab")(nHalf)
## Make vector of colors for values above threshold
rc2 <- colorRampPalette(colors = c("white", "orange"), space="Lab")(nHalf)
rampcols <- c(rc1, rc2)
## In your example, this line sets the color for values between 49 and 51.
rampcols[c(nHalf, nHalf+1)] <- rgb(t(col2rgb("green")), maxColorValue=256)
rb1 <- seq(Min, Thresh, length.out=nHalf+1)
rb2 <- seq(Thresh, Max, length.out=nHalf+1)[-1]
rampbreaks <- c(rb1, rb2)
heatmap(data_matrix, Rowv = NA, Colv = NA, scale="none",
col = rampcols, breaks = rampbreaks)
I found this thread very useful and also pulled some ideas from here, but for my purposes I needed to generalize some things and wanted to use the RColorBrewer package. While I was working on it Dr. Brewer (of Color Brewer fame) stopped in my office and told me I needed to interpolate within the smaller color breaks rather than just pick the end points. I thought others might find this useful so I am posting my function here for posterity.
The function takes in your data vector, the name of a diverging colorBrewer palette, and the center point for your color scheme (default is 0). It outputs a list containing 2 objects: a classIntervals object and a vector of colors: The function is set to interpolate a total of 100 colors but that can be modified with some care.
diverge.color <- function(data,pal_choice="RdGy",centeredOn=0){
nHalf=50
Min <- min(data,na.rm=TRUE)
Max <- max(data,na.rm=TRUE)
Thresh <- centeredOn
pal<-brewer.pal(n=11,pal_choice)
rc1<-colorRampPalette(colors=c(pal[1],pal[2]),space="Lab")(10)
for(i in 2:10){
tmp<-colorRampPalette(colors=c(pal[i],pal[i+1]),space="Lab")(10)
rc1<-c(rc1,tmp)
}
rb1 <- seq(Min, Thresh, length.out=nHalf+1)
rb2 <- seq(Thresh, Max, length.out=nHalf+1)[-1]
rampbreaks <- c(rb1, rb2)
cuts <- classIntervals(data, style="fixed",fixedBreaks=rampbreaks)
return(list(cuts,rc1))
}
in my work I am using this scheme to plot a raster layer (rs) using spplot like so:
brks<-diverge.color(values(rs))
spplot(rs,col.regions=brks[[2]],at=brks[[1]]$brks,colorkey=TRUE))