I want to plot a line chart. Depending on values it should change its color.
What I found is:
plot(sin(seq(from=1, to=10,by=0.1)),type="p",
col=ifelse(sin(seq(from=1, to=10,by=0.1))>0.5,"red","yellow"))
That works. But as soon as I change from type="p" to type="l" the conditional colouring disappears.
Is that behavior intended?
What is a solution with base graphics to plot a functional line with different colors?
Use segments instead of lines.
The segments function will only add to an existing plot. To create a blank plot with the correct axes and limits, first use plot with type="n" to draw "nothing".
x0 <- seq(1, 10, 0.1)
colour <- ifelse(sin(seq(from=1, to=10,by=0.1))>0.5,"red","blue")
plot(x0, sin(x0), type="n")
segments(x0=x0, y0=sin(x0), x1=x0+0.1, y1=sin(x0+0.1), col=colour)
See ?segments for more detail.
Here is a little different approach:
x <- seq(from=1, to=10, by=0.1)
plot(x,sin(x), col='red', type='l')
clip(1,10,-1,.5)
lines(x,sin(x), col='yellow', type='l')
Note that with this method the curve changes colors at exactly 0.5.
After you've drawn a line plot, you can color it with segments():
seq1 <- seq(from=1, to=10, by=0.1)
values <- sin(seq1)
s <- seq(length(seq1)-1)
segments(seq1[s], values[s], seq1[s+1], values[s+1], col=ifelse(values > 0.5, "red", "yellow"))
Related
Purpose
Create scatter plot with third dimension and multiple colors.
First:
- 3rd dimension with another scale in contrast to y-axis
- create two colors (this is done using col, see code)
Sketch simulating the purpose:
Code
Two "containers" of points plotted in this way:
plot(1:3, c(3,3,3))
points(1:3, c(2,2,2), col="blue")
Another nice plotting is done by:
#install.packages("hexbin")
library(hexbin)
x <- 1:1000#rnorm(1000)
y <- 1500:501#rnorm(1000)
bin<-hexbin(x, y, xbins=50)
plot(bin, main="Hexagonal Binning")
But I do not know how to use hexbin (I do not understand the functionality). There are needed two colors which I do not know how to generate.
Questions
How to create the 3rd axis with other scaling than the y-axis?
Can I use ´hexbin´ to get the result?
For some reason, using points() does not work, but using plot() does work:
#Set margin on right side to be a bit larger
par(mar = c(5,4.5,4,5))
#Plot first set of data
plot(1:3, rep(3,3), ylim=c(-5,5), xlab="X-Axis", ylab="Y-Axis 1")
#Plot second set of data on different axis.
par(new=T)
plot(1:3, rep(5,3), ylim=c(-10,10), col="blue", xlab="", ylab="", axes=FALSE)
#Add numbers and labels to the second y-axis
mtext("Y-Axis 2",side=4,line=3)
axis(4, ylim=c(-10,10))
I plot several lines on a graph using matplot:
matplot(cumsum(as.data.frame(daily.pnl)),type="l")
This gives me default colours for each line - which is fine,
But I now want to add a legend that reflects those same colours - how can I achieve that?
PLEASE NOTE - I am trying NOT to specify the colours to matplot in the first place.
legend(0,0,legend=spot.names,lty=1)
Gives me all the same colour.
The default color parameter to matplot is a sequence over the nbr of column of your data.frame. So you can add legend like this :
nn <- ncol(daily.pnl)
legend("top", colnames(daily.pnl),col=seq_len(nn),cex=0.8,fill=seq_len(nn))
Using cars data set as example, here the complete code to add a legend. Better to use layout to add the legend in a pretty manner.
daily.pnl <- cars
nn <- ncol(daily.pnl)
layout(matrix(c(1,2),nrow=1), width=c(4,1))
par(mar=c(5,4,4,0)) #No margin on the right side
matplot(cumsum(as.data.frame(daily.pnl)),type="l")
par(mar=c(5,0,4,2)) #No margin on the left side
plot(c(0,1),type="n", axes=F, xlab="", ylab="")
legend("center", colnames(daily.pnl),col=seq_len(nn),cex=0.8,fill=seq_len(nn))
I have tried to reproduce what you are looking for using the iris dataset. I get the plot with the following expression:
matplot(cumsum(iris[,1:4]), type = "l")
Then, to add a legend, you can specify the default lines colour and type, i.e., numbers 1:4 as follows:
legend(0, 800, legend = colnames(iris)[1:4], col = 1:4, lty = 1:4)
Now you have the same in the legend and in the plot. Note that you might need to change the coordinates for the legend accordingly.
I like the #agstudy's trick to have a nice legend.
For the sake of comparison, I took #agstudy's example and plotted it with ggplot2:
The first step is to "melt" the data-set
require(reshape2)
df <- data.frame(x=1:nrow(cars), cumsum(data.frame(cars)))
df.melted <- melt(df, id="x")
The second step looks rather simple in comparison to the solution with matplot
require(ggplot2)
qplot(x=x, y=value, color=variable, data=df.melted, geom="line")
Interestingly #agstudy solution does the trick, but only for n ≤ 6
Here we have a matrix with 8 columns. The colour of the first 6 labels are correct.
The 7th and 8th are wrong. The colour in the plots restarts from the beginning (black, red ...) , whereas in the label it continues (yellow, grey, ...)
Still haven't figured out why this is the case. I'll maybe update this post with my findings.
matplot(x = lambda, y = t(ridge$coef), type = "l", main="Ridge regression", xlab="λ", ylab="Coefficient-value", log = "x")
nr = nrow(ridge$coef)
legend("topright", rownames(ridge$coef), col=seq_len(nr), cex=0.8, lty=seq_len(nr), lwd=2)
Just discovered that matplot uses linetypes 1:5 and colors 1:6 to establish the appearance of the lines. If you want to create a legend try the following approach:
## Plot multiple columns of the data frame 'GW' with matplot
cstart = 10 # from column
cend = cstart + 20 # to column
nr <- cstart:cend
ltyp <- rep(1:5, times=length(nr)/5, each=1) # the line types matplot uses
cols <- rep(1:6, times=length(nr)/6, each=1) # the cols matplot uses
matplot(x,GW[,nr],type='l')
legend("bottomright", as.character(nr), col=cols, cex=0.8, lty=ltyp, ncol=3)
Scatter plots can be hard to interpret when many points overlap, as such overlapping obscures the density of data in a particular region. One solution is to use semi-transparent colors for the plotted points, so that opaque region indicates that many observations are present in those coordinates.
Below is an example of my black and white solution in R:
MyGray <- rgb(t(col2rgb("black")), alpha=50, maxColorValue=255)
x1 <- rnorm(n=1E3, sd=2)
x2 <- x1*1.2 + rnorm(n=1E3, sd=2)
dev.new(width=3.5, height=5)
par(mfrow=c(2,1), mar=c(2.5,2.5,0.5,0.5), ps=10, cex=1.15)
plot(x1, x2, ylab="", xlab="", pch=20, col=MyGray)
plot(x1, x2, ylab="", xlab="", pch=20, col="black")
However, I recently came across this article in PNAS, which took a similar a approach, but used heat-map coloration as opposed to opacity as an indicator of how many points were overlapping. The article is Open Access, so anyone can download the .pdf and look at Figure 1, which contains a relevant example of the graph I want to create. The methods section of this paper indicates that analyses were done in Matlab.
For the sake of convenience, here is a small portion of Figure 1 from the above article:
How would I create a scatter plot in R that used color, not opacity, as an indicator of point density?
For starters, R users can access this Matlab color scheme in the install.packages("fields") library, using the function tim.colors().
Is there an easy way to make a figure similar to Figure 1 of the above article, but in R? Thanks!
One option is to use densCols() to extract kernel densities at each point. Mapping those densities to the desired color ramp, and plotting points in order of increasing local density gets you a plot much like those in the linked article.
## Data in a data.frame
x1 <- rnorm(n=1E3, sd=2)
x2 <- x1*1.2 + rnorm(n=1E3, sd=2)
df <- data.frame(x1,x2)
## Use densCols() output to get density at each point
x <- densCols(x1,x2, colramp=colorRampPalette(c("black", "white")))
df$dens <- col2rgb(x)[1,] + 1L
## Map densities to colors
cols <- colorRampPalette(c("#000099", "#00FEFF", "#45FE4F",
"#FCFF00", "#FF9400", "#FF3100"))(256)
df$col <- cols[df$dens]
## Plot it, reordering rows so that densest points are plotted on top
plot(x2~x1, data=df[order(df$dens),], pch=20, col=col, cex=2)
You can get a similar effect by doing hexagonal binning, divide the region into hexagons, color each hexagon based on the number of points in the hexagon. The hexbin package has functions to do this and there are also functions in the ggplot2 package.
You can use smoothScatter for this.
colramp = colorRampPalette(c('white', 'blue', 'green', 'yellow', 'red'))
smoothScatter(x1, x2, colramp=colramp)
This is probably a simple question, but I´m not able to find the solution for this.
I have the following plot (I´m using plot CI since I´m not able to fill the points with plot()).
leg<-c("1","2","3","4","5","6","7","8")
Col.rar1<-c(rgb(1,0,0,0.7), rgb(0,0,1,0.7), rgb(0,1,1,0.7),rgb(0.6,0,0.8,0.7),rgb(1,0.8,0,0.7),rgb(0.4,0.5,0.6,0.7),rgb(0.2,0.3,0.2,0.7),rgb(1,0.3,0,0.7))
library(plotrix)
plotCI(test$size,test$Mean,
pch=c(21), pt.bg=Col.rar1,xlab="",ylab="", ui=test$Mean,li= test$Mean)
legend(4200,400,legend=leg,pch=c(21),pt.bg=Col.rar1, bty="n", cex=1)
I want to creat the same effect but with lines, instead of points (continue line)
Any suggestion?
You have 2 solutions :
Use The lines() function draws lines between (x, y) locations.
Use plot with type = "l" like line
hard to show it without a reproducible example , but you can do for example:
Col.rar1<-c(rgb(1,0,0,0.7), rgb(0,0,1,0.7), rgb(0,1,1,0.7),rgb(0.6,0,0.8,0.7),rgb(1,0.8,0,0.7),rgb(0.4,0.5,0.6,0.7),rgb(0.2,0.3,0.2,0.7),rgb(1,0.3,0,0.7))
x <- seq(0, 5000, length.out=10)
y <- matrix(sort(rnorm(10*length(Col.rar1))), ncol=length(Col.rar1))
plot(x, y[,1], ylim=range(y), ann=FALSE, axes=T,type="l", col=Col.rar1[1])
lapply(seq_along(Col.rar1),function(i){
lines(x, y[,i], col=Col.rar1[i])
points(x, y[,i]) # this is optional
})
When it comes to generating plots where you want lines connected according to some grouping variable, you want to get away from base-R plots and check out lattice and ggplot2. Base-R plots don't have a simple concept of 'groups' in an xy plot.
A simple lattice example:
library( lattice )
dat <- data.frame( x=rep(1:5, times=4), y=rnorm(20), gp=rep(1:4,each=5) )
xyplot( y ~ x, dat, group=gp, type='b' )
You should be able to use something like this if you have a variable in test similar to the color vector you define.
I used the information from this post to create a histogram with logarithmic scale:
Histogram with Logarithmic Scale
However, the output from plot looks nothing like the output from hist. Does anyone know how to configure the output from plot to resemble the output from hist? Thanks for the help.
A simplified, reproducible version of the linked answer is
x <- rlnorm(1000)
hx <- hist(x, plot=FALSE)
plot(hx$counts, type="h", log="y", lwd=10, lend="square")
To get the axes looking more "hist-like", replace the last line with
plot(hx$counts, type="h", log="y", lwd=10, lend="square", axes = FALSE)
Axis(side=1)
Axis(side=2)
Getting the bars to join up is going to be a nightmare using this method. I suggest using trial and error with values of lwd (in this example, 34 is somewhere close to looking right), or learning to use lattice or ggplot.
EDIT:
You can't set a border colour, because the bars aren't really rectangles – they are just fat lines. We can fake the border effect by drawing slightly thinner lines over the top. The updated code is
par(lend="square")
bordercol <- "blue"
fillcol <- "pink"
linewidth <- 24
plot(hx$counts, type="h", log="y", lwd=linewidth, col=bordercol, axes = FALSE)
lines(hx$counts, type="h", lwd=linewidth-2, col=fillcol)
Axis(side=1)
Axis(side=2)
How about using ggplot2?
x <- rnorm(1000)
qplot(x) + scale_y_log10()
But I agree with Hadley's comment on the other post that having a histogram with a log scale seems weird to me =).