I want to add the following x-axis label to my bar plot but unfortunately R does not recognize the character '!' and prints dots instead of whitespaces:
I want: I get:
!src x.x.x.x X.src.x.x.x.x
!TCP X.TCP
!udp && !src x.x.x.x X.udp.....src.x.x.x.x
Additionally a would like to increase the margin because the text is to long and when setting the size over 'cex.names=0.6' then it just vanishes!?
There are two reason I can think of that R will have substituted X. for instances of !.
I suspect that the labelling you are seeing is due to R's reading of your data. Those column names aren't really syntactically valid and the erroneous character has been replaced by X.. This happens at the data import stage, so I presume you didn't check how R had read your data in?, or
You have a vector and the names of that vector are similarly invalid and R has done the conversion.
However, as you haven't made this reproducible it could be anything.
To deal with case 1 above, either edit your data file to contain valid names or pass check.names = FALSE in your read.table() call used to read in the data. Although doing the latter will make it difficult for you to select variable by name without quoting the name fully.
If you have a vector, then you can reset the names again:
> vec <- 1:5
> names(vec) <- paste0("!",LETTERS[1:5])
> vec
!A !B !C !D !E
1 2 3 4 5
> barplot(vec)
Also note that barplot() has a names.arg argument that you can use to pass it the labels to draw beneath each bar. For example:
> barplot(vec, names.arg = paste0("!", letters[1:5]))
which means you don't need to rely on what R has read in/converted for you as you tell it exactly what to label the plot with.
To increase the size of the margin, there are several ways to specify the size but I find setting it in terms of number of lines most useful. You change this via graphical parameter mar, which has the defaults c(5,4,4,2) + 0.1 which correspond to the bottom, left, top, and right margins respectively. Use par() to change the defaults, for example in the code below the defaults are store in op and a much larger bottom margin specified
op <- par(mar = c(10,4,4,2) + 0.1)
barplot(vec, names.arg = paste0("!", letters[1:5]), las = 2)
par(op) ## reset
The las = 2 will rotate the bar labels 90 degrees to be perpendicular to the axis.
One option is to use ann=F and add anotation to the plot using mtext.
x <- 1:2
y <- runif(2, 0, 100)
par(mar=c(4, 4, 2, 4))
plot(x, y, type="l", xlim=c(0.5, 2.5), ylim=c(-10, 110),
axes=TRUE, ann=FALSE)
Then add annotation:
mtext("!udp && !src x.x.x.x ", side=1, line=2)
Edit It is a question of a barplot and not simple plot.
as said in Gavin solution, the names argument can be setted. Here I show an example.
barplot(VADeaths[1:2,], angle = c(45, 135),
density = 20, col = "grey",
names=c("!src x.x.x.x", "!TCP", "!udp && !src x.x.x.x", "UF"),
horiz=FALSE)
Related
control=data.frame(replicate(16,sample(0:1,111,rep=TRUE)))
control.tendon <- control[,seq(2, 16, 1)]
index <- rep(1, 15) %x% seq(1, 111, 1)
plot(index, c(10, rep(35, 1664)), xlab="Minutes",
ylab="Temperature (Degrees Celcius)", pch=subject, type="n")
for(i in 1:15) {
lines(seq(1, 111, 1), unlist(control.tendon[i]), col="red",
lty=i)
}
I should get an empty figure but nothing pops out. If anyone can help I would be very glad.
Mistake is in type = "n". That suggest to draw just frame where line or graph can be added in next statement.
If you dig into help you will find:
"n" for no plotting.
The type = "n" is used when you want to draw multiple graphs (say lines) on same plot then type = "n" can be used prepare the layout with limits and label of axis. This will follow with line and other statement to add graphs.
Modified code like:
plot(index, c(10, rep(35, 1664)), xlab="Minutes",
ylab="Temperature (Degrees Celcius)", pch="1", type="l")
It will work.
Note:
The value of pch is not correctly used in OP. Help suggests pch to be an integer or single character.
*pch
Either an integer specifying a symbol or a single character to be
used as the default in plotting points. See points for possible values
and their interpretation. Note that only integers and single-character
strings can be set as a graphics parameter (and not NA nor NULL).
Some functions such as points accept a vector of values which are recycled.*
I am quite new to programming/R and I'm having a very unusual problem. I've made a scatterplot and I would like to simply put the x y axis at 0 on the plot. However, when I use abline they are slightly off. I managed to get them to 0 using trial and error, but trying to plot other lines becomes impossible.
library('car')
scatterplot(cost~qaly, reg.line=FALSE, smooth=FALSE, spread=FALSE,
boxplots='xy', span=0.5, xlab="QALY", ylab="COST", main="Bootstrap",
cex=0.5, data=scat2, xlim=c(-.05,.05), grid=FALSE)
abline(v = 0, h = 0)
This gives lines which are slightly to the left and below 0.
here is an image of what this returns:
(I can't post an image since I'm new apparently)
I found that these values put the lines on 0:
abline(v=0.003)
abline(h=3000)
Thanks in advance for the help!
Using #Laterow's example, reproduce the issue
require(car)
set.seed(10)
x <- rnorm(1000); y <- rnorm(1000)
scatterplot(y ~ x)
abline(v=0, h=0)
scatterplot seems to be resetting the par settings on exit. You can sort of check this with locator(1) around some point, eg, for {-3,-3} I get
# $x
# [1] -2.469414
#
# $y
# [1] -2.223922
Option 1
As #joran points out, reset.par = FALSE is the easiest way
scatterplot(y ~ x, reset.par = FALSE)
abline(v=0, h=0)
Option 2
In ?scatterplot, it says that ... is passed to plot meaning you can use plot's very useful panel.first and panel.last arguments (among others).
scatterplot(y ~ x, panel.first = {grid(); abline(v = 0)}, grid = FALSE)
Note that if you were to do the basic
scatterplot(y ~ x, panel.first = abline(v = 0))
you would be unable to see the line because the default scatterplot grid covers it up, so you can turn that off, plot a grid first then do the abline.
You could also do the abline in panel.last, but this would be on top of your points, so maybe not as desirable.
Is it possible to use a mix of character and number as plotting symbols in R legend?
plot(x=c(2,4,8),y=c(5,4,2),pch=16)
points(x=c(3,5),y=c(2,4),pch="+")
legend(7,4.5,pch=c("+",16),legend=c("A","B")) #This is the problem
Use the numerical equivalent of the "+" character:
plot(x=c(2,4,8),y=c(5,4,2),pch=16)
points(x=c(3,5),y=c(2,4),pch="+")
legend(7,4.5,pch=c(43,16),legend=c("A","B"))
There are actually numerical equivalents for all symbols!
Source: Dave Roberts
The pch code is the concatenation of the Y and X coordinates of the above plot.
For example, the + symbol is in row (Y) 4 and column (X) 3, and therefore can be drawn using pch = 43.
Example:
plot(x=c(2,4,8),y=c(5,4,2),pch=16)
points(x=c(3,5),y=c(2,4),pch="+")
legend(7,4.5,pch=c(43,16),legend=c("A","B"))
My first thought is to plot the legend twice, once to print the character symbols and once to print the numeric ones:
plot(x=c(2,4,8),y=c(5,4,2),pch=16)
points(x=c(3,5),y=c(2,4),pch="+")
legend(7,4.5,pch=c(NA,16),legend=c("A","B")) # NA means don't plot pt. character
legend(7,4.5,pch=c("+",NA),legend=c("A","B"))
NOTE: Oddly, this works in R's native graphical device (on Windows) and in pdf(), but not in bmp() or png() devices ...
I bumped to this issue several time, so I wrote a tiny function below. You can use to specify the pch value, e.g.
pch=c(15:17,s2n("|"))
String to Numeric
As noted in previous answers, you can simply add the numerical equivalent of the numeric and character symbols you want to plot.
However, just a related aside: if you want to plot larger numbers (e.g., > 100) or strings (e.g., 'ABC') as symbols, you need to use a totally different approach based on using text().
`Plot(x,y,dat,type='n') ; text(x,y,labels = c(100,'ABC')
Creating a legend in this case is more complicated, and the best approach I've ever come up with is to stack legends on top of each other and using the legend argument for both the pch symbol and the description:
pchs <- c(100,'ABC','540',sum(13+200),'SO77')
plot(1:5,1:5,type='n',xlim=c(1,5.1))
text(1:5,1:5,labels = pchs)
legend(3.5,3,legend = pchs,bty='n',title = '')
legend(3.5,3,legend = paste(strrep(' ',12),'ID#',pchs),bty='n',title='Legend')
rect(xleft = 3.7, ybottom = 1.5, xright = 5.1, ytop = 3)
This uses strrep to concatenate spaces in order to shift the text over from the "symbols", and it uses rect to retroactively fit a box around the printed legend text.
library(gplots)
shades= c(seq(-1,0.8,length=64),seq(0.8,1.2,length=64),seq(1.2,3,length=64))
heatmap.2(cor_mat, dendrogram='none', Rowv=FALSE, Colv=FALSE, col=redblue(64),
breaks=shades, key=TRUE, cexCol=0.7, cexRow=1, keysize=1)
There is some problem with breaks. Wish to receive help on it.
After running the code I get this error message
Error in image.default(1:nc, 1:nr, x, xlim = 0.5 + c(0, nc), ylim = 0.5 + : must have one more break than colour
Thank you for your time and consideration.
Well, we don't have cor_mat so we can't try this ourselves, but the problem seems to be what it says on the tin, isn't it? The way heatmap (and generally all functions based on image) works with breaks and a vector of colours, is that the breaks define the points where changes in the value of your data matrix means the colour changes. In short, if break = c(1,2,3), and your col = c("red", "blue"):
values < 1 will be transparent
values >= 1, <= 2 will be plotted as red
values > 2, <= 3 will be plotted as blue
values > 3 will be transparent
What's going on in your code is that with 'shade' you've supplied a length 3*64 vector to break, while redblue(64) only gives you 64 colours. Try replacing redblue(64) with, say, redblue(3*64-1).
Say, I have a variable rv which has some numerical value. Now, I want to plot the value of this variable on a base plot but preceded by a nicely formatted symbol e.g., r subscript m, using expression. To write on the plot I use mtext.
However, what I get is either the value of the variable, but no nicely formatted symbol (left annotation), or a nicely formatted symbol, but not the value of the variable, but the variable name...
I tried to play around with eval, but didn't get what I wanted. Here is my code:
plot(1:10, rep(10,10), ylim=c(0,12))
rv <- 0.43
#left annotation:
mtext(paste(expression(italic(r[M])), " = ", rv), side = 1, line = -1.5, adj = 0.1)
#right annotation:
mtext(expression(paste(italic(r[M]), " = ", rv)), side = 1, line = -1.5, adj = 0.9)
This is the result:
How do i get both, nice format and value of the variable? Thanks.
btw: I know that I can get it, if I use two times mtext and play around with adj and stuff. But I would really like to get it in one call or without playing around with the position of two annotations.
The bquote function will create an expression and alow substitution of values using .(var) syntax. for your case do something like:
text( 5,1, bquote( italic(r[M]) == .(rv) ) )
Just combine what you have and plot two pieces, joined by using adj:
R> plot(1:10, rep(10,10), ylim=c(0,12))
R> text(2,12, expression(paste(italic(r[M]))), adj=1)
R> text(2,12, paste("=", rv), adj=0)