In trying to tweak the output of geom_bar(), I come across some solutions that include function(x) as here where the x-axis labels get wrapped by use of stringr:
scale_x_discrete(labels = function(x) str_wrap(x, width = 10)) +
What exactly is function(x) doing? It is simply saying that it's followed by a function to apply to x? But then what's x exactly, since it's the data defined in the original ggplot aes call, while here it's the label text?
Just when I think I understand something in R...
TIA.
The entire expression goes together
function(x) str_wrap(x, width = 10)
is an anonymous function in R (a function with no name). The labels= parameter of scale_x_discrete can take a vector of characters, or a function. If a function, ggplot will pass in the default breaks as input It doesn't have to be named "x", you can call it whatever you want. So this would work as well
scale_x_discrete(labels = function(default_breaks) str_wrap(default_breaks, width = 10)) +
This is documented on the ?ggplot2::scale_x_discrete help page.
Related
If I want to use paste0 inside expression to label the x-axis, it's not working as intended. But paste works.
library(ggplot2)
ggplot(mtcars, aes(mpg, disp)) +
geom_point() +
labs(
x = expression(paste0("It's","mpg")^("paste0 is not working")),
y = expression(paste("It's ", "disp")^("paste is working")), # had to give extra space
)
Please Note that
I have used paste0 in x-axis, which is not showing as intended
used paste in y-axis which is working, but had to give extra space to separate It's and disp, which should be the default behavior of paste (AFAIK). But that default behavior is not working inside the expression.
Also same happens with base-plot
plot(mtcars$disp ~ mtcars$mpg,
xlab = expression(paste0("mpg")^("paste0 is not working")),
ylab = expression(paste("disp")^("paste is working"))
)
Can anyone please break this out for me,
why paste0 is not working?
and also why pastes default behavior is not preserved here??
What's going on here?
Thanks.
It seems you can only use the operations listed in ?plotmath.
https://stat.ethz.ch/R-manual/R-devel/library/grDevices/html/plotmath.html
paste is there, but paste0 is not.
Also, the function features are not equivalent to the R's functions of the same name.
My objective is pass lists as arguments to the function geom_point2 using lapply or analogously mapply. In similar situations, I had success passing a list (or lists) to geom_cladelabel as in:
mapply(function (x,y,z,w,v,u,t,s) geom_cladelabel(node=x, label=y,
align=F, etc. # Where x y z etc are lists.
Problem is related to the use of aes inside geom_point2. (not in geom_cladelabel):
In the case of geom_point2, the node info is inside aes, and I could't do it. Normally I do not get any error message, but it does not work.
The objective is to make this example work, but using mapply instead of writting geom_point2 two times.
# source("https://bioconductor.org/biocLite.R")
# biocLite("ggtree")
library(ggtree)
library(ape)
#standard code
newstree<-rtree(10)
node1<-getMRCA(newstree,c(1,2))
node2<-getMRCA(newstree,c(3,4))
ggtree(newstree)+
geom_point2(aes(subset=(node ==node1) ), fill="black", size=3, shape=23)+
geom_point2(aes(subset=(node ==node2) ), fill="green", size=3, shape=23)
#desire to substitute the geom_point2 layers with mapply or lapply:
#lapply(c(node1,node2), function (x) geom_point2(aes(subset=(node=x)))))
Here is a solution calling geom_point2 usig mapply:
library(ggtree)
ggtree(rtree(10)) +
mapply(function(x, y, z)
geom_point2(
aes_string(subset=paste("node ==", x)),
fill=y,
size=10,
shape=z
),
x=c(11,12),
y=c("green", "firebrick"),
z=c(23,24)
) +
geom_text2(aes(subset=!isTip, label=node))
The solution is in the aes_string(), which writes the value of x directly in the aesthetics. The default aes() does not pass on the value of x, but just the string "x". When plotting, ggtree then looks for a node called "x", and ends with an empty node list.
I guess this has to do with the variable being stored in the mapply-environment and not being passed on to the plot.
PS: Sorry for my too quick answer with do.call() earlier. It is useful, but off-topic here.
I'm trying to make the legend of this plot pretty, so I need there the be an actual superscript, which is why I am using the pretty10exp() function from the sfsmisc library. It works when I use the c() function.
However, I am also trying to keep the string and the scientific notation number on the same line. The legend() is broken into two lines, which I think is due to c(). I thought I could use paste(), but for some reason the output is now incorrect.
plot(1:12)
pVal <- 4
legend("topright", legend = c("P value:", sfsmisc::pretty10exp(pVal)), cex = 1.5)
legend("topright", legend = paste("P value:", sfsmisc::pretty10exp(pVal)), cex = 1.5)
pVal being an arbitrary number represented in scientific notation. The second line results in output like this: "P value: (significand) %*% 10^-4". The first line also doesn't give me what I want. How can I fix this problem?
pretty10exp returns an expression which allows it to use the ?plotmath features for making nice looking numbers. When working with expressions, you can't just paste values in like strings. You need to manipulate them with a special set of functions. One such function is substitute. You can do
plot(1:12)
pVal <- 4
legend("topright", cex = 1.5,
legend = substitute("P value: "*x, list(x=sfsmisc::pretty10exp(pVal)[[1]])) )
We use substitute() to take the value contained in the expression from pretty10exp and prefix it with the label you want. (We use * to concatenate rather than paste() since plotmath allows it)
This is what I would do:
fun <- function(text, pVal) {
y <- floor(log10(pVal))
x <- pVal / 10^y
bquote(.(text)*":" ~ .(x) %.% 10 ^ .(y))
}
plot.new()
text(0.5,0.7,fun("P value", 0.4))
text(0.5, 0.3, fun("P value", signif(1/pi, 1)))
No package is needed.
I am going to plot a boxplot from a 4-column matrix pl1 using ggplot with dots on each box. The instruction for plotting is like this:
p1 <- ggplot(pl1, aes(x=factor(Edge_n), y=get(make.names(y_label)), ymax=max(get(make.names(y_label)))*1.05))+
geom_boxplot(aes(fill=method), outlier.shape= NA)+
theme(text = element_text(size=20), aspect.ratio=1)+
xlab("Number of edges")+
ylab(y_label)+
scale_fill_manual(values=color_box)+
geom_point(aes(x=factor(Edge_n), y=get(make.names(true_des)), ymax=max(get(make.names(true_des)))*1.05, color=method),
position = position_dodge(width=0.75))+
scale_color_manual(values=color_pnt)
Then, I use print(p1) to print it on an opened pdf. However, this does not work for me and I get the below error:
Error in make.names(true_des) : object 'true_des' not found
Does anyone can help?
Your example is not very clear because you give a call but you don't show the values of your variables so it's really hard to figure out what you're trying to do (for instance, is method the name of a column in the data frame pl1, or is it a variable (and if it's a variable, what is its type? string? name?)).
Nonetheless, here's an example that should help set you on the way to doing what you want:
Try something like this:
pl1 <- data.frame(Edge_n = sample(5, 20, TRUE), foo = rnorm(20), bar = rnorm(20))
y_label <- 'foo'
ax <- do.call(aes, list(
x=quote(factor(Edge_n)),
y=as.name(y_label),
ymax = substitute(max(y)*1.05, list(y=as.name(y_label)))))
p1 <- ggplot(pl1) + geom_boxplot(ax)
print(p1)
This should get you started to figuring out the rest of what you're trying to do.
Alternately (a different interpretation of your question) is that you may be running into a problem with the environment in which aes evaluates its arguments. See https://github.com/hadley/ggplot2/issues/743 for details. If this is the issue, then the answer might to override the default value of the environment argument to aes, for instance: aes(x=factor(Edge_n), y=get(make.names(y_label)), ymax=max(get(make.names(y_label)))*1.05, environment=environment())
I have a family of functions that are all the same except for one adjustable parameter, and I want to plot all these functions on one set of axes all superimposed on one another. For instance, this could be sin(n*x), with various values of n, say 1:30, and I don't want to have to type out each command individually -- I figure there should be some way to do it programatically.
library(ggplot2)
define trig functions as a function of frequency: sin(x), sin(2x), sin(3x) etc.
trigf <- function(i)(function(x)(sin(i*x)))
Superimpose two function plots -- this works manually of course
ggplot(data.frame(x=c(0,pi)), aes(x)) + stat_function(fun=trigf(1)) + stat_function(fun=trigf(2))
now try to generalize -- my idea was to make a list of the stat_functions using lapply
plotTrigf <- lapply(1:5, function(i)(stat_function(fun=function(x)(sin(i*x))) ))
try using the elements of the list manually but it doesn't really work -- only the i=5 plot is shown and I'm not sure why when that's not what I referenced
ggplot(data.frame(x=c(0,pi)), aes(x)) +plotTrigf[[1]] + plotTrigf[[2]]
I Thought this Reduce might handle the 'generalized sum' to add to a ggplot() but it doesn't work -- it complains of a non-numeric argument to binary operator
Reduce("+", plotTrigf)
So I'm kind of stuck both in executing this strategy, or perhaps there's some other way to do this.
Are you using version R <3.2? The problem is that you actually need to evaluate your i parameter in your lapply call. Right now it's being left as a promise and not getting evaulated till you try to plot and at that point i has the last value it had in the lapply loop which is 5. Use:
plotTrigf <- lapply(1:5, function(i) {force(i);stat_function(fun=function(x)(sin(i*x))) })
You can't just add stat_function calls together, even without Reduce() you get the error
stat_function(fun=sin) + stat_function(fun=cos)
# Error in stat_function(fun = sin) + stat_function(fun = cos) :
# non-numeric argument to binary operator
You need to add them to a ggplot object. You can do this with Reduce() if you just specify the init= parameter
Reduce("+", plotTrigf, ggplot(data.frame(x=c(0,pi)), aes(x)))
And actually the special + operator for ggplot objects allows you to add a list of objects so you don't even need the Reduce at all (see code for ggplot2:::add_ggplot)
ggplot(data.frame(x=c(0,pi)), aes(x)) + plotTrigf
The final result is
You need to use force in order to make sure the parameter is being evaluated at the right time. It's a very useful technique and a common source of confusion in loops, you should read about it in Hadley's book http://adv-r.had.co.nz/Functions.html
To solve your question: you just need to add force(i) when defining all the plots, inside the lapply function, before making the call to stat_function. Then you can use Reduce or any other method to combine them. Here's a way to combine the plots using lapply (note that I'm using the <<- operator which is discouraged)
p <- ggplot(data.frame(x=c(0,pi)), aes(x))
lapply(plotTrigf, function(x) {
p <<- p + x
return()
})