DISCLOSURE: I'm not sure how to make a reproducible example for this question.
I'm trying to plot a list of grobs using the gridExtra package.
I have some code that looks like this:
## Make Graphic Objects for Spec and raw traces
for (i in 1:length(morletPlots)){
gridplots_Spec[[i]]=ggplotGrob(morletPlots[[i]])
gridplots_Raw[[i]]=ggplotGrob(rawPlot[[i]])
gridplots_Raw[[i]]$widths=gridplots_Spec[[i]]$widths
}
names(gridplots_Spec)=names(morletPlots)
names(gridplots_Raw)=names(rawPlot)
## Combine spec and Raw traces
g=list()
for (i in 1:length(rawPlot)){
g[[i]]=arrangeGrob(gridplots_Spec[i],gridplots_Raw[i],heights=c(4/5,1/5))
}
numPlots = as.numeric(length(g))
##Plot both
for (i in 1:numPlots){
grid.draw(g[i],ncol=2)
}
Let me walk through the code.
morletPlots = a list of ggplots
rawplot = A list of ggplots
gridplots_spec and gridplots_Raw = list of grobs from the ggplots made above.
g = a list of the two grobs above combined so combining gridplots_spec[1] and gridplots_raw[1] so on and so on for the length of the list.
now my goal would be two plot all of those into 2 columns. But whenever I pass the gridplots_spec[i] through the grid.draw loop I get an error:
Error in UseMethod("grid.draw") :
no applicable method for 'grid.draw' applied to an object of class "list"
I can't unlist it becasue it just turns into a long character vector. any ideas?
If it's absolutely crucial I can spend the time to make an reproducible example but I'm more likely just missing a simple step.
Here's my interpretation of your script, if it's not the intended result you may want to use some bits and pieces to make your question reproducible.
library(grid)
library(gridExtra)
library(ggplot2)
morletPlots <- replicate(5, ggplot(), simplify = FALSE)
rawplot <- replicate(5, ggplot(), simplify = FALSE)
glets <- lapply(morletPlots, ggplotGrob)
graws <- lapply(rawplot, ggplotGrob)
rawlet <- function(raw, let, heights=c(4,1)){
g <- rbind(let, raw)
panels <- g$layout[grepl("panel", g$layout$name), ]
# g$heights <- grid:::unit.list(g$heights) # not needed
g$heights[unique(panels$t)] <- lapply(heights, unit, "null")
g
}
combined <- mapply(rawlet, raw = graws, let=glets, SIMPLIFY = FALSE)
grid.newpage()
grid.arrange(grobs=combined, ncol=2)
Edit I can't resist this mischievous hack to colour the plots for illustration; feel free to ignore it.
palette(RColorBrewer::brewer.pal(8, "Pastel1"))
ggplot.numeric = function(i) ggplot2::ggplot() +
theme(panel.background=element_rect(fill=i))
morletPlots <- lapply(1:5, ggplot)
rawplot <- lapply(1:5, ggplot)
Related
I have data I'm plotting using ggplot's facet_grid:
My data:
species <- c("spcies1","species2")
conditions <- c("cond1","cond2","cond3")
batches <- 1:6
df <- expand.grid(species=species,condition=conditions,batch=batches)
set.seed(1)
df$y <- rnorm(nrow(df))
df$replicate <- 1
df$col.fill <- paste(df$species,df$condition,df$batch,sep=".")
My plot:
integerBreaks <- function(n = 5, ...)
{
library(scales)
breaker <- pretty_breaks(n, ...)
function(x){
breaks <- breaker(x)
breaks[breaks == floor(breaks)]
}
}
library(ggplot2)
p <- ggplot(df,aes(x=replicate,y=y,color=col.fill))+
geom_point(size=3)+facet_grid(~col.fill,scales="free_x")+
scale_x_continuous(breaks=integerBreaks())+
theme_minimal()+theme(legend.position="none",axis.title=element_text(size=8))
which gives:
Obviously the labels are long and come out pretty messed up in the figure so I was wondering if there's a way edit these labels in the ggplot object (p) or the gtable/gTree/grob/gDesc object (ggplotGrob(p)).
I am aware that one way of getting better labels is to use the labeller function when the ggplot object is created but in my case I'm specifically looking for a way to edit the facet labels after the ggplot object has been created.
As I mentioned in the comments, the facet names are nested quite deeply within the gtable that ggplotGrob() gives you. However, this is still possible and since the OP explicitly wants to edit them after being plotted, you can do this with:
library(grid)
gg <- ggplotGrob(p)
edited_grobs <- mapply(FUN = function(x, y) {
x[["grobs"]][[1]][["children"]][[2]][["children"]][[1]][["label"]] <- y
return(x)
},
gg$grobs[which(grepl("strip-t",gg$layout$name))],
unique(gsub("cond","c", df$condition)),
SIMPLIFY = FALSE)
gg$grobs[which(grepl("strip-t",gg$layout$name))] <- edited_grobs
grid.draw(gg)
Note that this extracts all the strips using gg$grobs[which(grepl("strip-t",gg$layout$name))] and passes them to the mapply to be reset with the gsub(...) that OP specified in their comment.
In general, if you want to access just one of the text labels, there is a very similar structure which I made use of in my mapply:
num_to_access <- 1
gg$grobs[which(grepl("strip-t",gg$layout$name))][[num_to_access]][["grobs"]][[1]][["children"]][[2]][["children"]][[1]]$label
So to access the 4th label for example all you would need to do is change num_to_acces to be 4. Hope this helps!
I have named lists of entities (objects, lists, grobs?), from qplot or ggplot, that render or save just fine on their own, but I can't figure out how to pass them as a list or vector for arrangement. I believe my issue is with extracting list objects generally rather than with ggplot2.
library(ggplot2)
library(grid)
library(gridExtra)
# Generate a named list of ggplots
plotlist <- list()
for(a in c(1:4)) {
for(b in c(1:4)) {
plotlist[[paste0("Z",a,b)]] <-
qplot(rnorm(40,a,b),
geom="histogram",
xlab=paste0("Z",a,b))
}
}
# Arrange and display them
# The following two lines work fine, so the plots are in there:
plotlist[["Z12"]]
ggsave(plot=plotlist[["Z12"]], filename="deletable.png")
# The following two lines complain about not being 'grobs'
grid.arrange(plotlist, widths=c(1,1), ncol=2)
grid.arrange(unlist(plotlist), widths=c(1,1), ncol=2)
Can I somehow cast them as grobs without naming them explicitly, or find an alternative to unlist that lets the grob out?
Use lapply(plotlist, ggplot2::ggplotGrob) to generate a list of ggplot2 grobs. This list of grobs can then be passed to gridExtra::grid.arrange.
For example:
library(ggplot2)
library(gridExtra)
plotlist <- list()
for(a in c(1:4)) {
for(b in c(1:4)) {
plotlist[[paste0("Z",a,b)]] <-
qplot(rnorm(40,a,b),
geom="histogram",
xlab=paste0("Z",a,b))
}
}
grid.arrange(grobs = lapply(plotlist, ggplotGrob), widths = c(1, 1), ncol = 2)
I have several subjects for which I need to generate a plot, as I have many subjects I'd like to have several plots in one page rather than one figure for subject.
Here it is what I have done so far:
Read txt file with subjects name
subjs <- scan ("ListSubjs.txt", what = "")
Create a list to hold plot objects
pltList <- list()
for(s in 1:length(subjs))
{
setwd(file.path("C:/Users/", subjs[[s]])) #load subj directory
ifile=paste("Co","data.txt",sep="",collapse=NULL) #Read subj file
dat = read.table(ifile)
dat <- unlist(dat, use.names = FALSE) #make dat usable for ggplot2
df <- data.frame(dat)
pltList[[s]]<- print(ggplot( df, aes(x=dat)) + #save each plot with unique name
geom_histogram(binwidth=.01, colour="cyan", fill="cyan") +
geom_vline(aes(xintercept=0), # Ignore NA values for mean
color="red", linetype="dashed", size=1)+
xlab(paste("Co_data", subjs[[s]] , sep=" ",collapse=NULL)))
}
At this point I can display the single plots for example by
print (pltList[1]) #will print first plot
print(pltList[2]) # will print second plot
I d like to have a solution by which several plots are displayed in the same page, I 've tried something along the lines of previous posts but I don't manage to make it work
for example:
for (p in seq(length(pltList))) {
do.call("grid.arrange", pltList[[p]])
}
gives me the following error
Error in arrangeGrob(..., as.table = as.table, clip = clip, main = main, :
input must be grobs!
I can use more basic graphing features, but I d like to achieve this by using ggplot. Many thanks for consideration
Matilde
Your error comes from indexing a list with [[:
consider
pl = list(qplot(1,1), qplot(2,2))
pl[[1]] returns the first plot, but do.call expects a list of arguments. You could do it with, do.call(grid.arrange, pl[1]) (no error), but that's probably not what you want (it arranges one plot on the page, there's little point in doing that). Presumably you wanted all plots,
grid.arrange(grobs = pl)
or, equivalently,
do.call(grid.arrange, pl)
If you want a selection of this list, use [,
grid.arrange(grobs = pl[1:2])
do.call(grid.arrange, pl[1:2])
Further parameters can be passed trivially with the first syntax; with do.call care must be taken to make sure the list is in the correct form,
grid.arrange(grobs = pl[1:2], ncol=3, top=textGrob("title"))
do.call(grid.arrange, c(pl[1:2], list(ncol=3, top=textGrob("title"))))
library(gridExtra) # for grid.arrange
library(grid)
grid.arrange(pltList[[1]], pltList[[2]], pltList[[3]], pltList[[4]], ncol = 2, main = "Whatever") # say you have 4 plots
OR,
do.call(grid.arrange,pltList)
I wish I had enough reputation to comment instead of answer, but anyway you can use the following solution to get it work.
I would do exactly what you did to get the pltList, then use the multiplot function from this recipe. Note that you will need to specify the number of columns. For example, if you want to plot all plots in the list into two columns, you can do this:
print(multiplot(plotlist=pltList, cols=2))
I have a function which manipulates a ggplot object, by converting it to a grob and then modifying the layers. I would like the function to return a ggplot object not a grob. Is there a simple way to convert a grob back to gg?
The documentation on ggplotGrob is awfully sparse.
Simple example:
P <- ggplot(iris) + geom_bar(aes(x=Species, y=Petal.Width), stat="identity")
G <- ggplotGrob(P)
... some manipulation to G ...
## DESIRED:
P2 <- inverse_of_ggplotGrob(G)
such that, we can continue to use basic ggplot syntax, ie
`P2 + ylab ("The Width of the Petal")`
UPDATE:
To answer the question in the comment, the motivation here is to modify the colors of facet labels programmatically, based on the value of label name in each facet. The functions below work nicely (based on input from baptise in a previous question).
I would like for the return value from colorByGroup to be a ggplot object, not simply a grob.
Here is the code, for those interested
get_grob_strips <- function(G, strips=grep(pattern="strip.*", G$layout$name)) {
if (inherits(G, "gg"))
G <- ggplotGrob(G)
if (!inherits(G, "gtable"))
stop ("G must be a gtable object or a gg object")
strip.type <- G$layout[strips, "name"]
## I know this works for a simple
strip.nms <- sapply(strips, function(i) {
attributes(G$grobs[[i]]$width$arg1)$data[[1]][["label"]]
})
data.table(grob_index=strips, type=strip.type, group=strip.nms)
}
refill <- function(strip, colour){
strip[["children"]][[1]][["gp"]][["fill"]] <- colour
return(strip)
}
colorByGroup <- function(P, colors, showWarnings=TRUE) {
## The names of colors should match to the groups in facet
G <- ggplotGrob(P)
DT.strips <- get_grob_strips(G)
groups <- names(colors)
if (is.null(groups) || !is.character(groups)) {
groups <- unique(DT.strips$group)
if (length(colors) < length(groups))
stop ("not enough colors specified")
colors <- colors[seq(groups)]
names(colors) <- groups
}
## 'groups' should match the 'group' in DT.strips, which came from the facet_name
matched_groups <- intersect(groups, DT.strips$group)
if (!length(matched_groups))
stop ("no groups match")
if (showWarnings) {
if (length(wh <- setdiff(groups, DT.strips$group)))
warning ("values in 'groups' but not a facet label: \n", paste(wh, colapse=", "))
if (length(wh <- setdiff(DT.strips$group, groups)))
warning ("values in facet label but not in 'groups': \n", paste(wh, colapse=", "))
}
## identify the indecies to the grob and the appropriate color
DT.strips[, color := colors[group]]
inds <- DT.strips[!is.na(color), grob_index]
cols <- DT.strips[!is.na(color), color]
## Fill in the appropriate colors, using refill()
G$grobs[inds] <- mapply(refill, strip = G$grobs[inds], colour = cols, SIMPLIFY = FALSE)
G
}
I would say no. ggplotGrob is a one-way street. grob objects are drawing primitives defined by grid. You can create arbitrary grobs from scratch. There's no general way to turn a random collection of grobs back into a function that would generate them (it's not invertible because it's not 1:1). Once you go grob, you never go back.
You could wrap a ggplot object in a custom class and overload the plot/print commands to do some custom grob manipulation, but that's probably even more hack-ish.
You can try the following:
p = ggplotify::as.ggplot(g)
For more info, see https://cran.r-project.org/web/packages/ggplotify/vignettes/ggplotify.html
It involves a little bit of a cheat annotation_custom(as.grob(plot),...), so it may not work for all circumstances: https://github.com/GuangchuangYu/ggplotify/blob/master/R/as-ggplot.R
Have a look at the ggpubr package: it has a function as_ggplot(). If your grob is not too complex it might be a solution!
I would also advise to have a look at the patchwork package which combine nicely ggplots... it is likely to not be what you are looking for but... have a look.
This is cross-posted on the ggplot2 google group
My situation is that I'm working on a function that outputs an arbitrary number of plots (depending upon the input data supplied by the user). The function returns a list of n plots, and I'd like to lay those plots out in 2 x 2 formation. I'm struggling with the simultaneous problems of:
How can I allow the flexibility to be handed an arbitrary (n) number of plots?
How can I also specify I want them laid out 2 x 2
My current strategy uses grid.arrange from the gridExtra package. It's probably not optimal, especially since, and this is key, it totally doesn't work. Here's my commented sample code, experimenting with three plots:
library(ggplot2)
library(gridExtra)
x <- qplot(mpg, disp, data = mtcars)
y <- qplot(hp, wt, data = mtcars)
z <- qplot(qsec, wt, data = mtcars)
# A normal, plain-jane call to grid.arrange is fine for displaying all my plots
grid.arrange(x, y, z)
# But, for my purposes, I need a 2 x 2 layout. So the command below works acceptably.
grid.arrange(x, y, z, nrow = 2, ncol = 2)
# The problem is that the function I'm developing outputs a LIST of an arbitrary
# number plots, and I'd like to be able to plot every plot in the list on a 2 x 2
# laid-out page. I can at least plot a list of plots by constructing a do.call()
# expression, below. (Note: it totally even surprises me that this do.call expression
# DOES work. I'm astounded.)
plot.list <- list(x, y, z)
do.call(grid.arrange, plot.list)
# But now I need 2 x 2 pages. No problem, right? Since do.call() is taking a list of
# arguments, I'll just add my grid.layout arguments to the list. Since grid.arrange is
# supposed to pass layout arguments along to grid.layout anyway, this should work.
args.list <- c(plot.list, "nrow = 2", "ncol = 2")
# Except that the line below is going to fail, producing an "input must be grobs!"
# error
do.call(grid.arrange, args.list)
As I am wont to do, I humbly huddle in the corner, eagerly awaiting the sagacious feedback of a community far wiser than I. Especially if I'm making this harder than it needs to be.
You're ALMOST there! The problem is that do.call expects your args to be in a named list object. You've put them in the list, but as character strings, not named list items.
I think this should work:
args.list <- c(plot.list, 2,2)
names(args.list) <- c("x", "y", "z", "nrow", "ncol")
as Ben and Joshua pointed out in the comments, I could have assigned names when I created the list:
args.list <- c(plot.list,list(nrow=2,ncol=2))
or
args.list <- list(x=x, y=y, z=x, nrow=2, ncol=2)
Try this,
require(ggplot2)
require(gridExtra)
plots <- lapply(1:11, function(.x) qplot(1:10,rnorm(10), main=paste("plot",.x)))
params <- list(nrow=2, ncol=2)
n <- with(params, nrow*ncol)
## add one page if division is not complete
pages <- length(plots) %/% n + as.logical(length(plots) %% n)
groups <- split(seq_along(plots),
gl(pages, n, length(plots)))
pl <-
lapply(names(groups), function(g)
{
do.call(arrangeGrob, c(plots[groups[[g]]], params,
list(main=paste("page", g, "of", pages))))
})
class(pl) <- c("arrangelist", "ggplot", class(pl))
print.arrangelist = function(x, ...) lapply(x, function(.x) {
if(dev.interactive()) dev.new() else grid.newpage()
grid.draw(.x)
}, ...)
## interactive use; open new devices
pl
## non-interactive use, multipage pdf
ggsave("multipage.pdf", pl)
I'm answering a bit late, but stumbled on a solution at the R Graphics Cookbook that does something very similar using a custom function called multiplot. Perhaps it will help others who find this question. I'm also adding the answer as the solution may be newer than the other answers to this question.
Multiple graphs on one page (ggplot2)
Here's the current function, though please use the above link, as the author noted that it's been updated for ggplot2 0.9.3, which indicates it may change again.
# Multiple plot function
#
# ggplot objects can be passed in ..., or to plotlist (as a list of ggplot objects)
# - cols: Number of columns in layout
# - layout: A matrix specifying the layout. If present, 'cols' is ignored.
#
# If the layout is something like matrix(c(1,2,3,3), nrow=2, byrow=TRUE),
# then plot 1 will go in the upper left, 2 will go in the upper right, and
# 3 will go all the way across the bottom.
#
multiplot <- function(..., plotlist=NULL, file, cols=1, layout=NULL) {
require(grid)
# Make a list from the ... arguments and plotlist
plots <- c(list(...), plotlist)
numPlots = length(plots)
# If layout is NULL, then use 'cols' to determine layout
if (is.null(layout)) {
# Make the panel
# ncol: Number of columns of plots
# nrow: Number of rows needed, calculated from # of cols
layout <- matrix(seq(1, cols * ceiling(numPlots/cols)),
ncol = cols, nrow = ceiling(numPlots/cols))
}
if (numPlots==1) {
print(plots[[1]])
} else {
# Set up the page
grid.newpage()
pushViewport(viewport(layout = grid.layout(nrow(layout), ncol(layout))))
# Make each plot, in the correct location
for (i in 1:numPlots) {
# Get the i,j matrix positions of the regions that contain this subplot
matchidx <- as.data.frame(which(layout == i, arr.ind = TRUE))
print(plots[[i]], vp = viewport(layout.pos.row = matchidx$row,
layout.pos.col = matchidx$col))
}
}
}
One creates plot objects:
p1 <- ggplot(...)
p2 <- ggplot(...)
# etc.
And then passes them to multiplot:
multiplot(p1, p2, ..., cols = n)