ggplot2 : printing multiple plots in one page with

问题:

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

回答1:

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"))))

回答2:

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)

回答3:

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))