CODEWITHSUNDEEP
rggplot2gridextrado.call
Samy Geronymos
Samy Geronymos

Reputation: 405

calling grid.arrange on a list of ggplots

This question is linked to this one here:

Related post

My present question is: In my code I generate a list of ggplots in a list by calling lapply. I use lapply in the first place because I execute a fairly big amount of similar ggplots and it would be too cumbersome to generate each ggplots manually. how can I generalize my code?

p <- qplot(rnorm(30))

plist <- lapply(c(1:10),FUN=function(x){

  qplot(rnorm(30))

})

#works
year.plots <- list(p,p)
do.call(grid.arrange, c(year.plots))

#works
plist[[1]]

#works
grid.arrange(p,plist[[1]])

#does not work
year.plots <- list(p,plist[[1]])
do.call(grid.arrange, c(year.plots))

#How to generalize with the following idea?
year.plots <- list(p,plist[[1]],plist[[2]],...)
do.call(grid.arrange, c(year.plots))

Upvotes: 2

Views: 4632

Answers (3)

baptiste
baptiste

Reputation: 77116

With gridExtra v>=2.0.0, you can now do,

grid.arrange(grobs = year.plots)

Upvotes: 10

Jaap
Jaap

Reputation: 83265

It's not necessary to wrap the list in c(), both do.call(grid.arrange, year.plots) and do.call(grid.arrange, c(year.plots)) work.

However, if you want to include extra arguments, you will need to wrap them together with the list in the c() part like this:

do.call(grid.arrange, c(year.plots, ncol=2))

Upvotes: 6

Samy Geronymos
Samy Geronymos

Reputation: 405

Ok I'm silly I found the mistake in my code after 40 minutes.

do.call(grid.arrange, c(year.plots,plist,nrow=3))

Upvotes: 0

Related Questions