在ggplots列表上调用grid.arrange [英] calling grid.arrange on a list of ggplots

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我现在的问题是:在我的代码中，我通过调用lapply在列表中生成ggplots列表.我首先使用lapply，因为我执行了大量的类似ggplots，并且手动生成每个ggplots太麻烦了.如何概括我的代码?

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))

推荐答案

使用gridExtra v> = 2.0.0，您现在可以这样做，

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

grid.arrange(grobs = year.plots)

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