问题
I am building a table with a mixture of numbers, text, and plots. I constructed my plots with ggplot, and added them to the table afterwards (please see code below). Because I will (eventually) have many plots, I need to use a loop to efficiently create them all. However, because ggplot seems to require printing to generate image links for each plot, I am unable to use invisible(), and subsequently get the pesky '[1] [[2]] [[3]]' output at the top of the image below.
How can I compile the document without printing any visible output from ggplot?
```{r score_table, fig.show = "hide", echo = FALSE, fig.height=.75, fig.width=2.5}
#Load libraries
library(knitr)
library(ggplot2)
#Item data
items <- data.frame(text = sapply(1:3, FUN = function(x){
paste0(sample(x = LETTERS, size = 60, replace = T), collapse = "")}))
#Score data
score_set = replicate(n = 3, expr = {data.frame(other = rep("other", 4),
score=sample(1:7,4,TRUE))}, simplify = F)
#Plot function
plotgen<-function(score_set,other,score){
p <- ggplot(score_set, aes(factor(other), score))
p + geom_violin(fill = "#99CCFF") + coord_flip() + scale_x_discrete(name=NULL) +
scale_y_continuous(breaks = round(seq(1, 7, by = 1),1), limits = c(1,7), name=NULL) +
theme(axis.text.y=element_blank(),axis.title.y=element_blank(),axis.ticks.y=elemen t_blank(),
panel.grid.major.y = element_line(colour = "black"),
panel.grid.minor = element_blank(),
panel.background = element_rect(fill = "white"),
panel.border = element_rect(colour = "black", fill=NA, size=1)) +
geom_hline(yintercept=sample(1:7,1,TRUE), size = 1.5, colour = "#334466")
}
#Generate plots
print(lapply(seq_along(score_set), FUN = function(x){plotgen(score_set[[x]],other,score)}))
out <- cbind(row.names(items), as.character(items$text), sprintf("",
opts_current$get("fig.path"), opts_current$get("label"), 1:nrow(items)))
#Build table
kable(out, col.names = c("ID", "Text", "Scores"))
```
回答1:
lapply returns a list. When you print a list, regardless of it's contents, it also prints the list indices, [[1]], [[2]], [[3]], .... If you instead save the list,
plot_list <- lapply(seq_along(score_set), FUN = function(x){plotgen(score_set[[x]],other,score)})
and then print each plot in the list instead of printing the whole list (and this we can wrap in invisible() so the returned list isn't printed)
invisible(lapply(plot_list, print))
it won't print the indices of the list. Because you will be printing each plot individually, not printing a list which happens to contain plots.
Demonstrating on a simple list:
x = list(1, 2, 3)
print(x)
# [[1]]
# [1] 1
#
# [[2]]
# [1] 2
#
# [[3]]
# [1] 3
invisible(lapply(x, print))
# [1] 1
# [1] 2
# [1] 3
An alternate solution, not requiring invisible because it doesn't return anything, is just a for loop:
for (i in seq_along(plot_list)) print(plot_list[[i]])
I'll leave it to you to see which you prefer.
Addressing the worry that a for loop would be slower:
p = ggplot(mtcars, aes(x = hp, y = mpg)) + geom_point()
plist = list(p, p)
library(microbenchmark)
microbenchmark(
forloop = {for (i in seq_along(plist)) print(plist[[i]])},
lapply = invisible(lapply(plist, print)),
times = 10L
)
# Unit: milliseconds
# expr min lq mean median uq max neval cld
# forloop 260.4532 271.2784 295.8415 276.1587 289.7507 402.1792 10 a
# lapply 258.8032 269.5915 296.2268 287.9524 294.8860 398.6803 10 a
The difference is a few milliseconds.
来源：https://stackoverflow.com/questions/39401789/producing-ggplots-from-a-loop-and-generating-the-files-without-printing-any-vi