CODEWITHSUNDEEP
r
neversaint
neversaint

Reputation: 63984

How use lapply on list of data frames and using a function that takes the ID name of the list

I have the following lists of data frame:

d1 <- data.frame(y1=c(1,2,3),y2=c(4,5,6))
d2 <- data.frame(y1=c(3,2,1),y2=c(6,5,4))
df <- list(FOO=d1, BAR=d2)

Which looks like this:

> df
$FOO
  y1 y2
1  1  4
2  2  5
3  3  6

$BAR
  y1 y2
1  3  6
2  2  5
3  1  4

And a function that plot each individual data frame and named them based on list ID:

plot_df_func <- function(basename="QUX",in_df) {
    outfile <- paste(basename,".png",sep="")

    png(outfile)
    plot(in_df,main=basename)
    dev.off()
}

How can I use lapply in this case?

At the end of the day it will have "FOO.png" and "BAR.png".

I tried this but failed:

> lapply(df, plot_df_func) 
Error in plot(in_df, main = basename) : 
  argument "in_df" is missing, with no default

Upvotes: 2

Views: 78

Answers (1)

Roland
Roland

Reputation: 132696

I would use a for loop. In my opinion, a for loop is always preferable if you want only side effects (like plots or files) and no return value.

d1 <- data.frame(y1=c(1,2,3),y2=c(4,5,6))
d2 <- data.frame(y1=c(3,2,1),y2=c(6,5,4))
df <- list(FOO=d1, BAR=d2)

plot_df_func <- function(basename="QUX",in_df) {
  #outfile <- paste(basename,".png",sep="")

  #png(outfile)
  plot(in_df,main=basename)
  #dev.off()
}

layout(1:2)
for (i in names(df)) plot_df_func(i, df[[i]])

resulting plot

Upvotes: 1

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