CODEWITHSUNDEEP
rhistogramgraphing
MCalvi
MCalvi

Reputation: 70

How to add titles from a list to a series of histograms?

Is it possibile to attach titles via an apply-family function to a series of histograms where the titles are picked from a list?

The code below creates three histograms. I want to give each a different name from a list (named "list"), in order. How do I do this?

data <- read.csv("outcome-of-care-measures.csv")
outcome <-data[,c(11,17,23)]
out <- apply(outcome, 2,as.data.frame) 
par(mfrow=c(3,1))
apply(outcome,2, hist, xlim=range(out,na.rm=TRUE), xlab="30 day death rate")

Upvotes: 2

Views: 1445

Answers (2)

Paul Hiemstra
Paul Hiemstra

Reputation: 60944

I'd use facet_wrap from ggplot2 to get this done. ggplot2 supports this kind of plots very painlessly:

library(ggplot2)
theme_set(theme_bw())

df = data.frame(values = rnorm(3000), ID = rep(LETTERS[1:3], each = 1000))
ggplot(df, aes(x = values)) + geom_histogram() + facet_wrap(~ ID)

enter image description here

To change the text in the box above each facet, simply replace the text in the ID variable:

id_to_text_translator = c(A = 'Facet text for A',
                          B = 'Facet text for B',
                          C = 'Facet text for C')
df$ID = id_to_text_translator[df$ID]

I would recommend taking a close look at what happens in these two lines. Using vectorized subsetting to perform this kind of replacement has compact syntax and high performance. Replacing this code would require a for or apply loop combined with a set of if statements would make the code much longer and slower.

Another option is to directly influence the levels of ID, which is a factor:

levels(df$ID) = id_to_text_translator[levels(df$ID)]

This is a lot faster, especially on datasets with a lot of rows. In addition, it keeps ID a factor, while the previous solution makes ID a character vector.

The resulting plot:

ggplot(df, aes(x = values)) + geom_histogram() + facet_wrap(~ ID)

enter image description here

Upvotes: 2

Mark Heckmann
Mark Heckmann

Reputation: 11431

As it is not only the columns but another argument that changes, you may use mapply to have a function from the apply family.

args <- list(xlim=range(data, na.rm=TRUE), xlab="30 day death rate")
titles <- list("Title 1", "Title 2", "Title 3")
par(mfrow=c(3,1))
mapply(hist, data, main=titles, MoreArgs=args)

If you wrap the invisible function around the last line you can avoid the console output.

enter image description here

Note: I think using loops here is far more straightforward.

Upvotes: 1

Related Questions