Writing an R-function that works with multiple existing dataframes

Question

I'm working on a R-Script where I have to create multiple variables for multiple existing dataframes. For example: I got the three dataframes AAA, BBB and CCC, and each one of them has a column named "height". I would like to get this:

AAA_Skal <- mean(AAA$height) / 6
BBB_Skal <- mean(BBB$height) / 6
CCC_Skal <- mean(CCC$height) / 6

without actually writing a entire line of code for each dataframe. (The reason for this is that I got more than just 3 dataframes and I have to proceed a lot of code with them).

What I tried is the following:

dfs <- c("AAA", "BBB", "CCC")

Skal <- function(x) {
sprintf("%s_Skal", dfs[x]) <- mean(sprintf("%s_$height", dfs[x])) / 6
}

I should then be able to type Skal(1) to get

AAA_Skal <- mean(AAA$height) / 6

First I thought it isnt working because sprintf gives "AAA" as an output, with quotation marks. So I tried as.name(sprintf()), but it doesent work either. I hope someone can help me on this issue, and sorry for my bad english.

Jake Burkhead · Accepted Answer · 2014-02-16 17:24:54Z

3

You can use assign and get

AAA <- data.frame(x = letters[1:10], height = rnorm(10))
BBB <- data.frame(y = sample(20:60, 10), height = rnorm(10, 5))
CCC <- data.frame(z = rbinom(10, 5, .2), height = rnorm(10, 50))

dfs <- c("AAA", "BBB", "CCC")

x <- 1
assign(sprintf("%s_Skal", dfs[x]), mean(get(dfs[x])$height) / 6)
AAA_Skal
## [1] -0.02807432

But I would recommend storing your data.frames in a list so you can do something like

l <- list(AAA, BBB, CCC)
sapply(l, function(x) mean(x$height) / 6 )
## [1] -0.02807432  0.84785430  8.38655123

answered Feb 16, 2014 at 17:24

Jake Burkhead

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3 Comments

Duck Over a year ago

Dear @JakeBurkhead Maybe could it get the same result using lapply(l, function(x) mean(x$height) / 6 ) what is the difference between lapply() and sapply()

Jake Burkhead Over a year ago

@Duck lapply will return a list of length 1 numeric vectors (try it out and see!). sapply tries to coerce its output into a vector or matrix and names it output, lapply will always return a list. sapply(..., simplify = FALSE, USE.NAMES = FALSE) is equivalent to lapply. See ?sapply for more info.

Duck Over a year ago

Ok @JakeBurkhead sapply() is more practical than lapply() and it gives me a more clean solution than lapply that gives a list as result.

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