CODEWITHSUNDEEP
rnestedlapplylm
wsen
wsen

Reputation: 33

Nested lapply with substitute in R

I've created code to run the lm command multiple times with different variables:

DVs <- c('mpg', 'wt')

lapply(DVs, function(x)
  lm(substitute(scale(DV) ~ scale(disp)+scale(qsec), list(DV = as.name(x))), data=mtcars))

However, I would like to be able to create a nested loop, to iterate through another variable as well. The code I have used to attempt that is:

DVs <- c('mpg', 'wt')
IVs <- c('disp', 'hp')

    lapply(DVs, function(x) lapply(IVs, function(y)
      lm(substitute(substitute(scale(DV) ~ scale(IV)+scale(qsec), list(DV = as.name(x)), list(IV = as.name(y)), data=mtcars)))))

However this leads to an error:

Error in substitute(scale(DV) ~ scale(IV) + scale(qsec), list(DV = as.name(x)), : unused arguments (list(IV = as.name(X[[i]])), data = mtcars)

How can I make this nested loop work?

Upvotes: 2

Views: 559

Answers (1)

wsen
wsen

Reputation: 33

Thanks for the suggestion Josh O'Brien! This worked:

DVs <- c('mpg', 'wt')
IVs <- c('disp', 'hp')

lapply(DVs, function(x) lapply(IVs, function(y) {lm(reformulate(response=x, termlabels=y), data=mtcars)}))

I couldn't figure out how to scale my data within the call, but I can just scale my whole dataframe:

mtcarsSC <- as.data.frame(scale(mtcars))

DVs <- c('mpg', 'wt')
IVs <- c('disp', 'hp')

lapply(DVs, function(x) lapply(IVs, function(y) {lm(reformulate(response=x, termlabels=y), data=mtcarsSC)}))

Upvotes: 1

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