# Vectorize a Function

This article is originally published at http://www.beardedanalytics.com

I was recently working and decided to write a function to assist in the process. It assigns a label to a number based upon the value. My first attempt worked, but only for one value at a time.

KMO_adequacy2 <- function(x){ if ( x > .90 ) {result <- "Marvelous"} if (x <=.90 & x >.80 ) { result <-"Meritorious" } if (x <= .8 & x >.70 ) { result <- "Middling" } if (x <=.7 & x>.60) { result <-"Mediocre" } if (x <=.6 & x>.50) {result <-"Miserable" } if (x <=.5) { result <- "Unacceptable" } return(result) }

This works.

KMO_adequacy2(.95) #[1] "Marvelous"

This does not work.

KMO_adequacy2(c(.95,.50)) #[1] "Marvelous" # Warning messages: # 1: In if (x > 0.9) { : # the condition has length > 1 and only the first element will be used # 2: In if (x <= 0.9 & x > 0.8) { : # the condition has length > 1 and only the first element will be used # 3: In if (x <= 0.8 & x > 0.7) { : # the condition has length > 1 and only the first element will be used # 4: In if (x <= 0.7 & x > 0.6) { : # the condition has length > 1 and only the first element will be used # 5: In if (x <= 0.6 & x > 0.5) { : # the condition has length > 1 and only the first element will be used # 6: In if (x <= 0.5) { : # the condition has length > 1 and only the first element will be used

I should have thought more about the end goal of the function before I started coding, but I didn't. I started searching for good ways to vectorize a function in R. I found there is a function in base R called '*Vectorize*'. All I needed was to create a new function using '*Vectorize*' and I was done.

KMO_adequacy <- Vectorize(KMO_adequacy2) KMO_adequacy(c(.95,.50)) #[1] "Marvelous" "Unacceptable"

This allowed me to easily add a column to my data frame containing individual KMO measures and associate a label. I reached out to my fellow blogger Jeremy and he gave me a quick re-write of my original function. Here is his approach.

# Jeremy's re-write KMO_J <- function(x){ names(x) <- x names(x) <- ifelse(x > .90, "Marvelous", names(x)) names(x) <- ifelse(x <=.9 & x >.80, "Meritorious", names(x)) names(x) <- ifelse(x <=.8 & x >.70, "Middling", names(x)) names(x) <- ifelse(x <=.7 & x>.60, "Mediocre", names(x)) names(x) <- ifelse(x <=.6 & x>.50, "Miserable", names(x)) names(x) <- ifelse(x <=.5, "Unacceptable", names(x)) return(names(x)) } KMO_J(c(.95,.50)) #[1] "Marvelous" "Unacceptable"

We can do a quick check to make sure that we are getting the same output.

test <- seq(0,1,.1) all.equal ( KMO_adequacy(test) , KMO_J(test) ) #[1] TRUE

My next question, is there a major performance difference between the two?Â I ran a short simulation which is summarized in the plot below and shows that there is not a large difference in performance for the samples tested.

Have a better way to solve this problem? Post it in the comments below. If you are wondering what this KMO thing is all about, it is a measure of sampling accuracy (MSA) for conducting exploratory factor analysis (EFA). The cutoffs and names were taken from:

Barbara A. Cerny , Henry F. Kaiser

Multivariate Behavioral Research

Vol. 12, Iss. 1, 1977

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This article is originally published at http://www.beardedanalytics.com

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