Using CUT and Quartile to generate breaks in R function

r cut
mikebmassey picture mikebmassey · Jul 30, 2012 · Viewed 52.5k times · Source

Following some great advice from before, I'm now writing my 2nd R function and using a similar logic. However, I'm trying to automate a bit more and may be getting too smart for my own good.

I want to break the clients into quintiles based on the number of orders. Here's my code to do so:

# sample data
clientID <- round(runif(200,min=2000, max=3000),0)
orders <- round(runif(200,min=1, max=50),0)

df <- df <- data.frame(cbind(clientID,orders))

#function to break them into quintiles
ApplyQuintiles <- function(x) {
  cut(x, breaks=c(quantile(df$orders, probs = seq(0, 1, by = 0.20))), 
      labels=c("0-20","20-40","40-60","60-80","80-100"))
}

#Add the quintile to the dataframe
df$Quintile <- sapply(df$orders, ApplyQuintiles)

table(df$Quintile)

0-20   20-40   40-60    60-80   80-100 
40     39      44       38      36

You'll see here that in my sample data, I created 200 observations, yet only 197 are listed via table. The 3 left off are NA

Now, there are some clientIDs that have an 'NA' for quintile. It seems if they were at the lowest break, in this case, 1, then they were not included in the cut function.

Is there a way to make cut inclusive of all observations?

Answer

Edward picture Edward · Jul 30, 2012

Try the following:

set.seed(700)

clientID <- round(runif(200,min=2000, max=3000),0)
orders <- round(runif(200,min=1, max=50),0)

df <- df <- data.frame(cbind(clientID,orders))

ApplyQuintiles <- function(x) {
  cut(x, breaks=c(quantile(df$orders, probs = seq(0, 1, by = 0.20))), 
      labels=c("0-20","20-40","40-60","60-80","80-100"), include.lowest=TRUE)
}
df$Quintile <- sapply(df$orders, ApplyQuintiles)
table(df$Quintile)

0-20  20-40  40-60  60-80 80-100 
  40     41     39     40     40 

I included include.lowest=TRUE in your cut function, which seems to make it work. See ?cut for more details.