Translating Stata to R - recoding

Question

I am trying to translate the following from Stata

clear
set obs 1000
generate y = floor((10-0+1)*runiform() +0)
recode y (7=0) (8=0) (9=1) (10=2)

I thought I had it with the following code:

library(dplyr)
mydata <- y ~ floor((10-0+1)*runif(1000)+0)
recode (mydata, '7'=0, '8'=0, '9'=1, '10'=2)

However, the last line keeps giving me an error:

Error in UseMethod("recode"):no applicable method for 'recode' applied to an object of class "formula".

Any ideas?

• JasonAizkalns: Try mydata <- floor((10-0+1)*runif(1000)+0)
• Nick Cox: Minor details: 10 - 0 + 1 is 11, always and in both Stata and R. Similarly, adding 0 is useless. In both cases, if these are steps towards something more general, then fine, but that is not at all obvious.

Answer 1

You can use cut:

n = 1000L
y = cut(runif(n, 0, 11), c(-Inf, 9, 10, Inf), right = FALSE, ordered = TRUE)

You can see how it worked with table:

# y
# [-Inf,9) [9,10) [10, Inf) 
# 813 91 96

If you really want the codes, you can use as.integer(y)-1L. Read ?cut and ?factor for more details on ordinal data in R.

---

I've been assuming so far that there is some rhyme or reason to the recoding rule. If there is not, best to store it in a separate table and draw values from there (which is the same thing I would do in Stata):

rec = data.frame(old = c(7,8,9,10), new = c(0,0,1,2))
n = 1000L
y = floor(runif(n, 0, 11))
DF = data.frame(id = 1:10, y)
library(data.table)
setDT(DF)
DF[rec, on=c(y = "old"), y := new]
DF[, .N, keyby=y]
# y N
# 1: 0 288
# 2: 1 179
# 3: 2 174
# 4: 3 101
# 5: 4 82
# 6: 5 93
# 7: 6 83

You'd need to install the data.table package for this to work, though.

• Álvaro A. Gutiérrez-Vargas: Cool answer, @Frank, but any clue how to perform the recoding using only base R?Thxs
• Frank: Thanks @ÁlvaroA.Gutiérrez-Vargas I think something involving the match function would work, maybe m = match(DF$y, rec$old); DF$y = ifelse(is.na(m), DF$y, rec$new[m]) You could also post a bounty to see if someone else has a better answer stackoverflow.com/help/bounty