Find and replace without backsearching

Question

R novice here with what is probably a silly/ simple question.

I am trying to use R to find and replace within a vector "census.new" (basically trying to use R to do what excel does with find/ replace). However, because the original data (what I am searching for) overlaps partially with what I am trying to replace, I am ending up with some funky results.

I am trying to replace 2 with 2008b, 3 with 2009a, etc.

census.new<-sub("2","2008b", census.new, fixed=TRUE)
census.new<-sub("3","2009a", census.new, fixed=TRUE)
census.new<-sub("4","2009b", census.new, fixed=TRUE)
census.new<-sub("5","2010a", census.new, fixed=TRUE)
census.new<-sub("8","2011b", census.new, fixed=TRUE)
census.new<-sub("10","2012a", census.new, fixed=TRUE)
census.new<-sub("11","2012b", census.new, fixed=TRUE)
census.new<-sub("12","2013a", census.new, fixed=TRUE)
census.new<-sub("13","2013b", census.new, fixed=TRUE)
table(census.new)

The resulting table is:

2002020202013babbb    2009a     2009b     202013ba00202012bbb    202013ba009a
              6268     3129      3129                    3129            3129

20202013baa         20202013bab      2020202013baaa   2020202013babb
       3129                3129                3129             3129 

Thanks in advance!

Answer 1

Your choice to use sub is handicapped by its use of regular expressions: for example, it is replacing every instance of "2" by "2008b", so "2012" becomes "2008b012008b". Since you subsequently look for other (similar) strings, it cascades (snow-balls) quickly. As an example:

gsub('2', '2008b', c('2', '22'), fixed = TRUE)
## [1] "2008b"      "2008b2008b"

There are several ways around it, some more elegant than others. First, I'll generate some unique data, since I don't know exactly how yours looks:

set.seed(42)
dat <- sample(as.character(c(2, 3, 4, 5, 8, 10, 11, 12, 13)),
              size = 31300, replace = TRUE)
table(dat)
## dat
##    2    3    4    5    8   10   11   12   13 
## 3577 3573 3275 3499 3453 3481 3487 3439 3516 

sub with regular expressions

This is by far not the best or optimal solution. If you want/need to continue down the road of using sub (or gsub), you should read up on regular expressions (many many more websites and tutorials exist for regexps).

dat1 <- sub('^2$', '2008b', dat)
dat1 <- sub('^3$', '2009a', dat1)
dat1 <- sub('^4$', '2009b', dat1)
dat1 <- sub('^5$', '2010a', dat1)
dat1 <- sub('^8$', '2011b', dat1)
dat1 <- sub('^10$', '2012a', dat1)
dat1 <- sub('^11$', '2012b', dat1)
dat1 <- sub('^12$', '2013a', dat1)
dat1 <- sub('^13$', '2013b', dat1)
table(dat1)
## dat1
## 2008b 2009a 2009b 2010a 2010b 2012a 2012b 2013a 2013b 
##  3577  3573  3275  3499  3453  3481  3487  3439  3516 

You can nest these (sub('^2$', '2008b', sub('^3$', '2009a', dat))), but it doesn't make it more readable or extensible. It's ugly, too (IMHO).

car::recode

This function is meant for (what I'm inferring is) your purpose.

library(car)
dat2 <- recode(dat, "
    2 = '2008b'; 3 = '2009a'; 4 = '2009b';
    5 = '2010a'; 8 = '2011b'; 10 = '2012a';
    11 = '2012b'; 12 = '2013a'; 13 = '2013b'")
table(dat2)
## dat2
## 2008b 2009a 2009b 2010a 2011b 2012a 2012b 2013a 2013b 
##  3577  3573  3275  3499  3453  3481  3487  3439  3516 
identical(dat1, dat2)
## [1] TRUE

Relative Performance

And the performance of car::recode isn't bad:

library(microbenchmark)
microbenchmark(
  re = {
    dat1 <- sub('^2$', '2008b', dat)
    dat1 <- sub('^3$', '2009a', dat1)
    dat1 <- sub('^4$', '2009b', dat1)
    dat1 <- sub('^5$', '2010a', dat1)
    dat1 <- sub('^8$', '2011b', dat1)
    dat1 <- sub('^10$', '2012a', dat1)
    dat1 <- sub('^11$', '2012b', dat1)
    dat1 <- sub('^12$', '2013a', dat1)
    dat1 <- sub('^13$', '2013b', dat1)
  },
  car = {
    recode(dat, "
        2 = '2008b'; 3 = '2009a'; 4 = '2009b';
        5 = '2010a'; 8 = '2011b'; 10 = '2012a';
        11 = '2012b'; 12 = '2013a'; 13 = '2013b'")
  }, times = 50)
## Unit: milliseconds
##  expr      min       lq     mean   median       uq       max neval cld
##    re 74.44709 75.97057 78.10516 77.20569 78.43732 124.42665   100   b
##   car 23.21131 24.11697 26.08724 25.74997 26.25260  77.97541   100  a