Idiom for ifelse-style recoding for multiple categories

孤街醉人 提交于 2019-11-26 17:33:17

One could use a list as an associative array to define the brand -> product code mapping, i.e.:

brands <- list(Tylenol=1:3, Advil=4:6, Bayer=7:9, Generic=10:12)

Once you have this, you can then either invert this to create a product code -> brand list (could take a lot of memory), or just use a search function:

find.key <- function(x, li, default=NA) {
    ret <- rep.int(default, length(x))
    for (key in names(li)) {
        ret[x %in% li[[key]]] <- key
    }
    return(ret)
}

I'm sure there are better ways of writing this function (the for loop is annoying me!), but at least it is vectorised, so it only requires a single pass through the list.

Using it would be something like:

> dat$brand <- find.key(dat$product, brands)
> dat
   product   brand
1       11 Generic
2       11 Generic
3        9   Bayer
4        9   Bayer
5        6   Advil
6        1 Tylenol
7       11 Generic
8        5   Advil
9        7   Bayer
10      11 Generic
11       5   Advil
12      11 Generic
13       4   Advil
14       3 Tylenol
15      10 Generic
16       7   Bayer
17      10 Generic
18       5   Advil
19       9   Bayer
20       8   Bayer

The recode and levels<- solutions are very nice, but they are also significantly slower than this one (and once you have find.key this is easier-for-humans than recode and on par with the levels<-):

> microbenchmark(
     recode=recode(dat$product,recodes="1:3='Tylenol';4:6='Advil';7:9='Bayer';10:12='Generic'"), 
     find.key=find.key(dat$product, brands),
     levels=`levels<-`(factor(dat$product),brands))
Unit: microseconds
      expr      min        lq    median        uq      max
1 find.key   64.325   69.9815   76.8950   83.8445  221.748
2   levels  240.535  248.1470  274.7565  306.8490 1477.707
3   recode 1636.039 1683.4275 1730.8170 1855.8320 3095.938

(I can't get the switch version to benchmark properly, but it appears to be faster than all of the above, although it is even worse-for-humans than the recode solution.)

Marek

You could convert your variable to a factor and change its levels by levels<- function. In one command it could be like:

`levels<-`(
    factor(dat$product),
    list(Tylenol=1:3, Advil=4:6, Bayer=7:9, Generic=10:12)
)

In steps:

brands <- factor(dat$product)
levels(brands) <- list(Tylenol=1:3, Advil=4:6, Bayer=7:9, Generic=10:12)

I like the recode function in the car package:

library(car)

dat$brand <- recode(dat$product,
  recodes="1:3='Tylenol';4:6='Advil';7:9='Bayer';10:12='Generic'")

# > dat
#    product   brand
# 1       11 Generic
# 2       11 Generic
# 3        9   Bayer
# 4        9   Bayer
# 5        6   Advil
# 6        1 Tylenol
# 7       11 Generic
# 8        5   Advil
# 9        7   Bayer
# 10      11 Generic
# 11       5   Advil
# 12      11 Generic
# 13       4   Advil
# 14       3 Tylenol
# 15      10 Generic
# 16       7   Bayer
# 17      10 Generic
# 18       5   Advil
# 19       9   Bayer
# 20       8   Bayer

I often use the technique below:

key <- c()
key[1:3] <- "Tylenol"
key[4:6] <- "Advil"
key[7:9] <- "Bayer"
key[10:12] <- "Generic"

Then,

> key[dat$product]
 [1] "Generic" "Generic" "Bayer"   "Bayer"   "Advil"   "Tylenol" "Generic" "Advil"   "Bayer"   "Generic"
[11] "Advil"   "Generic" "Advil"   "Tylenol" "Generic" "Bayer"   "Generic" "Advil"   "Bayer"   "Bayer"  

The "database approach" is to keep a separate table (a data.frame) for your product keys definitions. It makes even more sense since you say your product keys translate into not only a brand, but also a size:

product.keys <- read.table(textConnection("

product brand   size
1       Tylenol small
2       Tylenol medium
3       Tylenol large
4       Advil   small
5       Advil   medium
6       Advil   large
7       Bayer   small
8       Bayer   medium
9       Bayer   large
10      Generic small
11      Generic medium
12      Generic large

"), header = TRUE)

Then, you can join your data using merge:

merge(dat, product.keys, by = "product")
#    product   brand   size
# 1        1 Tylenol  small
# 2        3 Tylenol  large
# 3        4   Advil  small
# 4        5   Advil medium
# 5        5   Advil medium
# 6        5   Advil medium
# 7        6   Advil  large
# 8        7   Bayer  small
# 9        7   Bayer  small
# 10       8   Bayer medium
# 11       9   Bayer  large
# 12       9   Bayer  large
# 13       9   Bayer  large
# 14      10 Generic  small
# 15      10 Generic  small
# 16      11 Generic medium
# 17      11 Generic medium
# 18      11 Generic medium
# 19      11 Generic medium
# 20      11 Generic medium

As you notice, the order of the rows is not preserved by merge. If this is a problem, the plyr package has a join function that does preserve the order:

library(plyr)
join(dat, product.keys, by = "product")
#    product   brand   size
# 1       11 Generic medium
# 2       11 Generic medium
# 3        9   Bayer  large
# 4        9   Bayer  large
# 5        6   Advil  large
# 6        1 Tylenol  small
# 7       11 Generic medium
# 8        5   Advil medium
# 9        7   Bayer  small
# 10      11 Generic medium
# 11       5   Advil medium
# 12      11 Generic medium
# 13       4   Advil  small
# 14       3 Tylenol  large
# 15      10 Generic  small
# 16       7   Bayer  small
# 17      10 Generic  small
# 18       5   Advil medium
# 19       9   Bayer  large
# 20       8   Bayer medium

Finally, if your tables are large and speed is an issue, consider using data.tables (from the data.table package) instead of data.frames.

This one takes some typing but if you really do have a huge data set this may be the way to go. Bryangoodrich and Dason at talkstats.com taught me this one. It's using a hash table or creating a environment that contains a look up table. I actually keep this one on my .Rprofile (the hash function that is) for dictionary type look ups.

I replicated your data 1000 times to make it a bit larger.

#################################################
# THE HASH FUNCTION (CREATES A ENW ENVIRONMENT) #
#################################################
hash <- function(x, type = "character") {
    e <- new.env(hash = TRUE, size = nrow(x), parent = emptyenv())
    char <- function(col) assign(col[1], as.character(col[2]), envir = e)
    num <- function(col) assign(col[1], as.numeric(col[2]), envir = e)
    FUN <- if(type=="character") char else num
    apply(x, 1, FUN)
    return(e)
}
###################################
# YOUR DATA REPLICATED 1000 TIMES #
###################################
dat <- dat <- structure(list(product = c(11L, 11L, 9L, 9L, 6L, 1L, 11L, 5L, 
    7L, 11L, 5L, 11L, 4L, 3L, 10L, 7L, 10L, 5L, 9L, 8L)), .Names = "product", row.names = c(NA, 
    -20L), class = "data.frame")
dat <- dat[rep(seq_len(nrow(dat)), 1000), , drop=FALSE]
rownames(dat) <-NULL
dat
#########################
# CREATE A LOOKUP TABLE #
#########################
med.lookup <- data.frame(val=as.character(1:12), 
    med=rep(c('Tylenol', 'Advil', 'Bayer', 'Generic'), each=3))  

########################################
# USE hash TO CREATE A ENW ENVIRONMENT #
########################################  
meds <- hash(med.lookup)  

##############################
# CREATE A RECODING FUNCTION #
##############################          
recoder <- function(x){
    x <- as.character(x) #turn the numbers to character
    rc <- function(x){
       if(exists(x, env = meds))get(x, e = meds) else NA 
    }  
    sapply(x, rc, USE.NAMES = FALSE) 
}
#############
# HASH AWAY #
#############
recoder(dat[, 1])    

In this case hashing is slow but if you have more levels to recode then it will increase in speed over others.

Somewhat more readable than nested ifelse's:

unlist(lapply(as.character(dat$product), switch,
              `1`=,`2`=,`3`='tylenol',
              `4`=,`5`=,`6`='advil',
              `7`=,`8`=,`9`='bayer',
              `10`=,`11`=,`12`='generic'))

Caveat: not very efficient.

I tend to use this function:

recoder <- function (x, from = c(), to = c()) {
  missing.levels <- unique(x)
  missing.levels <- missing.levels[!missing.levels %in% from]
  if (length(missing.levels) > 0) {
    from <- append(x = from, values = missing.levels)
    to <- append(x = to, values = missing.levels)
  }
  to[match(x, from)]
}

As in:

recoder(x = dat$product, from = 1:12, to = c(rep("Product1", 3), rep("Product2", 3), rep("Product3", 3), rep("Product4", 3)))

If you have codes in sequential groups like in the example, this may cut the mustard:

cut(dat$product,seq(0,12,by=3),labels=c("Tylenol","Advil","Bayer","Generic"))
 [1] Generic Generic Bayer   Bayer   Advil   Tylenol Generic Advil   Bayer  
[10] Generic Advil   Generic Advil   Tylenol Generic Bayer   Generic Advil  
[19] Bayer   Bayer  
Levels: Tylenol Advil Bayer Generic

There's also arules:discretize, but I like it less because it makes you separate the labels from the range of values:

library(arules)
discretize( dat$product, method = "fixed", categories = c( 1,3,6,9,12 ), labels = c("Tylenol","Advil","Bayer","Generic") )

[1] Generic Generic Generic Generic Bayer   Tylenol Generic Advil   Bayer   Generic Advil   Generic Advil   Advil   Generic Bayer   Generic Advil   Generic Bayer  
Levels: Tylenol Advil Bayer Generic

Another version, that would work in this case:

c("Tylenol","Advil","Bayer","Generic")[(dat$product %/% 3.1) + 1]
imsc

For completeness (and probably fastest and simplest solution) one can create and named vector and use it for lookup. Credit: http://adv-r.had.co.nz/Subsetting.html#applications

product.code <- c(1='Tylenol', 2='Tylenol', 3='Tylenon', 4='Advil', 5 ='Advil', 6='Advil', 7='Bayer', 8='Bayer', 9='Bayer', 10='Generic', 11='Generic', 12='Generic')

To get the output

$unname(product.code[dat$product])

Bench-marking for speed with the top solutions

$microbenchmark(
 named_vector = unname(product.code[dat$product]), 
 find.key = find.key(dat$product, brands),
 levels = `levels<-`(factor(dat$product),brands))
Unit: microseconds
         expr     min       lq      mean   median       uq     max neval
 named_vector  11.777  20.4810  26.12832  23.0410  28.1610 207.360   100
     find.key  34.305  55.8090  58.75804  59.1370  65.5370 130.049   100
       levels 143.361 224.7685 234.02545 247.5525 255.7445 338.944   100

This solution is very similar to @kohske's solution but would work for non-numerical lookup.

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