How to replace column with strings with look-up codes in R

问题

Imagine that I have a dataframe or datatable with strings column where one row looks like this:

a1; b: b1, b2, b3; c: c1, c2, c3; d: d1, d2, d3, d4

and a look-up table with codes for mapping each of these strings. For example:

string code
a1     10
b1     20
b2     30
b3     40
c1     50
c2     60
...

I would like to have a mapping function that maps this string to code:

10; b: 20, 30, 40; c: 50, 60, 70; d: 80, 90, 100

I have a column of these strings in data.table/data.frame (more tha 100k) so any quick solution would be very appreciated. Note that this string length is not always the same... for example in one row i can have strings a to d, in other a to f.

EDIT:

We got the solution for case above, however imagine I have a string like this:

a; b: peter, joe smith, john smith; c: luke, james, john smith

How to replace these knowning that john smith can have two different codes depending on whether it belongs to b or c category? Also, string can contain words with space in between them.

EDIT 2:

   string     code
    a          10
    peter      20
    joe smith  30
    john smith 40
    luke       50
    james      60
    john smith 70
...

The final solution is:

10; b: 20, 30, 40; c: 50, 60, 70

EDIT 3 As suggested, I have opened a new question for next issue: How to replace repeated strings and space in-between with look-up codes in R

回答1:

We can use gsubfn

library(gsubfn)
gsubfn("([a-z]\\d+)", setNames(as.list(df1$code), df1$string), str1)
#[1] "10; b: 20, 30, 40; c: 50, 60, 70; d: 80, 90, 100, 110"

For the edited version

gsubfn("(\\w+ ?\\w+?)",  setNames(as.list(df2$code), df2$string), str2)
#[1] "a; b: 20, 30, 40; c: 50, 60, 40"

data

str1 <- "a1; b: b1, b2, b3; c: c1, c2, c3; d: d1, d2, d3, d4"
df1 <- structure(list(string = c("a1", "b1", "b2", "b3", "c1", "c2", 
 "c3", "d1", "d2", "d3", "d4"), code = c(10L, 20L, 30L, 40L, 50L, 
 60L, 70L, 80L, 90L, 100L, 110L)), class = "data.frame",
  row.names = c(NA, -11L))

str2 <- "a; b: peter, joe smith, john smith; c: luke, james, john smith"

df2 <- structure(list(string = c("a", "peter", "joe smith", "john smith", 
"luke", "james", "john smith"), code = c(10L, 20L, 30L, 40L, 
50L, 60L, 70L)), class = "data.frame", row.names = c(NA, -7L))

回答2:

A much faster alternative would be to use stringr::str_replace_all():

library(stringr)
library(gsubfn)

mystring <- "a1; b: b1, b2, b3; c: c1, c2, c3; d: d1, d2, d3, d4"
mystrings <- rep(mystring, 10000)

str_replace_all(mystrings, setNames(as.character(df$code), df$string))

microbenchmark::microbenchmark(gsubfn = gsubfn("([a-z]\\d+)", setNames(as.list(df$code), df$string), mystrings),
                               stringr = str_replace_all(mystrings, setNames(as.character(df$code), df$string)), check = "equal", times = 50)

Unit: milliseconds
    expr        min         lq      mean     median         uq        max neval cld
  gsubfn 4846.19633 5584.54845 5923.5042 5939.49794 6261.29821 7479.04022    50   b
 stringr   29.01798   29.94274   31.6118   30.80002   31.72871   50.57533    50  a

回答3:

Here are some base R solutions.

Approach 1: use Reduce

res <- Reduce(function(x,k) gsub(df$string[k],df$code[k],x),
              c(s,as.list(1:nrow(df))))

such that

> res
[1] "10; b: 20, 30, 40; c: 50, 60, c3; d: d1, d2, d3, d4"

Approach 2: define custom recursive function f to make it

f <- function(k) ifelse(k==0,s,gsub(df$string[k],df$code[k],f(k-1)))
res <- f(nrow(df))

such that

> res
[1] "10; b: 20, 30, 40; c: 50, 60, c3; d: d1, d2, d3, d4"

DATA

s <- "a1; b: b1, b2, b3; c: c1, c2, c3; d: d1, d2, d3, d4"
df <-structure(list(string = c("a1", "b1", "b2", "b3", "c1", "c2"), 
    code = c(10L, 20L, 30L, 40L, 50L, 60L)), class = "data.frame", row.names = c(NA, 
-6L))

来源：https://stackoverflow.com/questions/60765697/how-to-replace-column-with-strings-with-look-up-codes-in-r

标签

dataframe

data.table

str-replace

gsub