R: gsub, pattern = vector and replacement = vector

Solution 1:

Lot's of solutions already, here are one more:

The qdap package:

library(qdap)
names(x1) <- mgsub(a,b,names(x1))

Solution 2:

From stringr documentation of str_replace_all, "If you want to apply multiple patterns and replacements to the same string, pass a named version to pattern."

Thus using a, b, and names(x1) from above

stringr::str_replace_all(names(x1), setNames(b, a))

EDIT

stringr::str_replace_all calls stringi::stri_replace_all_regex, which can be used directly and is quite a bit quicker.

x <- names(x1)
pattern <- a
replace <- b

microbenchmark::microbenchmark(
  str  = stringr::str_replace_all(x, setNames(replace, pattern)),
  stri = stringi::stri_replace_all_regex(x, pattern, replace, vectorize_all = FALSE)
  )

Unit: microseconds
 expr    min      lq     mean  median   uq    max neval cld
  str 1022.1 1070.45 1286.547 1175.55 1309 2526.8   100   b
 stri  145.2  150.45  190.124  160.55  178  457.9   100  a

Solution 3:

New Answer

If we can make another assumption, the following should work. The assumption this time is that you are really interested in substituting the first 10 characters from each value in names(x1).

Here, I've stored names(x1) as a character vector named "X1". The solution essentially uses substr to separate the values in X1 into 2 parts, match to figure out the correct replacement option, and paste to put everything back together.

a <- c("2110027599", "2110025622", "2110028045",
       "2110034716", "2110069349", "2110023264")
b <- c("Inv1","Inv2","Inv3","Inv4","Inv5","Inv6")

X1pre <- substr(X1, 1, 10)
X1post <- substr(X1, 11, max(nchar(X1)))

paste0(b[match(X1pre, a)], X1post)
#  [1] "Inv6A.Ms.Amp"        "Inv6A.Ms.Vol"        "Inv6A.Ms.Watt"      
#  [4] "Inv6A1.Ms.Amp"       "Inv6A2.Ms.Amp"       "Inv6A3.Ms.Amp"      
#  [7] "Inv6A4.Ms.Amp"       "Inv6A5.Ms.Amp"       "Inv6B.Ms.Amp"       
# [10] "Inv6B.Ms.Vol"        "Inv6B.Ms.Watt"       "Inv6B1.Ms.Amp"      
# [13] "Inv6Error"           "Inv6E-Total"         "Inv6GridMs.Hz"      
# [16] "Inv6GridMs.PhV.phsA" "Inv6GridMs.PhV.phsB" "Inv6GridMs.PhV.phsC"
# [19] "Inv6GridMs.TotPFPrc" "Inv6Inv.TmpLimStt"   "Inv6InvCtl.Stt"     
# [22] "Inv6Mode"            "Inv6Mt.TotOpTmh"     "Inv6Mt.TotTmh"      
# [25] "Inv6Op.EvtCntUsr"    "Inv6Op.EvtNo"        "Inv6Op.GriSwStt"    
# [28] "Inv6Op.TmsRmg"       "Inv6Pac"             "Inv6PlntCtl.Stt"    
# [31] "Inv6Serial Number"   "Inv2A.Ms.Amp"        "Inv2A.Ms.Vol"       
# [34] "Inv2A.Ms.Watt"       "Inv2A1.Ms.Amp"       "Inv2A2.Ms.Amp"      
# [37] "Inv2A3.Ms.Amp"       "Inv2A4.Ms.Amp"       "Inv2A5.Ms.Amp"      
# [40] "Inv2B.Ms.Amp"        "Inv2B.Ms.Vol"        "Inv2B.Ms.Watt"      
# [43] "Inv2B1.Ms.Amp"       "Inv2Error"           "Inv2E-Total"        
# [46] "Inv2GridMs.Hz"       "Inv2GridMs.PhV.phsA" "Inv2GridMs.PhV.phsB"

Old Answer

If we can assume that names(x1) is in the same order as the pattern and replacement and that it is basically a one-for-one replacement, you might be able to get away with just sapply.

Here's an example of that particular situation:

Imagine "names(x)" looks something like this:

X1 <- paste0("A2", a, sequence(length(a)))
X1
# [1] "A221100275991" "A221100256222" "A221100280453" 
# [4] "A221100347164" "A221100693495" "A221100232646"

Here's our pattern and replacement vectors:

a <- c("2110027599", "2110025622", "2110028045", 
       "2110034716", "2110069349", "2110023264")
b <- c("Inv1","Inv2","Inv3","Inv4","Inv5","Inv6")

This is how we might use sapply if these assumptions are valid.

sapply(seq_along(a), function(x) gsub(a[x], b[x], X1[x]))
# [1] "A2Inv11" "A2Inv22" "A2Inv33" "A2Inv44" "A2Inv55" "A2Inv66"

Solution 4:

Try mapply.

names(x1) <- mapply(gsub, a, b, names(x1), USE.NAMES = FALSE)

Or, even easier, str_replace from stringr.

library(stringr)
names(x1) <- str_replace(names(x1), a, b)