CODEWITHSUNDEEP
r
K.I.N
K.I.N

Reputation: 129

How to replace a specific word in dataframe

category.1 <- c("TM","TM","CPA","TM","CPC")
category.2 <- c("LS","LS","DSP","DSP","AF")
platform <- c("facebook","facebook","yahoo","google","google")

dat <- data.frame(platform,category.1,category.2)
dat
  platform category.1 category.2
1 facebook         TM         LS
2 facebook         TM         LS
3    yahoo        CPA        DSP
4   google         TM        DSP
5   google        CPC         AF

when category.1 is 'TM' and category.2 'LS', I wanna replace 'LS' to 'LS1'

      platform category.1 category.2
    1 facebook         TM         LS1
    2 facebook         TM         LS1
    3    yahoo        CPA        DSP
    4   google         TM        DSP
    5   google        CPC         AF

I tried this way, its return error.

 dat$category.1[dat$category.1=='TM'& dat$category.2=='LS',] <- 'LS1'

thanks for your reading.

Upvotes: 2

Views: 109

Answers (4)

David Arenburg
David Arenburg

Reputation: 92282

If you want a really efficient way of doing replacements by condition, check out the data.table package and its binary search/replacement by reference

library(data.table)
setkey(setDT(dat), category.1, category.2)
dat[J("TM", "LS"), category.2 := "LS1"][]
#    platform category.1 category.2
# 1:    yahoo        CPA        DSP
# 2:   google        CPC         AF
# 3:   google         TM        DSP
# 4: facebook         TM        LS1
# 5: facebook         TM        LS1

setDT converts to data.table object. setkey keys the data in order to perform a binary join. J() performs the actual binary join. := performs assigment by reference and updates category.1 in place.

Though if your data set isn't big, you could just do 

dat[category.1 == "TM" & category.2 == "LS", category.2 := "LS1"][]

Some benchmarks on a slightly bigger data set (I didn't test base because you need to convert to character class in order for this to work)

library(data.table)
library(dplyr)
library(microbenchmark)

dat2 <- data.frame(lapply(dat, rep, 1e5))
dat3 <- copy(dat2)
dat4 <- copy(dat2)

dplyrfunc <- function(x) {
  x <- x %>%
    mutate(category.2 = 
          ifelse(category.1 == "TM" & category.2 == "LS",
          "LS1", as.character(category.2)))
  x
}

data.tablefunc1 <- function(x){
  setkey(setDT(x), category.1, category.2)
  x[J("TM", "LS"), category.2 := "LS1"][]
}

data.tablefunc2 <- function(x){
  setDT(x)[category.1 == "TM" & category.2 == "LS", category.2 := "LS1"][]
}

## Unit: milliseconds
##                  expr        min         lq      mean     median         uq       max neval
##       dplyrfunc(dat2) 277.261833 291.647719 313.76279 302.337902 335.703250 401.38212   100
## data.tablefunc1(dat3)   5.371047   5.905744   8.12169   6.904871   8.266383  59.83116   100
## data.tablefunc2(dat4)  31.980348  32.870719  38.26239  34.745612  39.309186  88.91202   100

Upvotes: 3

Mamoun Benghezal
Mamoun Benghezal

Reputation: 5314

You can set stringsAsFactors = FALSE when creating your data set

dat <- data.frame(platform,category.1,category.2, stringsAsFactors = FALSE)

Then you can use your code, just remove the comma like this

dat$category.2[dat$category.1=='TM'& dat$category.2=='LS'] <- "LS1"

Upvotes: 4

Bart
Bart

Reputation: 759

Another approach; using dplyr and the ifelse base function.

> library(dplyr)
> dat <-
    dat %>%
    mutate(category.2 = ifelse(category.1 == "TM" & category.2 == "LS",
           "LS1",
           as.character(category.2)))


> dat
  platform category.1 category.2
1 facebook         TM        LS1
2 facebook         TM        LS1
3    yahoo        CPA        DSP
4   google         TM        DSP
5   google        CPC         AF

Upvotes: 6

lawyeR
lawyeR

Reputation: 7654

You can use revalue from the plyr package:

library(plyr)
dat$category.2 <- revalue(dat$category.2, c("LS" = "LS1"))
dat

  platform category.1 category.2
1 facebook         TM        LS1
2 facebook         TM        LS1
3    yahoo        CPA        DSP
4   google         TM        DSP
5   google        CPC         AF

Upvotes: 1

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