r - Replace values in columns

This should work as long as your data are strings and not factors:

for(i in 3:7){
  j = data[,2]==data[,i]
  data[j,i] = data[j,1]
}

Using a combination of lapply and ifelse, you can do:

mydf[,3:7] <- lapply(mydf[,3:7], function(x) ifelse(x==mydf$V2, mydf$V1, x))

which gives:

> mydf
  V1 V2 V3 V4 V5 V6 V7
a  F  B  C  D  F  A  T
b  R  D  C  R  F  A  T
c  A  C  A  R  F  A  T

Or:

newdf <- data.frame(sapply(mydf[,3:7], function(x) ifelse(x==mydf$V2, mydf$V1, x)))

which gives:

> newdf
  V3 V4 V5 V6 V7
1  C  D  F  A  T
2  C  R  F  A  T
3  A  R  F  A  T

Here is another method using lapply:

df[, 3:7] <- lapply(df[,3:7], function(i) {i[i == df$V2] <- df$V1[i == df$V2]; i})

df
  V1 V2 V3 V4 V5 V6 V7
a  F  B  C  D  F  A  T
b  R  D  C  R  F  A  T
c  A  C  A  R  F  A  T

For each variable, matches are substituted using subsetting.

This same method may be used the the replace function:

df[, 3:7] <- lapply(df[,3:7],
                    function(i) replace(i, i == df$V2, df$V1[i == df$V2]))

As with the solution of @mr-rip, these variables must be stored as character and not factor for this to work.

This also works with data.table:

library(data.table)
setDT(df)[, lapply(.SD, function(col) ifelse(col == V2, V1, col))][, V3:V7, with=F]
#    V3 V4 V5 V6 V7
# 1:  C  D  F  A  T
# 2:  C  R  F  A  T
# 3:  A  R  F  A  T