replace some column values from a data.frame based on another data.frame

Question

I have two data.frames, (df1, df2) and I would like to replace the values in columns P1-P10 the letters with the values of df1$V2 but keeping the first two colu

Answer 1

Try some *pply magic:

lookup<-tapply(df1$V2, df1$V1, unique) #Creates a lookup table
lookup.function<-function(x) as.numeric(lookup[as.character(x)]) #The function
df4<-data.frame(df2[,1:2], apply(df2[,3:12], 2,lookup.function )) #Builds the output

Update:

The *pply family is much faster than merge, at least an order of magnitude. Check this out

num<-1000
df1 = data.frame(V1=LETTERS, V2=rnorm(26))
df2<-data.frame(cbind(first=1:num,second=1:num, matrix(sample(LETTERS, num^2, replace=T), nrow=num, ncol=num)))


start<-Sys.time()
lookup<-tapply(df1$V2, df1$V1, unique)
lookup.function<-function(x) as.numeric(lookup[as.character(x)])
df4<-data.frame(cbind(df2[,1:2], data.frame(apply(df2[,3:(num+2)], 2, lookup.function ))))
(difftime(Sys.time(),start))


start<-Sys.time()
df4.merge <- "[<-"(df2, 3:num, value = df1[match(as.character(unlist(df2[3:num])), as.character(df1[[1]])), 2])
(difftime(Sys.time(),start))

sum(df4==df4.merge)==num^2

For 3000 columns and rows the *pply combination needs 4.3s whereas merge needs about 22s on my slow Intel. And it scales nicely. For 4000 columns and rows the respective times are 7.4 sec and 118 sec.