Thanks for the complete reproducible example. Here's one way to approach the problem; there are many others.
recodeDat <- function(x, invals, outvals) { x <- as.character(x) invals <- as.character(invals) outvals <- as.character(outvals) # a loop is the most understandable approach for(i in seq_along(invals)) { x[x == invals[i]] <- outvals[i] } # I would change the zeros to NA values x[x == "0"] <- NA factor(x, levels=sort(unique(outvals))) } dat1.recode <- dat1 dat1.recode[, 3:ncol(dat1.recode)] <- apply(dat1.recode[, 3:ncol(dat1.recode)], 2, recodeDat, invals=dat2[,1], outvals=dat2[,2]) On Tue, Mar 8, 2016 at 4:26 PM, Marna Wagley <marna.wag...@gmail.com> wrote: > Hi R Users, > I have been struggling to replace texts in a table by new text. but it > seems crazy as of I am doing it manually in R. I have very big files and > some of the text has to be replaced by another class based on another file > if the name corresponds. I was able to perform following example but it > should be easier if there is a loop. Any suggestions on making a loop for > this example? > > Here is the example how I did. I want to assign class into dat1 based on > dat2 table. > > dat1<-structure(list(ID = structure(1:8, .Label = c("X127", "X128", > "X129", "X130", "X131", "X132", "X133", "X134"), class = "factor"), > Name = structure(1:8, .Label = c("Site1", "Site2", "Site3", > "Site4", "Site5", "Site6", "Site7", "Site8"), class = "factor"), > Time1 = structure(c(4L, 2L, 3L, 5L, 1L, 1L, 4L, 5L), .Label = c("0", > "GT", "R", "Tr", "W2"), class = "factor"), Time2 = structure(c(2L, > 1L, 4L, 2L, 1L, 3L, 4L, 1L), .Label = c("0", "GT", "MA", > "UA"), class = "factor"), Time3 = structure(c(5L, 1L, 4L, > 4L, 2L, 3L, 3L, 1L), .Label = c("0", "GT", "R", "Tr", "Y7" > ), class = "factor")), .Names = c("ID", "Name", "Time1", > "Time2", "Time3"), class = "data.frame", row.names = c(NA, -8L > )) > > dat1 > > dat2<-structure(list(site = structure(c(4L, 2L, 5L, 3L, 6L, 7L, 8L, > 1L), .Label = c("GT", "MA", "R", "Tr", "UA", "W1", "W2", "Y7" > ), class = "factor"), To.be.assinged = structure(c(1L, 2L, 3L, > 1L, 4L, 4L, 1L, 2L), .Label = c("A", "B", "C", "D"), class = "factor")), > .Names = c("site", > "To.be.assinged"), class = "data.frame", row.names = c(NA, -8L > )) > dat2 > > A2 <- as.data.frame(lapply(dat1,function(x) > if(is.character(x)|is.factor(x)) gsub("Tr","A",x) else x)) > A3 <- as.data.frame(lapply(A2,function(x) if(is.character(x)|is.factor(x)) > gsub("MA","B",x) else x)) > A4 <- as.data.frame(lapply(A3,function(x) if(is.character(x)|is.factor(x)) > gsub("UA","C",x) else x)) > A5 <- as.data.frame(lapply(A4,function(x) if(is.character(x)|is.factor(x)) > gsub("R","A",x) else x)) > A6 <- as.data.frame(lapply(A5,function(x) if(is.character(x)|is.factor(x)) > gsub("W1","D",x) else x)) > A7 <- as.data.frame(lapply(A6,function(x) if(is.character(x)|is.factor(x)) > gsub("W2","D",x) else x)) > A8 <- as.data.frame(lapply(A7,function(x) if(is.character(x)|is.factor(x)) > gsub("Y7","A",x) else x)) > A9 <- as.data.frame(lapply(A8,function(x) if(is.character(x)|is.factor(x)) > gsub("GT","B",x) else x)) > A9 > > Your help is highly appreciated. > Thanks > -- Sarah Goslee http://www.functionaldiversity.org ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.