i doing little function in order change levels of variables(factors) data set other levels come data set (both same name of variables).
the main code line solve problem be:
data1$variable <- factor(data1$variable, levels= levels(data2$variable)) a<- factor(c(1,2,3,4)) b<- factor(c(1,2,3)) a<- factor(a, levels=levels(b)) [1] 1 2 3 <na> and adding loop going through columns done with:
list_levels_d2 <- sapply(d2, levels) names_d1 <- colnames(d1) for(i in 1:length(names_d1)){ d1[, names_d1[i]] <- factor(d1[, names_d1[i]], levels= list_levels_d2[[i]]) } basically wanted ask if can figure out how make same avoiding loop "for" using sapply? apply?
i can´t use this:
function(x, y){ sapply(x, factor, levels= levels(y[, names(x)])) } because in part levels(y[, names(x)]), x not taking 1 name takes whole array of names , value null
initial levels of first dataset ('d1')
lapply(d1, levels) #$v1 #[1] "a" "c" "e" #$v2 #[1] "a" "c" "d" "e" #$v3 #[1] "c" "d" "e" we can try map loop on corresponding columns of both datasets ('d1' , 'd2') , change levels of each column corresponding column level of second dataset.
d1[] <- map(function(x,y) {factor(x, levels=levels(y))}, d1, d2[names(d1)]) or option using lapply loop on column names
d1[] <- lapply(names(d1), function(x) {factor(x, levels=levels(d2[,x])) }) after change
lapply(d1, levels) #$v1 #[1] "b" "c" "d" "e" "f" "g" "h" #$v2 #[1] "a" "b" "c" "d" "f" "g" #$v3 #[1] "b" "c" "d" "e" "f" "g" "h" which same levels of second dataset
lapply(d2[names(d1)], levels) #$v1 #[1] "b" "c" "d" "e" "f" "g" "h" #$v2 #[1] "a" "b" "c" "d" "f" "g" #$v3 #[1] "b" "c" "d" "e" "f" "g" "h" data
set.seed(28) d1 <- as.data.frame(matrix(sample(letters[1:5], 5*3, replace=true), ncol=3)) set.seed(34) d2 <- as.data.frame(matrix(sample(letters[1:8], 10*3, replace=true), ncol=3)) set.seed(429) d2 <- d2[sample(ncol(d2))] 
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