r - Getting column names of each row in order -

i have dataframe 3 columns of numeric values. want sort each row , print string column names in sorted order each row. here code:

> df <- data.frame(x = c(1,2,3), y = c(3,1,2), d = c(4,0, 5)) > df   x y d 1 1 3 4 2 2 1 0 3 3 2 5 > for(r in 1:nrow(df)) + print(paste(colnames(sort(df[r,])), collapse = " ")) [1] "x y d" [1] "d y x" [1] "y x d" 

this work, takes long time when have large dataframe. there more effective way perform sort? tried use apply, gave blank strings:

> apply(df, 1, function(row) paste( colnames( sort( row, decreasing = t )) , collapse = " " )) [1] "" "" "" 

when call function within apply on single row, works:

> paste( colnames( sort( df[1,], decreasing = t )) , sep = " " ) [1] "d" "y" "x" 

i looking faster way perform operation on each row of large dataframe loop. , think apply faster, cannot work.

t(apply(df, 1, function(x) names(x)[order(x)])) 

this question popped again in reading, thought shall edit add more ways of doing it.. might later :

library(data.table) setdt(df)[, paste(colnames(df)[order(.sd)], collapse = " "), = 1:nrow(df)] 

logic : groupby row index( means row-wise operation) - apply rank each group(which row). .sd means subset of data( columns)( can control using .sdcols = argument. simple paste column names accordingly

same logic above implemented in dplyr

library(dplyr) library(tidyr) df %>% rowwise() %>%         do(rank = paste(colnames(df)[order(unlist(.))], collapse = " ")) %>%         unnest() 

output :

#   nrow    v1 #1:    1 x y d #2:    2 d y x #3:    3 y x d