loops - R: Apply family that deletes columns as part of the function -
i trying iterate through each row in matrix, find column minimum value , column name , delete column after has been used new minimum can calculated. correct answer should this:
result 1/1 50 2/2 61 3/3 72 4/4 83 test_matrix <- matrix(c(50:149), ncol = 10 , byrow=false) names <- c(1:10) colnames(test_matrix) <- names rownames(test_matrix) <- names result <- t(sapply(seq(nrow(test_matrix)), function(i) { j <- which.min(test_matrix[i,]) c(paste(rownames(test_matrix)[i], colnames(test_matrix)[j], sep='/'), test_matrix[i,j]) drops <- colnames(test_matrix)[j] test_matrix[ , !(names(test_matrix) %in% drops)] })) result second question choose order of rows during iteration chooses go next row had same name column name. example, if column minimum named 5, column 5 deleted , minimum row named 5 calculated next.
wondering if possible , if loop needed these calculations.
as new r user, appreciate help. thanks!
for first part of question:
test_matrix <- matrix(c(50:149), ncol = 10 , byrow=false) names <- c(1:10) colnames(test_matrix) <- names rownames(test_matrix) <- names result <- matrix(nrow=0, ncol=2) (i in 1:nrow(test_matrix)) { test_matrix <- as.matrix(test_matrix) #when test_matrix has 1 column r converts vector j <- which.min(test_matrix[i, ]) result <- rbind(result, c(paste(rownames(test_matrix)[i], colnames(test_matrix)[j], sep='/'), as.numeric(test_matrix[i,j]))) test_matrix <- test_matrix[, -j] #remove column j } result ## [,1] [,2] ## [1,] "1/1" "50" ## [2,] "2/2" "61" ## [3,] "3/3" "72" ## [4,] "4/4" "83" ## [5,] "5/5" "94" ## [6,] "6/6" "105" ## [7,] "7/7" "116" ## [8,] "8/8" "127" ## [9,] "9/9" "138" ##[10,] "10/" "149" edit: second part, instead of for loop, can use this:
i <- 1 while(length(test_matrix)>0) { test_matrix <- as.matrix(test_matrix) j <- which.min(test_matrix[i, ]) result <- rbind(result, c(paste(rownames(test_matrix)[i], colnames(test_matrix)[j], sep='/'), as.numeric(test_matrix[i,j]))) test_matrix <- test_matrix[, -j] <- as.numeric(names(j))+1 } 