如何在矩阵的行上复制相同的函数

Question

I am trying to write a loop that determines which cell has the greatest value and select that cell as a result with a high medium or low string. Here is the data for try out.

data <- matrix(c(0.3000003,0.3299896,0.3700101,
                 0.3299896,0.3700101,0.3000003,
                 0.3700101,0.3000003,0.3299896,
                 0.3000003,0.3299896,0.3700101,
                 0.3299896,0.3700101,0.3000003,
                 0.3700101,0.3000003,0.3299896),6,3)
colnames(data) <- c("Low","Medium","High")
rownames(data) <- paste("case",1:6)

> data
             Low    Medium      High
case 1 0.3000003 0.3700101 0.3299896
case 2 0.3299896 0.3000003 0.3700101
case 3 0.3700101 0.3299896 0.3000003
case 4 0.3299896 0.3000003 0.3700101
case 5 0.3700101 0.3299896 0.3000003
case 6 0.3000003 0.3700101 0.3299896

I am using this function but it seems like it is only calculating the first row.

assign.levels <- function(data) {

  for (i in nrow(data)) {

    scored.thetas.1 <- names(which.max(data[i,1:3])) ## I wrote 1:3 here because I have multiple columns in the original dataset.
    return(scored.thetas.1)

  }
}


> assign.levels(data)
[1] "Medium"

Any thoughts?

Thanks in advance!

Answer 1

Here's a vectorized solution that you may prefer:

colnames(data)[apply(data, 1, which.max)]
# [1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

That's a concise version of your attempt: apply the function which.max to each row (dimension 1 ) of data and get a corresponding column name.

In terms of your attempt, here's a corrected version:

assign.levels <- function(data) {
  scored.thetas.1 <- rep(NA, nrow(data))
  for (i in 1:nrow(data))
    scored.thetas.1[i] <- names(which.max(data[i, ]))
  scored.thetas.1
}
assign.levels(data)
# [1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

Several things to mention about your attempt: 1) you were iterating with i in nrow(data) , while nrow(data) is just a number. So basically you were looking only at the last row; 2) you kept redefining the same variable scored.thetas.1 in every iteration (in this case there was only one iteration, but the tendency was bad); 3) a loop is not a function, you don't need to return anything from it and instead you most likely want to store somewhere your newly obtained values.

In comparison, note that first I define an empty vector scored.thetas.1 of length nrow(data) . Then I iterate over all the rows ( 1:nrow(data) ) and store a value for each row/iteration to scored.thetas.1[i] .

Answer 2

This should be fast

colnames(data)[max.col(data)]
#[1] "Medium" "High"   "Low"    "High"   "Low"    "Medium"

---

Here is a little benchmark.

n <- 1e6
set.seed(1)
data <- matrix(runif(n * 3), ncol = 3)
colnames(data) <- c("Low","Medium","High")

library(microbenchmark)

benchmark <- microbenchmark(
  OP = assign.levels(data), # as defined in Julius's answer
  Julius = colnames(data)[apply(data, 1, which.max)],
  markus = colnames(data)[max.col(data)], times = 20
)

autoplot(benchmark)