问lapply vs for loop - Performance R

EN

# prepare input
data(iris)
irissubdf <- iris[1:100, c(1, 3, 5)]
names(irissubdf) <- c("sepal", "petal", "species")
head(irissubdf)
irissubdf$y <- 1
irissubdf[irissubdf[, 3] == "setosa", 4] <- -1
x <- irissubdf[, c(1, 2)]
y <- irissubdf[, 4]

# perceptron function with for
perceptron <- function(x, y, eta, niter) {

  # initialize weight vector
  weight <- rep(0, dim(x)[2] + 1)
  errors <- rep(0, niter)


  # loop over number of epochs niter
  for (jj in 1:niter) {

    # loop through training data set
    for (ii in 1:length(y)) {

      # Predict binary label using Heaviside activation
      # function
      z <- sum(weight[2:length(weight)] * as.numeric(x[ii, 
        ])) + weight[1]
      if (z < 0) {
        ypred <- -1
      } else {
        ypred <- 1
      }

      # Change weight - the formula doesn't do anything
      # if the predicted value is correct
      weightdiff <- eta * (y[ii] - ypred) * c(1, 
        as.numeric(x[ii, ]))
      weight <- weight + weightdiff

      # Update error function
      if ((y[ii] - ypred) != 0) {
        errors[jj] <- errors[jj] + 1
      }

    }
  }

  # weight to decide between the two species

  return(errors)
}

err <- perceptron(x, y, 1, 10)

### my rewriting in functional form auxiliary
### function
faux <- function(x, weight, y, eta) {
  err <- 0
  z <- sum(weight[2:length(weight)] * as.numeric(x)) + 
    weight[1]
  if (z < 0) {
    ypred <- -1
  } else {
    ypred <- 1
  }

  # Change weight - the formula doesn't do anything
  # if the predicted value is correct
  weightdiff <- eta * (y - ypred) * c(1, as.numeric(x))
  weight <<- weight + weightdiff

  # Update error function
  if ((y - ypred) != 0) {
    err <- 1
  }
  err
}

weight <- rep(0, 3)
weightdiff <- rep(0, 3)

f <- function() {
  t <- replicate(10, sum(unlist(lapply(seq_along(irissubdf$y), 
    function(i) {
      faux(irissubdf[i, 1:2], weight, irissubdf$y[i], 
        1)
    }))))
  weight <<- rep(0, 3)
  t
}

                                                      expr       min         lq       mean     median         uq
                                                        f() 48670.878 50600.7200 52767.6871 51746.2530 53541.2440
  perceptron(as.matrix(irissubdf[1:2]), irissubdf$y, 1, 10)  4184.131  4437.2990  4686.7506  4532.6655  4751.4795
 perceptronC(as.matrix(irissubdf[1:2]), irissubdf$y, 1, 10)    95.793   104.2045   123.7735   116.6065   140.5545
        max neval
 109715.673   100
   6513.684   100
    264.858   100