r - N 处平均精度的更快 R 实现

Question

优秀的Metrics包提供了计算平均精度的函数： apk.

问题是，它基于for循环，而且速度很慢：

require('Metrics')
require('rbenchmark')
actual <- 1:20000
predicted <- c(1:20, 200:600, 900:1522, 14000:32955)
benchmark(replications=10,
          apk(5000, actual, predicted),
          columns= c("test", "replications", "elapsed", "relative"))

                          test replications elapsed relative
1 apk(5000, actual, predicted)           10   53.68        1

我不知道如何向量化这个函数，但我想知道是否有更好的方法在 R 中实现它。

Answer 1

我不得不同意实现看起来很糟糕......试试这个：

apk2 <- function (k, actual, predicted)  {

    predicted <- head(predicted, k)

    is.new <- rep(FALSE, length(predicted))
    is.new[match(unique(predicted), predicted)] <- TRUE

    is.relevant <- predicted %in% actual & is.new

    score <- sum(cumsum(is.relevant) * is.relevant / seq_along(predicted)) /
             min(length(actual), k)
    score
}

benchmark(replications=10,
          apk(5000, actual, predicted),
          apk2(5000, actual, predicted),
          columns= c("test", "replications", "elapsed", "relative"))

#                            test replications elapsed relative
# 1  apk(5000, actual, predicted)           10  62.194 2961.619
# 2 apk2(5000, actual, predicted)           10   0.021    1.000

identical(apk(5000, actual, predicted),
          apk2(5000, actual, predicted))
# [1] TRUE

Answer 2

I happen to have average precision code written using for loop. I think it is fast enough.

ap <- function(prediction) {
    #prediction is a two column matrix. The first one is the true label and the second one is the prediction value
    result = 0
    ranklist <- prediction[sort(prediction[,2],decreasing=TRUE, index.return=TRUE)$ix,]
    numpos <- length(which(ranklist[,1]==1))
    deltaRecall <- 1/numpos
    pcount <- 0

    for(i in 1:nrow(ranklist)) {
        if(ranklist[i,1] == 1) {
            pcount <- pcount + 1
            precision <- pcount/i
            result <- result + precision*deltaRecall
        }
    }
    return(result)
}

ap_at_N <- function(prediction, N=20) {
    #average precision at N
    result = 0
    ranklist <- prediction[sort(prediction[,2],decreasing=TRUE, index.return=TRUE)$ix,]
    numpos <- length(which(ranklist[,1]==1))
    numpos <- min(N, numpos)
    deltaRecall <- 1/numpos
    pcount <- 0

    for(i in 1:(min(nrow(ranklist),N))) {
        if(ranklist[i,1] == 1) {
            pcount <- pcount + 1
            precision <- pcount/i
            result <- result + precision*deltaRecall
        }
    }
    return(result)
}