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我正在 R 中编写朴素贝叶斯的自定义修改版本，并且由于正在处理的数据的大小而遇到运行时问题。我需要处理约 145k 行，每行有 95 个元素。我目前正在使用以下函数来获得朴素贝叶斯的第一步。

probGen <- function(x, i)
{
  return(1/(sqrt(2*pi*sdBreakdown[i,]^2)
            *exp(-((x - meanBreakdown[i,])^2)/(2*(sdBreakdown[i,]^2)))))
}

在此函数中，sdBreakdown 和 meanBreakdown 是每个可能解决方案的聚合值。每次运行应用时，我们都会获得每个给定列的概率。应用程序在矩阵上按如下方式运行，其中每一行是我们试图分类的另一个元素。

test.1 <- t(apply(temp,MARGIN=1,FUN=probGen, 1))
test.2 <- t(apply(temp,MARGIN=1,FUN=probGen, 2))
test.3 <- t(apply(temp,MARGIN=1,FUN=probGen, 3))
test.4 <- t(apply(temp,MARGIN=1,FUN=probGen, 4))
test.5 <- t(apply(temp,MARGIN=1,FUN=probGen, 5))
test.6 <- t(apply(temp,MARGIN=1,FUN=probGen, 6))
test.7 <- t(apply(temp,MARGIN=1,FUN=probGen, 7))
test.8 <- t(apply(temp,MARGIN=1,FUN=probGen, 8))
test.9 <- t(apply(temp,MARGIN=1,FUN=probGen, 9))

这是我目前如何调用每个应用程序。这为每个可能的分类 1-9 给出了每个元素的概率。我不想使用开箱即用的朴素贝叶斯，因为我试图更好地理解 R 并且想要尝试一些潜在的准确性改进。

我不确定如何以更及时的方式运行它，尽管按照编码它需要几个小时，如果我在运行时积极从事其他项目，可能需要 7 或 8 个小时。

编辑：

为了澄清这个例子中的数据。

temp 是 145kx95 矩阵，其中每一行是要分类的项目，每一列是用数字表示的质量。

meanBreakdown 是一个 9x95 矩阵，每一行是一个不同的分类，每一列对应于该分类的平均平均质量。

sdBreakdown 与 meanBreakdown 相同，只是存储标准偏差而不是平均平均值。

并行处理似乎可以工作，但我不认为（显然我错了）数据集足够大以至于有必要。

编辑2：这是完整的代码。如果它是非常糟糕的 R 代码，请原谅我。我一直是 C 开发人员，所以 R 在思想上发生了很大变化，我只在 R 中完成了少数几个小项目来了解其中的来龙去脉。

training <- read.csv(file = 'data\\train.csv', sep=',', header=T)

negativeOne <- function(x)
{
  x <- pmin(1, x)
  return(1-mean(x))
}

pullZeros <- function(x)
{
  x <- ifelse(x == 0, 1, 0)
  return(mean(x))
}

trainingSet <- function(x)
{
  x <- ifelse(x == 0, NA, x)
  return(mean(x, na.rm=T))
}
trainingSetSd <- function(x)
{
  x <- ifelse(x == 0, NA, x)
  return(sd(x, na.rm=T))
}

positiveBreakDown <- aggregate(x=training[,colnames(training)[grepl("feat",colnames(training))]],
                         by=list(training$target), FUN=trainingSet)

positiveBreakDownSd <- aggregate(x=training[,colnames(training)[grepl("feat",colnames(training))]],
                               by=list(training$target), FUN=trainingSetSd)

negativeBreakDown <- aggregate(x=training[,colnames(training)[grepl("feat",colnames(training))]],
                     by=list(training$target), FUN=negativeOne)

meanBreakdown <- positiveBreakDown[,colnames(positiveBreakDown)[grepl("feat",colnames(positiveBreakDown))]]

sdBreakdown <- positiveBreakDownSd[,colnames(positiveBreakDownSd)[grepl("feat",colnames(positiveBreakDownSd))]]

probGen <- function(x, i)
{
  return(1/(sqrt(2*pi*sdBreakdown[i,]^2)
            *exp(-((x - meanBreakdown[i,])^2)/(2*(sdBreakdown[i,]^2)))))
}

test <-  read.csv(file = 'data\\test.csv', sep=',', header=T)

PosTest <- test[,colnames(test)[grepl("feat",colnames(test))]]


NegTest <- aggregate(x=test[,colnames(test)[grepl("feat",colnames(test))]],
                  by=list(test$id), FUN=pullZeros)

NegTest$Group.1 <- NULL
temp <- PosTest

sweepTest.1 <- t(apply(temp,MARGIN=1,FUN=probGen, 1))
sweepTest.2 <- t(apply(temp,MARGIN=1,FUN=probGen, 2))
sweepTest.3 <- t(apply(temp,MARGIN=1,FUN=probGen, 3))
sweepTest.4 <- t(apply(temp,MARGIN=1,FUN=probGen, 4))
sweepTest.5 <- t(apply(temp,MARGIN=1,FUN=probGen, 5))
sweepTest.6 <- t(apply(temp,MARGIN=1,FUN=probGen, 6))
sweepTest.7 <- t(apply(temp,MARGIN=1,FUN=probGen, 7))
sweepTest.8 <- t(apply(temp,MARGIN=1,FUN=probGen, 8))
sweepTest.9 <- t(apply(temp,MARGIN=1,FUN=probGen, 9))

temp <- NegTest
temp$Group.1 <- NULL

N.sweepTest.1 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[1, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.2 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[2, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.3 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[3, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.4 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[4, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.5 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[5, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.6 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[6, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.7 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[7, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.8 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[8, grepl("feat",colnames(positiveBreakDown))]),`*`)
N.sweepTest.9 <- sweep(as.matrix(temp),MARGIN=2,
                       as.numeric(negativeBreakDown[9, grepl("feat",colnames(positiveBreakDown))]),`*`)


sweepTest.1 <- (-1*(N.sweepTest.1 - 1)*sweepTest.1) + N.sweepTest.1
sweepTest.2 <- (-1*(N.sweepTest.2 - 1)*sweepTest.2) + N.sweepTest.2
sweepTest.3 <- (-1*(N.sweepTest.3 - 1)*sweepTest.3) + N.sweepTest.3
sweepTest.4 <- (-1*(N.sweepTest.4 - 1)*sweepTest.4) + N.sweepTest.4
sweepTest.5 <- (-1*(N.sweepTest.5 - 1)*sweepTest.5) + N.sweepTest.5
sweepTest.6 <- (-1*(N.sweepTest.6 - 1)*sweepTest.6) + N.sweepTest.6
sweepTest.7 <- (-1*(N.sweepTest.7 - 1)*sweepTest.7) + N.sweepTest.7
sweepTest.8 <- (-1*(N.sweepTest.8 - 1)*sweepTest.8) + N.sweepTest.8
sweepTest.9 <- (-1*(N.sweepTest.9 - 1)*sweepTest.9) + N.sweepTest.9

rm(N.sweepTest.1,N.sweepTest.2,N.sweepTest.3,N.sweepTest.4,N.sweepTest.5,N.sweepTest.6,N.sweepTest.7,N.sweepTest.8,N.sweepTest.9)

dist <- 1:9

for(i in 1:9)
{
  dist[i] <- nrow(training[training$target == paste0("Class_",i),])
}

res1 <- dist[1]*apply(t(sweepTest.1), MARGIN=2, FUN=prod)
res2 <- dist[2]*apply(t(sweepTest.2), MARGIN=2, FUN=prod)
res3 <- dist[3]*apply(t(sweepTest.3), MARGIN=2, FUN=prod)
res4 <- dist[4]*apply(t(sweepTest.4), MARGIN=2, FUN=prod)
res5 <- dist[5]*apply(t(sweepTest.5), MARGIN=2, FUN=prod)
res6 <- dist[6]*apply(t(sweepTest.6), MARGIN=2, FUN=prod)
res7 <- dist[7]*apply(t(sweepTest.7), MARGIN=2, FUN=prod)
res8 <- dist[8]*apply(t(sweepTest.8), MARGIN=2, FUN=prod)
res9 <- dist[9]*apply(t(sweepTest.9), MARGIN=2, FUN=prod)

rm(sweepTest.1,sweepTest.2,sweepTest.3,sweepTest.4,sweepTest.5,sweepTest.6,sweepTest.7,sweepTest.8,sweepTest.9)

interRes <- data.frame(Class_1 = res1, Class_2 = res2,Class_3 = res3,
                       Class_4 = res4,Class_5 = res5,Class_6 = res6,
                       Class_7 = res7,Class_8 = res8,Class_9 = res9)


rm(res1,res2,res3,res4,res5,res6,res7,res8,res9)

temp <- apply(t(interRes), MARGIN=2, FUN=sum)

tempRes <- interRes/temp

data<- data.frame(id=test$id)

data <- cbind(data,tempRes)

fname <- file.choose()
write.table(data, fname, row.names=FALSE, sep=",")
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2 回答 2

2

您需要正确矢量化您的代码。有了这么简单的函数，就不需要使用applywhich 基本上只是一个for循环。

首先我们生成一些虚假数据：

rm(list = ls())
set.seed(1)
# Dimensions of data and some faux data
n <- 144000
m <- 95
temp <- matrix(rnorm(n*m), nrow = n, ncol = m)

meanBreakdown <- matrix(seq(-1, 1, l = 9*m), 9, m)  # Matrix of means
sdBreakdown <- matrix(seq(1, 2, l = 9*m), 9, m)  # Matrix of std. deviations

让我们为您的版本计时一个i = 1. 我冒昧地使它更具可读性。另外，我想我发现了一个错误（如果函数只是高斯密度）。反正，

probGen <- function(x, means, sds) { # NOTE THAT THIS HAS CHANGED
   return(1/sqrt(2*pi*sds^2)*exp(-(1/(2*sds^2))*(x - means)^2) )
}

i <- 1
t1 <- system.time({
  res1 <- t(apply(temp, 1, probGen, mean = meanBreakdown[i,], 
                                    sds = sdBreakdown[i,]))
})
print(res1[1:5, 1:7])
#          [,1]       [,2]        [,3]       [,4]       [,5]      [,6]           [,7]
#[1,] 0.3720575 0.38038806 0.385805475 0.36747185 0.32253028 0.3008070 0.37473829
#[2,] 0.1980087 0.02837476 0.019424716 0.03520653 0.25872889 0.2223151 0.05506068
#[3,] 0.3935892 0.24920567 0.116377580 0.13580043 0.07012818 0.1682480 0.35898510
#[4,] 0.0137505 0.37288236 0.002338961 0.21928922 0.36341271 0.0250388 0.05103852
#[5,] 0.1648476 0.32981193 0.031723978 0.12681473 0.25509082 0.1959218 0.35277957
print(t1)
#   user  system elapsed 
#  3.452   0.205   3.662

这是一个替代版本，我们利用矩阵与 R 的复制规则一起以列为主的方式存储：

probGen2 <- function(x, means, sds) {    
  return(t(1/sqrt(2*pi*sds^2)*exp(-(1/(2*sds^2))*(t(x) - means)^2)))
}

i <- 1
t2 <- system.time({
  res2 <- probGen2(x = temp, means = meanBreakdown[i, ],
                             sds = sdBreakdown[i, ])
})
print(res2[1:5, 1:7])
#          [,1]       [,2]        [,3]       [,4]       [,5]      [,6]       [,7]
#[1,] 0.3720575 0.38038806 0.385805475 0.36747185 0.32253028 0.3008070 0.37473829
#[2,] 0.1980087 0.02837476 0.019424716 0.03520653 0.25872889 0.2223151 0.05506068
#[3,] 0.3935892 0.24920567 0.116377580 0.13580043 0.07012818 0.1682480 0.35898510
#[4,] 0.0137505 0.37288236 0.002338961 0.21928922 0.36341271 0.0250388 0.05103852
#[5,] 0.1648476 0.32981193 0.031723978 0.12681473 0.25509082 0.1959218 0.35277957
print(t2)
#   user  system elapsed 
#  0.499   0.014   0.515

如您所见，我们已经对一些非常简单的更改进行了相当大的加速。您显然可以将其与并行计算结合起来，以获得进一步的速度提升。

最后，让我们检查一下是否一切都一样：

 all.equal(res1, res2)
 # [1] TRUE
于 2015-03-18T18:13:12.937 回答
0

检查parallel包和mcmapply或并行mclapply运行apply调用。如所写，您的代码是按顺序运行的（即，您必须在转到 2 之前完成所有 1 分类，依此类推）。

据我了解，您在相同的数据上运行相同的功能，但参数不同。与其进行多次apply调用，不如重新构造函数以允许使用mcmapply- 它允许您使用该apply功能，但迭代多个参数。

于 2015-03-18T17:02:46.907 回答