R - 用 apply 系列中的函数替换双循环 [英] R - Replace a double loop by a function from the apply family

问题描述

我有这些循环:

xall = data.frame()
for (k in 1:nrow(VectClasses))
{
for (i in 1:nrow(VectIndVar))
  {
   xall[i,k] = sum(VectClasses[k,] == VectIndVar[i,])
  }
}

数据:

VectClasses = 包含每个类的特征的数据框

VectClasses = Data Frame containing the characteristics of each classes

VectIndVar = 包含数据库每条记录的数据帧

VectIndVar = Data Frame containing each record of the data base

这两个 for 循环工作并提供我可以使用的输出，但是，它花费的时间太长，因此我需要应用系列

The two for loops work and give an output I can work with, however, it takes too long, hence my need for the apply family

我正在寻找的输出是这样的:

The output I am looking for is as this:

    V1 V2 V3 V4
 1  3  3  2  2
 2  2  2  1  1
 3  3  4  3  3
 4  3  4  3  3
 5  4  4  3  3
 6  3  2  3  3

我尝试使用:

xball = data.frame()
xball = sapply(xball, function (i,k){
 sum(VectClasses[k,] == VectIndVar[i,])})

xcall = data.frame()
xcall = lapply(xcall, function (i, k){sum(VectClasses[k,] == VectIndVar[i,]} )

但似乎都没有填充数据框

but neither seems to be filling the dataframe

可重现的数据(缩短):

reproductible data (shortened):

VectIndVar <- data.frame(a=sample(letters[1:5], 100, rep=T), b=floor(runif(100)*25), 
 c = sample(c(1:5), 100, rep=T), 
 d=sample(c(1:2), 100, rep=T))

和:

> K1 = 4
VectClasses= VectIndVar [sample(1:nrow(VectIndVar ), K1, replace=FALSE), ]

你能帮我吗?

推荐答案

我会使用 outer 而不是 *apply:

I would use outer instead of *apply:

res <- outer( 
  1:nrow(VectIndVar), 
  1:nrow(VectClasses),
  Vectorize(function(i,k) sum(VectIndVar[i,-1]==VectClasses[k,-1]))
)

(感谢此问答阐明需要Vectorize.)

这给了

> head(res) # with set.seed(1) before creating the data
     [,1] [,2] [,3] [,4]
[1,]    1    1    2    1
[2,]    0    0    1    0
[3,]    0    0    0    0
[4,]    0    0    1    0
[5,]    1    0    0    1
[6,]    1    1    1    1

<小时>

至于速度，我建议使用矩阵而不是 data.frames:

---

As for speed, I would suggest using matrices instead of data.frames:

cmat <- as.matrix(VectClasses[-1]); rownames(cmat)<-VectClasses$a
imat <- as.matrix(VectIndVar[-1]);  rownames(imat)<-VectIndVar$a

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