R-比较两个分层列表之间共有元素的循环 [英] R - A loop comparing elements in common between two hierarchical lists

问题描述

一段时间以来，我一直在尝试构建一个矩阵，该矩阵由两个等级列表之间共有的元素数量组成.

I have been trying, for some time, to build a matrix populated by the counts of elements in common between two herarchical lists.

以下是一些虚拟数据:

site<-c('A','A','A','A','A','A','A','A','A','B','B','B','B','B','B')
group<-c('A1','A1','A2','A2','A2','A3','A3','A3','A3', 
'B1','B1','B2','B2','B2','B2')
element<-c("red","orange","blue","black","white", "black","cream","yellow","purple","red","orange","blue","white","gray","salmon")
d<-cbind(site,group,element)

我创建了一个列表结构，假设由于每个列表中os元素的数量不同，该结构将是程序性的.另外，由于我不想在组之间进行所有可能的比较，而只希望在站点之间进行比较.

I created a list structure, assuming it would be procedural due to the different number os elements in each list. Also, since I don´t want every possible comparison between groups, but only between sites.

#first level list - by site
sitelist<-split(nodmod, list(nodmod$site),drop = TRUE)
#list by group 
nestedlist <- lapply(sitelist, function(x) split(x, x[['mod']], drop = TRUE))

我的意图是创建一个表或矩阵，其中两个站点的组之间的元素数量相同(我的原始数据还有其他站点).像这样:

My intention is to create a table, or matrix with the number of element in common between groups from the two sites (my original data has additional sites). Like such:

    A1  A2  A3
B1  2   0   0
B2  0   2   0

这个问题的嵌套性质对我来说具有挑战性.我对列表不太熟悉，因为我主要使用数据框解决了问题.我的尝试归结为这一点.我觉得它已经接近了，但是在循环的正确语法上有很多缺点.

The nested nature of this problem is challenging to me. I am not as familiar with lists, as I´ve solved problems mostly using dataframes. My attempt boiled down to this. I felt it got close, but have many shortcomings with the correct syntax for loops.

t <- outer(1:length(d$A),
         1:length(d$B),
         FUN=function(i,j){
           sapply(1:length(i),
                  FUN=function(x) 
                    length(intersect(d$A[[i]]$element, d$B[[j]]$element)) )
         })

任何帮助将不胜感激.如果解决了类似的问题，我们深表歉意.我已经搜索了互联网，但是没有找到它，或者没有理解使它可以转让给我的解决方案.

Any help would be much appreciated. Apologies if a similar problem has been solved. I have scoured the internet, but have not found it, or did not comprehend the solution to make it transferable to mine.

推荐答案

类似于@Parfait使用矩阵乘法的方法.您可能需要尝试处理数据生成，以将其扩展到您的应用程序:

A similar approach to @Parfait's using matrix multiplication. You may need to play around with the data generation to extend it to your application:

site<-c('A','A','A','A','A','A','A','A','A','B','B','B','B','B','B')
group<-c('A1','A1','A2','A2','A2','A3','A3','A3','A3', 
         'B1','B1','B2','B2','B2','B2')
element<-c("red","orange","blue","black","white", "black","cream","yellow","purple","red","orange","blue","white","gray","salmon")

d<-data.frame(group, el = as.factor(element), stringsAsFactors = FALSE)


As <- d[group %in% paste0("A", 1:3), ]
Bs <- d[group %in% paste0("B", 1:2), ]

A_mat <- as.matrix(table(As))
B_mat <- as.matrix(table(Bs))

结果:

> A_mat
         el
group black blue cream gray orange purple red salmon white yellow
   A1     0    0     0    0      1      0   1      0     0      0
   A2     1    1     0    0      0      0   0      0     1      0
   A3     1    0     1    0      0      1   0      0     0      1


> B_mat
         el
group black blue cream gray orange purple red salmon white yellow
   B1     0    0     0    0      1      0   1      0     0      0
   B2     0    1     0    1      0      0   0      1     1      0


> B_mat %*% t(A_mat)
     group
group A1 A2 A3
   B1  2  0  0
   B2  0  2  0

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