Question

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。

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，它会成为程序性的。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。

#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：

    A1  A2  A3
B1  2   0   0
B2  0   2   0

这个问题的嵌套性质对我构成了挑战。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，但是循环的正确语法有很多缺点。

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。

Answer 1

通过创建唯一的 element 值的辅助矩阵（通过唯一的 group 值并在每个值中分配一个值）来创建矩阵，从而考虑矩阵乘法x %*% y（请参阅?matmult）相应的单元格。然后将矩阵乘法作为与自身的转置，然后是行和列的子集：

# EMPTY MATRIX
helper_mat <- matrix(0, nrow=length(unique(element)), ncol=length(unique(group)),
                     dimnames=list(unique(element), unique(group)))

# ASSIGN 1's AT SELECT LOCATIONS
for(i in seq_along(site)) {
  helper_mat[element[i], group[i]] <- 1
}

helper_mat
#        A1 A2 A3 B1 B2
# red     1  0  0  1  0
# orange  1  0  0  1  0
# blue    0  1  0  0  1
# black   0  1  1  0  0
# white   0  1  0  0  1
# cream   0  0  1  0  0
# yellow  0  0  1  0  0
# purple  0  0  1  0  0
# gray    0  0  0  0  1
# salmon  0  0  0  0  1

# MATRIX MULTIPLICATION WITH SUBSET
final_mat <- t(helper_mat) %*% helper_mat
final_mat <- final_mat[grep("B", rownames(final_mat)), grep("A", colnames(final_mat))]

final_mat
#    A1 A2 A3
# B1  2  0  0
# B2  0  2  0

感谢@Lamia，甚至版本更短：

helper_mat <- table(element, group)

final_mat <- t(helper_mat) %*% helper_mat # ALTERNATIVELY: crossprod(helper_mat)

final_mat <- final_mat[grep("B", rownames(final_mat)), grep("A", colnames(final_mat))]

final_mat
#      group
# group A1 A2 A3
#    B1  2  0  0
#    B2  0  2  0

Answer 2

# example dataset
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)

library(tidyverse)

# save as dataframe
d = data.frame(d)

expand.grid(groupA = unique(d$group[d$site=="A"]),
            groupB = unique(d$group[d$site=="B"])) %>%               # get all combinations of A and B columns
  rowwise() %>%                                                      # for each row
  mutate(counts = length(intersect(d$element[d$group==groupA], 
                                   d$element[d$group==groupB]))) %>% # count common elements
  spread(groupA, counts) %>%                                         # reshape data
  data.frame() %>%                                                   
  column_to_rownames("groupB")

#    A1 A2 A3
# B1  2  0  0
# B2  0  2  0

您可以使用向量化的函数代替rowwise，该函数将（自动）应用于每一行，如下所示：

# create a function and vectorise it
CountCommonElements = function(x, y) length(intersect(d$element[d$group==x], d$element[d$group==y]))
CountCommonElements = Vectorize(CountCommonElements)

expand.grid(groupA = unique(d$group[d$site=="A"]),
            groupB = unique(d$group[d$site=="B"])) %>%                                                              
  mutate(counts = CountCommonElements(groupA, groupB)) %>% 
  spread(groupA, counts) %>%                                       
  data.frame() %>%                                                   
  column_to_rownames("groupB")

#    A1 A2 A3
# B1  2  0  0
# B2  0  2  0

Answer 3

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

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

R-比较两个分层列表之间共有元素的循环

3 个答案: