Question

我有一个向量列表我想找到所有常见的向量。也就是说，那些包含完全相同元素的元素，将每个列表的位置编号保存在R中。如果可能，可以使用一个衬管命令。

这是mylist：

mylist<-list(c("yes", "no"), c("no", "other", "up", 
"down"), c("no", "yes"), c("no", 
"yes"), c("no", "yes", "maybe"), c("no", 
"yes", "maybe"), c("no", "yes", "maybe"))

期望的输出：

常见列表是：匹配1：1,3,4 比赛2：5,6,7

Answer 1

.interaction.show_B { max-height: 325px; transition: max-height 0.15s ease-in; }接受列表作为其主要参数。所以你可以使用

duplicated

第一个例子。请注意，这不会区分具有公共元素的列表元素组，但仅对具有相同元素的元素返回TRUE。

对于第二个示例数据集，您可以使用以下内容查找组的位置

which(duplicated(mylist1) | duplicated(mylist1, fromLast=TRUE))
[1] 3 4 5 6 7

数据

# get group values as integers groups <- as.integer(factor(sapply(mylist2, function(x) paste(sort(x), collapse="")))) # return list of groups lapply(seq_len(max(groups)), function(x) which(x == groups)) [[1]] [1] 2 [[2]] [1] 5 6 7 [[3]] [1] 1 3 4

Answer 2

以下是使用split

的选项

Filter(function(x) length(x) >1, split(seq_along(mylist),
                     sapply(mylist, function(x) toString(sort(x)))))
#$`maybe, no, yes`
#[1] 5 6 7

#$`no, yes`
#[1] 1 3 4

Answer 3

这是一个有趣的。您可以使用mtabulate包中的qdapTools来获取以下数据框

d1 <- qdapTools::mtabulate(mylist)
d1
#  down maybe no other up yes
#1    0     0  1     0  0   1
#2    1     0  1     1  1   0
#3    0     0  1     0  0   1
#4    0     0  1     0  0   1
#5    0     1  1     0  0   1
#6    0     1  1     0  0   1
#7    0     1  1     0  0   1

然后你可以通过粘贴来分割它，

l1 <- split(d1, do.call(paste, d1))

l1
#$`0 0 1 0 0 1`
#  down maybe no other up yes
#1    0     0  1     0  0   1
#3    0     0  1     0  0   1
#4    0     0  1     0  0   1

#$`0 1 1 0 0 1`
#  down maybe no other up yes
#5    0     1  1     0  0   1
#6    0     1  1     0  0   1
#7    0     1  1     0  0   1

#$`1 0 1 1 1 0`
#  down maybe no other up yes
#2    1     0  1     1  1   0

您可以随意使用该列表，即

lapply(l1, rownames)
#$`0 0 1 0 0 1`
#[1] "1" "3" "4"

#$`0 1 1 0 0 1`
#[1] "5" "6" "7"

#$`1 0 1 1 1 0`
#[1] "2"

甚至，

setNames(lapply(l1, rownames), lapply(l1, function(i)toString(names(i)[i[1,] == 1])))
#$`no, yes`
#[1] "1" "3" "4"

#$`maybe, no, yes`
#[1] "5" "6" "7"

#$`down, no, other, up`
#[1] "2"

Answer 4

这对我有用：

mylist<-list(c("yes", "no"), c("no", "other", "up", 
                               "down"), c("no", "yes"), c("no", 
                                                          "yes"), c("no", "yes", "maybe"), c("no", 
                                                                                             "yes", "maybe"), c("no", "yes", "maybe"))

library(dplyr)

# function to create a dataframe from your list. Might not be the most efficient way to do this.
f <- function(data) {
  nCol <- max(vapply(data, length, 0))
  data <- lapply(data, function(row) c(row, rep(NA, nCol-length(row))))
  data <- matrix(unlist(data), nrow=length(data), ncol=nCol, byrow=TRUE)
  data.frame(data)
}

# create a dataframe from the list, and add a 'key' column
df = f(mylist)
df$key = apply( df , 1 , paste , collapse = "-" )

# find the total times the key occurs
df_total = df %>% group_by(key) %>% summarise(n =n())

# find the indices that belong to the groups
result = lapply(df_total$key, function(x) which(df$key==x))

结果：

> result
[[1]]
[1] 2

[[2]]
[1] 5 6 7

[[3]]
[1] 3 4

[[4]]
[1] 1

希望这有帮助！

Answer 5

数据

mylist <- list(c("yes", "no"), c("no", "other", "up", "down"), c("no", "yes"), 
           c("no", "yes"), c("no", "yes", "maybe"), c("no", "yes", "maybe"), 
           c("no", "yes", "maybe"))

（长）单线

sapply(unique(unlist(lapply(mylist, function(x) paste(sort(x), collapse = " ")))), function(y) which(y == unlist(lapply(mylist, function(x) paste(sort(x), collapse = " ")))))

输出：

$`no yes`
[1] 1 3 4

$`down no other up`
[1] 2

$`maybe no yes`
[1] 5 6 7

在列表列表中查找所有匹配的列表

5 个答案: