找到列表中的所有匹配列表

Question

我有一个列表 我想找到所有常用向量。也就是说，那些包含完全相同元素的元素，在R中保留每个列表的位置编号。如果可能的话，使用一个班轮命令。

这里是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")) 

希望的输出：

共用列表是：匹配1：1,3,4 匹配2：5,6,7

Answer 1

下面是使用split

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 2

duplicated接受列表作为它的主要论据。所以你可以使用

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

为你的第一个例子。请注意，这不会区分带有公共元素的列表元素组，但只会为具有相同元素的元素返回TRUE。

对于第二个示例数据集，您可以使用以下方法来查找组的位置

# 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 

数据

mylist1 <- 
list(c("yes", "no"), c("no", "other", "up", "down"), c("no", 
"yes", "maybe"), c("no", "yes", "maybe"), c("no", "yes", "maybe" 
), c("no", "yes", "maybe"), c("no", "yes", "maybe")) 

mylist2 <- 
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")) 

Answer 3

这个工作对我来说：

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 

希望这会有所帮助！

Answer 4

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")) 

一个（长）的单行

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

输出一个选项：

$`no yes` 
[1] 1 3 4 

$`down no other up` 
[1] 2 

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

Answer 5

这是一个有趣的。您可以使用mtabulate从qdapTools包得到以下数据帧，

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 

然后你就可以通过粘贴把它分解，

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 

但是，您可以利用该列表中你想要的，即

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"