在DataFrame R / Compinatorics中找到可行的组合

发布于 2025-01-19 15:03:30 字数 647 浏览 1 评论 0原文

我有以下挑战：具有218个观测值（行）和218个变量（COLS）的数据框架。值是真或错误。现在，我需要找到至少2行出现（true）的变量（col）组合。

以下是一个示例：

data <- data.frame(matrix(FALSE, nrow = 3, ncol = 5))
colnames(data) = paste("item_", 1:5, sep = "")
rownames(data) = paste("Process_", 1:3, sep = "")
data["Process_1",c("item_1","item_2","item_3")] = TRUE
data["Process_2",c("item_2","item_3")] = TRUE
data["Process_3",c("item_1","item_2","item_3","item_4","item_5")] = TRUE

在示例中，可行组合（或发现的目标）是以下组合：

c1：item1，item2，item3 c2：item2

c2：item3

c3：item3 c3：item1，item2

c4：item2 c4：item1，item1，item3

谢谢你，非常感谢您回答或提示很多:)

欢呼

原文

I have the following challenge:
dataframe with 218 observations (rows) and 218 variables (cols).
The values are either TRUE or FALSE.
Now i need to find combinations of variables (cols) that appear (TRUE) in at least 2 rows.

Here is a little example:

data <- data.frame(matrix(FALSE, nrow = 3, ncol = 5))
colnames(data) = paste("item_", 1:5, sep = "")
rownames(data) = paste("Process_", 1:3, sep = "")
data["Process_1",c("item_1","item_2","item_3")] = TRUE
data["Process_2",c("item_2","item_3")] = TRUE
data["Process_3",c("item_1","item_2","item_3","item_4","item_5")] = TRUE

For the example the feasible combinations (or the goal to find out) are the following combinations:

c1: item1,item2,item3

c2: item2,item3

c3: item1, item2

c4: item1, item3

Thank you very much for an answer or a hint :)

Cheers

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但可醉心 2025-01-26 15:03:30

#all items that have TRUE in 2 or more rows
items <- names(which(colSums(data) >= 2))
# all possible combinations of 2 (or more) items
lapply(2:length(items), function(x) combn(items, x)
# [[1]]
#          [,1]     [,2]     [,3]    
# [1,] "item_1" "item_1" "item_2"
# [2,] "item_2" "item_3" "item_3"
# 
# [[2]]
#          [,1]    
# [1,] "item_1"
# [2,] "item_2"
# [3,] "item_3"

#all items that have TRUE in 2 or more rows
items <- names(which(colSums(data) >= 2))
# all possible combinations of 2 (or more) items
lapply(2:length(items), function(x) combn(items, x)
# [[1]]
#          [,1]     [,2]     [,3]    
# [1,] "item_1" "item_1" "item_2"
# [2,] "item_2" "item_3" "item_3"
# 
# [[2]]
#          [,1]    
# [1,] "item_1"
# [2,] "item_2"
# [3,] "item_3"

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