矩阵计算该矩阵的一行和列组合的次数是数据框中

Question

我仍然对R的新事物是新的，所以请与我一起露面:)

我需要创建一个矩阵，该矩阵计算数据框中存在该矩阵的行和列的组合的次数。

由于我的描述可能很含糊，因此我给了下面的示例集。实际上，我的数据集将在矩阵中包含更多的水果和数据框架中的更多果汁，因此我正在寻找一种有效的方法来解决此问题。

#Stackoverflow example
#Create empty matrix ----
newMatrix <- matrix(0, nrow = 5, ncol = 5)
colnames(newMatrix) <- c("Apple", "Pear", "Orange", "Mango", "Banana")
rownames(newMatrix) <- c("Apple", "Pear", "Orange", "Mango", "Banana")

#Create dataframe ----
newDf <- data.frame(c("Juice 1", "Juice 2", "Juice 3", "Juice 4","Juice 5"),
                    c("Banana", "Banana", "Orange", "Pear", "Apple"),
                    c("Pear", "Orange", "Pear", "Apple", "Pear"),
                    c("Orange", "Mango", NA, NA, NA))
colnames(newDf) <- c("Juice", "Fruit 1", "Fruit 2", "Fruit 3")

我想创建一个for循环，该循环遍历我的newmatrix中的每个元素，并添加+1，如果列和行的组合存在于newdf的行中。<<<<<<<<<<<<< br> 因此，从本质上讲，有多少果汁结合了苹果和梨，有多少果汁具有苹果和芒果的组合，等等。

输出应该看起来像这样：

       Apple Pear Orange Mango Banana
Apple      0    2      0     0      0
Pear       2    0      2     0      1
Orange     0    2      0     1      2
Mango      0    0      1     0      1
Banana     0    1      2     1      0

我首先尝试创建一个for循环，但我陷入了 part的：

for (i in 1:nrow(adj_matrix)){
  for (j in 1:ncol(adj_matrix)) {
    if (???)
      adj_matrix[i,j] <- adj_matrix[i,j] + 1
  }
}

有人可以帮助我吗？将不胜感激！

Answer 1

使用base r，您可以将值组合起来，然后使用igraph获得邻接矩阵：

library(igraph)

m <- do.call(cbind, apply(newDf[-1], 1, \(x) if(sum(complete.cases(x)) >= 2) combn(x, m = 2) else x, simplify = F))
g <- graph_from_data_frame(na.omit(t(m)), directed = F)
get.adjacency(g, sparse = F)

       Banana Pear Orange Apple Mango
Banana      0    1      2     0     1
Pear        1    0      2     2     0
Orange      2    2      0     0     1
Apple       0    2      0     0     0
Mango       1    0      1     0     0

---

它可能会有些复杂，但是您也可以将tidyverse软件包：

library(igraph)
library(tidyverse)

newDf %>% 
  pivot_longer(-Juice) %>% 
  group_by(Juice) %>% 
  summarise(new = ifelse(n() > 1, paste(combn(na.omit(value), 2), collapse = "-"), value)) %>% 
  separate_rows(new, sep = "(?:[^-]*(?:-[^-]*){1})\\K-") %>% 
  separate(new, into = c("X1", "X2")) %>% 
  select(-Juice) %>% 
  graph_from_data_frame(directed = FALSE) %>% 
  get.adjacency(sparse = FALSE)

       Banana Pear Orange Apple Mango
Banana      0    1      2     0     1
Pear        1    0      2     2     0
Orange      2    2      0     0     1
Apple       0    2      0     0     0
Mango       1    0      1     0     0

Answer 2

可以像这样编写循环的。

cb <- combn(2:4, 2)  ## cols combinations newDf 

## initialize adj_matrix
v <- c("Apple", "Pear", "Orange", "Mango", "Banana")
adj_matrix <- matrix(0, length(v), length(v), dimnames=list(v, v))

for (k in seq_len(nrow(newDf))) {
    for (l in seq_len(ncol(cb))) {
        x <- unlist(newDf[k, cb[, l]])
        if (length(x[!is.na(x)]) == 2) {
        adj_matrix[x[1], x[2]] <- adj_matrix[x[1], x[2]] + 1
        adj_matrix[x[2], x[1]] <- adj_matrix[x[2], x[1]] + 1
        }
    }
}

adj_matrix
#        Apple Pear Orange Mango Banana
# Apple      0    2      0     0      0
# Pear       2    0      2     0      1
# Orange     0    2      0     1      2
# Mango      0    0      1     0      1
# Banana     0    1      2     1      0

---

数据：

newDf <- structure(list(Juice = c("Juice 1", "Juice 2", "Juice 3", "Juice 4", 
"Juice 5"), `Fruit 1` = c("Banana", "Banana", "Orange", "Pear", 
"Apple"), `Fruit 2` = c("Pear", "Orange", "Pear", "Apple", "Pear"
), `Fruit 3` = c("Orange", "Mango", NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-5L))