r - 成对计算所有行之间给定字符的位置匹配数的快速方法

Question

我有一个矩阵，我想确定每个字符在所有成对之间出现在相同位置的次数。

下面是我正在做的一个例子，但是我的矩阵有 10,000 行，而且它花费的时间太长了。

# This code will generate a dataframe with one row for each pair and columns that 
# count the number of position match each letter have  
my_letters <- c("A", "B", "C", "D")
size_vector <- 175
n_vectors <- 10
indexes_vectors <- seq_len(n_vectors)

mtx <- sapply(indexes_vectors, 
              function(i) sample(my_letters, n_vectors, replace = TRUE))
rownames(mtx) <- indexes_vectors

df <- as.data.frame(t(combn(indexes_vectors, m = 2)))
colnames(df) <- c("index_1", "index_2")

for(l in my_letters){
  cat(l, "\n")
  df[,l] <- apply(df[,1:2], 1,
                  function(ids) {
                    sum(mtx[ids[1],] ==  mtx[ids[2],] & 
                          mtx[ids[1],] == l, na.rm = TRUE)
                  }) 

}

Answer 1

m1 <- t(sapply(1:nrow(df), function(i) 
  table(factor(mtx[df[i,1],][mtx[df[i,1],] == mtx[df[i,2],]],
               levels = my_letters))))
cbind(df, m1)

>  V1 V2 A B C D
1   1  2 0 0 1 1
2   1  3 1 0 1 1
3   1  4 1 0 2 1
4   1  5 0 0 1 0
5   1  6 2 0 2 0
6   1  7 0 0 1 0
7   1  8 1 0 1 1
8   1  9 0 0 1 0
9   1 10 1 0 1 1
10  2  3 0 0 1 1
11  2  4 1 1 1 2
12  2  5 0 0 0 1
13  2  6 1 0 2 1
14  2  7 1 0 0 1
15  2  8 1 0 0 0
16  2  9 2 0 0 0
17  2 10 1 0 1 0
18  3  4 0 0 0 0
19  3  5 0 2 1 0
20  3  6 1 1 2 1
21  3  7 0 1 0 0
22  3  8 1 1 0 0
23  3  9 0 1 2 0
24  3 10 0 0 1 0
25  4  5 1 1 0 1
26  4  6 2 1 1 0
27  4  7 1 0 1 1
28  4  8 0 1 0 0
29  4  9 1 0 0 0
30  4 10 2 0 0 0
31  5  6 0 2 0 0
32  5  7 0 1 3 1
33  5  8 0 1 2 0
34  5  9 1 0 2 0
35  5 10 0 0 2 0
36  6  7 0 0 0 0
37  6  8 1 1 0 0
38  6  9 0 0 1 0
39  6 10 3 0 1 0
40  7  8 0 1 1 0
41  7  9 1 0 1 0
42  7 10 0 0 1 0
43  8  9 1 1 1 1
44  8 10 0 0 1 0
45  9 10 0 0 0 0