2

我有一个矩阵，我想确定每个字符在所有成对之间出现在同一位置的次数。

下面是我正在做的一个示例，但是我的矩阵有 10,000 行，而且花费的时间太长。

# 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)
                  }) 

}
4

3 回答 3

3

我不知道这是否会表现良好，但这是一种选择：

library(data.table)
matchDT = setDT(melt(mtx))[, 
  CJ(row1 = Var1, row2 = Var1)[row1 < row2], by=.(value, col = Var2)]
]

dcast(matchDT, row1 + row2 ~ value)

这不包括没有匹配项的行组合。为了让他们回来，也许...

levs = seq_len(nrow(mtx))
dcast(matchDT, factor(row1, levels=levs) + factor(row2, levels = levs) ~ value, drop = FALSE)[as.integer(row1) < as.integer(row2)]

Aggregate function missing, defaulting to 'length'
    row1 row2 A B C D
 1:    1    2 1 0 2 0
 2:    1    3 1 0 1 1
 3:    1    4 1 1 0 1
 4:    1    5 0 1 1 0
 5:    1    6 1 0 1 1
 6:    1    7 0 0 1 0
 7:    1    8 0 2 1 0
 8:    1    9 1 2 2 1
 9:    1   10 0 1 1 0
10:    2    3 2 0 0 0
11:    2    4 1 0 1 0
12:    2    5 0 1 1 0
13:    2    6 1 0 1 1
14:    2    7 0 0 1 0
15:    2    8 2 0 1 0
16:    2    9 1 0 1 0
17:    2   10 1 0 1 0
18:    3    4 0 0 0 2
19:    3    5 0 0 0 0
20:    3    6 1 0 0 2
21:    3    7 1 1 1 0
22:    3    8 1 0 0 1
23:    3    9 1 1 0 0
24:    3   10 1 0 1 0
25:    4    5 0 2 1 0
26:    4    6 0 1 0 2
27:    4    7 0 0 0 0
28:    4    8 1 1 0 2
29:    4    9 0 2 0 0
30:    4   10 0 2 1 0
31:    5    6 0 1 1 0
32:    5    7 0 2 1 0
33:    5    8 0 1 0 1
34:    5    9 0 1 1 0
35:    5   10 0 2 1 1
36:    6    7 0 1 2 1
37:    6    8 0 0 0 1
38:    6    9 1 1 1 0
39:    6   10 0 1 0 0
40:    7    8 0 0 1 0
41:    7    9 0 0 1 0
42:    7   10 0 1 2 0
43:    8    9 1 2 1 0
44:    8   10 1 1 1 1
45:    9   10 0 2 1 0
    row1 row2 A B C D
于 2018-04-11T17:21:12.613 回答
3 
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
于 2018-04-11T20:03:28.923 回答
1

基于 R 的可能解决方案：

l1 <- lapply(split(df, 1:nrow(df)), as.integer)

l2 <- lapply(l1, function(x) {
  m <- mtx[x[1],] == mtx[x[2],]
  l <- lapply(my_letters, '==', mtx[x[1],])
  sapply(l, function(i) sum(i & m))
})

cbind(df, setNames(do.call(rbind.data.frame, l2), my_letters))

这使：

   index_1 index_2 A B C D
1        1       2 0 0 0 0
2        1       3 0 0 2 1
3        1       4 0 0 0 1
4        1       5 0 1 2 0
5        1       6 0 0 3 1
6        1       7 0 1 1 3
7        1       8 0 1 2 2
8        1       9 0 0 2 1
9        1      10 0 0 2 0
10       2       3 0 1 0 1
11       2       4 0 1 0 2
12       2       5 0 1 0 0
13       2       6 0 0 0 2
14       2       7 0 1 0 1
15       2       8 1 0 0 0
16       2       9 0 1 0 2
17       2      10 2 1 0 3
18       3       4 0 0 1 0
19       3       5 0 0 1 1
20       3       6 0 0 1 1
21       3       7 0 1 1 2
22       3       8 0 0 0 1
23       3       9 1 0 0 0
24       3      10 0 0 0 1
25       4       5 0 2 1 0
26       4       6 0 0 1 1
27       4       7 1 1 0 1
28       4       8 1 1 1 1
29       4       9 0 1 1 2
30       4      10 0 1 0 2
31       5       6 0 1 2 0
32       5       7 0 1 1 0
33       5       8 0 2 1 0
34       5       9 0 1 2 0
35       5      10 0 2 1 0
36       6       7 1 0 1 1
37       6       8 0 0 3 1
38       6       9 0 1 2 0
39       6      10 0 0 1 1
40       7       8 0 1 0 2
41       7       9 0 1 0 1
42       7      10 0 0 0 1
43       8       9 0 0 2 1
44       8      10 1 1 1 0
45       9      10 0 0 2 1
于 2018-04-11T18:08:13.967 回答