總是很有趣看到一個合理的答案被否決。雖然我不喜歡這個減得分,但我會保留我的答案。選民,你覺得怎麼樣?
讓我們先得到一些重複性玩具數據:
set.seed(0); x <- replicate(10, sample((0:3), 10, replace = TRUE))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 3 0 3 1 1 2 1 3 3 0
# [2,] 1 0 3 1 3 1 3 1 1 0
# [3,] 1 0 0 2 2 3 1 3 2 0
# [4,] 2 2 2 1 3 1 1 1 1 2
# [5,] 3 1 0 0 2 0 1 1 1 3
# [6,] 0 3 1 3 2 0 2 1 3 3
# [7,] 3 1 1 2 3 0 1 3 0 3
# [8,] 3 2 0 3 0 1 1 3 2 1
# [9,] 2 3 1 0 1 2 3 1 0 1
#[10,] 2 1 3 2 2 2 0 3 0 3
對於任何輸入矩陣x
,你可以使用:
y <- unlist(lapply(seq_len(ncol(x)-1L),
function(i) colSums(x[, (i+1):ncol(x), drop = FALSE] == x[, i])))
# [1] 1 2 3 2 4 1 4 2 3 3 1 0 0 3 1 3 5 1 3 1 2 4 1 4 3 4 2 3 5 1 1 3 2 1 2 2 3 3
#[39] 1 2 3 1 4 3 1
max(y)
# [1] 5
通過@大衛的意見基本上是做同樣的事情,但方法要慢:
y <- combn(ncol(x), 2, FUN = function(u) sum(x[, u[1]] == x[, u[2]]))
# [1] 1 2 3 2 4 1 4 2 3 3 1 0 0 3 1 3 5 1 3 1 2 4 1 4 3 4 2 3 5 1 1 3 2 1 2 2 3 3
#[39] 1 2 3 1 4 3 1
max(y)
# [1] 5
標杆
我們生成實驗有10×1000矩陣:
set.seed(0); x <- replicate(1e+3, sample((0:3), 10, replace = TRUE))
system.time(unlist(lapply(seq_len(ncol(x)-1L), function(i) colSums(x[, (i+1):ncol(x), drop = FALSE] == x[, i]))))
# user system elapsed
# 0.176 0.032 0.207
system.time(combn(ncol(x), 2, FUN = function(u) sum(x[, u[1]] == x[, u[2]])))
# user system elapsed
# 4.692 0.008 4.708
像一個距離矩陣?
有了這樣的想法,你也可以生成的所有列之間不等於元素數量「距離」矩陣(只是!=
更換==
):
y <- unlist(lapply(seq_len(ncol(x)-1L),
function(i) colSums(x[, (i+1):ncol(x), drop = FALSE] != x[, i])))
z <- matrix(0L, ncol(x), ncol(x))
z[lower.tri(z)] <- y
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 0 0 0 0 0 0 0 0 0 0
# [2,] 9 0 0 0 0 0 0 0 0 0
# [3,] 8 7 0 0 0 0 0 0 0 0
# [4,] 7 9 9 0 0 0 0 0 0 0
# [5,] 8 10 7 7 0 0 0 0 0 0
# [6,] 6 10 9 6 9 0 0 0 0 0
# [7,] 9 7 8 8 7 8 0 0 0 0
# [8,] 6 9 6 7 8 7 8 0 0 0
# [9,] 8 7 9 5 9 7 7 6 0 0
#[10,] 7 5 6 9 8 9 9 7 9 0
注意只由於對稱性,計算下三角矩陣。對角線全部爲零(或當然)。
試試'combn(ncol(test),2,FUN = function(x)sum(test [,x [1]] == test [,x [2]]))'也許 –