r - 如何检查矩阵的列是否遵循R中的预定顺序？

Question

矩阵看起来像......

      [,1]   [,2]   [,3]   [,4]   [,5]  
 [1,] "notB" "notB" "B"    "notB" "notB"
 [2,] "notB" "notB" "notB" "notB" "notB"
 [3,] "notB" "notB" "notB" "notB" "notB"
 [4,] "B"    "notB" "notB" "notB" "B"   
 [5,] "notB" "notB" "notB" "notB" "notB"
 [6,] "notB" "B"    "B"    "notB" "B"   
 [7,] "notB" "notB" "notB" "B"    "notB"
 [8,] "B"    "B"    "B"    "B"    "B"   
 [9,] "B"    "B"    "notB" "B"    "notB"

这个想法是计算(在数千列的设置中)有多少个“B”元素保持在一起，无论在哪个位置，例如B B B notB notB notB notB notB notB，或notB notB B B B notB notB notB notB。

在上面发布的矩阵中，只有[,4]列满足B“在一起”的标准。

<小时/>

这是生成矩阵的代码:

b=c(rep("B", 3), rep("notB", 6))
n = 1000
d = replicate(n,sample(b, replace=F))

Answer 1

另一种方法:

apply(mat,2, function(c) all(diff(which(c=="B")) == 1))

我们在这里apply每列的一个函数，用于检查是否 all diff对应于 "B" 的元素索引之间的间隔是 1 。当且仅当所有 "B" 时，此条件成立。在一起了。

使用您发布的数据可以得出:

##   V1    V2    V3    V4    V5 
##FALSE FALSE FALSE  TRUE FALSE

然后我们可以使用which提取符合条件的列:

which(apply(mat,2, function(c) all(diff(which(c=="B")) == 1)))
## V4
##  4

或者，正如 @IaroslavDomin 评论的那样，我们可以改为 apply功能

apply(mat, 2, function(c){w <- which(c == "B"); length(w) == diff(range(w)) + 1})

这样做的好处是，我们不必检查 "B" 的相邻索引的所有差异。是 1 。相反，我们只需检查最后一个和第一个之间的差异(即 diff(range(w)) 加 1)是否与 "B" 的数量匹配。在列中。

数据:

mat <- structure(c("notB", "notB", "notB", "B", "notB", "notB", "notB", 
"B", "B", "notB", "notB", "notB", "notB", "notB", "B", "notB", 
"B", "B", "B", "notB", "notB", "notB", "notB", "B", "notB", "B", 
"notB", "notB", "notB", "notB", "notB", "notB", "notB", "B", 
"B", "B", "notB", "notB", "notB", "B", "notB", "B", "notB", "B", 
"notB"), .Dim = c(9L, 5L), .Dimnames = list(NULL, c("V1", "V2", 
"V3", "V4", "V5")))
      V1     V2     V3     V4     V5    
 [1,] "notB" "notB" "B"    "notB" "notB"
 [2,] "notB" "notB" "notB" "notB" "notB"
 [3,] "notB" "notB" "notB" "notB" "notB"
 [4,] "B"    "notB" "notB" "notB" "B"   
 [5,] "notB" "notB" "notB" "notB" "notB"
 [6,] "notB" "B"    "B"    "notB" "B"   
 [7,] "notB" "notB" "notB" "B"    "notB"
 [8,] "B"    "B"    "B"    "B"    "B"   
 [9,] "B"    "B"    "notB" "B"    "notB"