r - 当行包含相同字符时删除列

Question

我有一个字符/数字组合矩阵，我需要删除那些列的两行中都出现相同字符的列。对于一个简化的例子:

> chars <- c("A1","A2","B1","B2")
> charsmat <- combn(chars, 2)
> charsmat
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,] "A1" "A1" "A1" "A2" "A2" "B1"
[2,] "A2" "B1" "B2" "B1" "B2" "B2"

当单列的两行包含相同的字符(在本例中为第 1 列和第 6 列)时，我需要删除该列。我觉得我有碎片:使用 gsub()或 str_extract()隔离字符，并测试行之间是否匹配，但我不知道如何制定它。提前感谢您提供的任何帮助。

Answer 1

首先，创建一个仅提取字母部分的新矩阵:

> (charsmat.alpha <- substr(charsmat, 0, 1))
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,] "A"  "A"  "A"  "A"  "A"  "B" 
[2,] "A"  "B"  "B"  "B"  "B"  "B"

然后，从 charsmat 中获取列的子集其中 charsmat.alpha 的两行不一样:

> charsmat[,(charsmat.alpha[1,] != charsmat.alpha[2,])]
     [,1] [,2] [,3] [,4]
[1,] "A1" "A1" "A2" "A2"
[2,] "B1" "B2" "B1" "B2"