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我一直在玩 R 的 gsub2 函数R: replace characters using gsub, 如何创建函数?创建密文:

from<-c('s','l','k','u','m','i','x','j','o','p','n','q','b','v','w','z','f','y','t','g','h','a','e','d','c','r')
  to<-c('z','e','b','r','a','s','c','d','f','g','h','i','j','k','l','m','n','o','p','q','t','u','v','w','x','y')

例如:

原文:《谁是1973》

密文:ptv ltn'm 1973

问题是,gsub2 替换了一些字母两次(o->f->n 和 s->z->n),这弄乱了我的密文并且几乎无法解码。谁能指出我犯的错误?谢谢!

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2 回答 2

7

一种方法是使用命名向量作为编码密码。创建此类命名向量的一种简单方法是使用setNames

cipher <- setNames(to, from)
cipher
  s   l   k   u   m   i   x   j   o   p   n   q   b   v   w   z   f   y   t   g   h   a   e   d   c   r 
"z" "e" "b" "r" "a" "s" "c" "d" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "t" "u" "v" "w" "x" "y" 

对于编码功能,您可以使用strsplitand paste

encode <- function(x){
  splitx <- strsplit(x, "")[[1]]
  xx <- cipher[splitx]
  xx[is.na(xx)] <- splitx[is.na(xx)]
  paste(xx, collapse="")
}

encode("the who's 1973")
[1] "ptv ltf'z 1973"
于 2012-11-22T00:06:36.430 回答
6

您还可以使用chartr(流行:12 ups)对您引用的问题的回答中提到的方法:

cipher <- function(x) 
    chartr( "slkumixjopnqbvwzfytghaedcr", "zebrascdfghijklmnopqtuvwxy", x )
于 2012-11-22T00:52:24.897 回答