如何比较一列中的两个相邻字符串并遍历所有字符串？

Question

我用来查找两个字符串之间差异的函数（found on the R-help website）：

X  <- "abcdefg" ; Y <- "aBcDEfg" 
diff <- function(X,Y){
  X0 <- unlist(strsplit(X,split=""))  ## Nasty but necessary! 
  Y0 <- unlist(strsplit(Y,split=""))  ## ... 
  ix <- which(X0 != Y0) 
  cbind(ix,X0[ix],Y0[ix])   
}
diff(X,Y)
     ix         
[1,] "2" "b" "B"
[2,] "4" "d" "D"
[3,] "5" "e" "E"

我需要比较的数据框状态中的列：

grint <- 
c("45CCBC44B", "45CCBC44B", "45CCBC44B", "45CCBC44B", "45CCBC44B", 
"45CCBC44B", "45CCBC44B", "45CCBC44B", "45CCBC44B", "<5CCBC:4B", 
"<5CCBC:4B", "<5CCBC:4B", "<<CCBC::B", "<<GGBG::E", "<<GGBG::E", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "CC11B1CCE", 
"CC11B1CCE", "CC55B1CCE", "55CCBC44B", "55CCBC44B", "55CCBC44B", 
"55CCBC44B", "55CCBC44B", "55CCBC44B", "G1CCBC1GB", "G1CCBC1GB", 
"G1CCBC1GB", "G1CCBC1GB", "G1CCBC1GB", "G1CCBC1GB", "G1CCBC1GB", 
"G1CCBC1GB", "G1CCBC1GB", "G1CCBC1GB", "G1CCBC1GB", "G1CCBC1GB", 
"G1CCBC1GB", "G1CCBC1GB", "91CCBC11B", "01CCBC11B", "01CCBC11B", 
"01CCBC11B", "01CCBC11B", "11CCBC11B", "11CCBC11B", "11CCBC11B", 
"15CCBC11B", "15CCBC11B", "15CCBC11B", "15CCBC11B", "15CCBC11B", 
"15CCBC11B", "15CCBC11B", "15CCBC11B", "15CCBC11B", "15CCBC11B", 
"55CCBC11B", "55CCBC11B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", 
"55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B", "55CCBC41B"
)

我需要连续比较列中的值，这意味着它可以找到发生的差异并比较两个字符串。例如grint[9] 和grint[10] 不同并显示出来。 我尝试使用lapply 函数来完成一个循环以应用于每个字符串以查找每个更改，但我失败了：

a <-grint[i]
b <-grint[i+1]

lapply(grint,diff(a,b))

错误：

Error in match.fun(FUN) : 
  'diff(a, b)' is not a function, character or symbol

所以我想知道我应该怎么做？非常感谢！

Answer 1

我认为你需要match，它返回第一个匹配的索引。删除第一个元素，

> ( m <- match(unique(x), x)[-1] )
 [1] 10 13 14 16 45 47 48 54 68 69 73 76 86

将匹配项与前一个元素进行比较，我们可以看到出现了差异。

> cbind(x[m-1], x[m])
      [,1]        [,2]       
 [1,] "45CCBC44B" "<5CCBC:4B"
 [2,] "<5CCBC:4B" "<<CCBC::B"
 [3,] "<<CCBC::B" "<<GGBG::E"
 [4,] "<<GGBG::E" "55CCBC41B"
 [5,] "55CCBC41B" "CC11B1CCE"
 [6,] "CC11B1CCE" "CC55B1CCE"
 [7,] "CC55B1CCE" "55CCBC44B"
 [8,] "55CCBC44B" "G1CCBC1GB"
 [9,] "G1CCBC1GB" "91CCBC11B"
[10,] "91CCBC11B" "01CCBC11B"
[11,] "01CCBC11B" "11CCBC11B"
[12,] "11CCBC11B" "15CCBC11B"
[13,] "15CCBC11B" "55CCBC11B"

Answer 2

比我下面的答案更好，只需按照@Andrie 在评论中建议的diff(grint[-1], grint[-length(grint)]) 进行操作即可。

这里有两种略有不同的方法，可以处理不同长度的字符串。如果所有字符串的长度相同，则不需要str_pad from stringr

samplestrings <- c("apple", "apple", "banana", "banana", "apple", "apple","aslkd;fa")
library(stringr)
samplestrings <- str_pad(samplestrings, max(nchar(samplestrings)) , side="right")

  X0 <- unlist(strsplit(samplestrings,split=""))  ## Nasty but necessary!
  Y0 <- unlist(strsplit(c(samplestrings[-1], rep(" ", max(nchar(samplestrings)))),split="")) ## ...
  ix <- which(X0[-length(X0):-(length(X0)-max(nchar(samplestrings))+1)] != 
              Y0[-length(X0):-(length(X0)-max(nchar(samplestrings))+1)])
  cbind(ix,X0[ix],Y0[ix])

      ix          
 [1,] "9"  "a" "b"
 [2,] "10" "p" "a"
 [3,] "11" "p" "n"
 [4,] "12" "l" "a"
 [5,] "13" "e" "n"
 [6,] "14" " " "a"
 [7,] "25" "b" "a"
 [8,] "26" "a" "p"
 [9,] "27" "n" "p"
[10,] "28" "a" "l"
[11,] "29" "n" "e"
[12,] "30" "a" " "
[13,] "42" "p" "s"
[14,] "43" "p" "l"
[15,] "44" "l" "k"
[16,] "45" "e" "d"
[17,] "46" " " ";"
[18,] "47" " " "f"
[19,] "48" " " "a"

---

方法 2： 在我意识到 OP 正在寻找什么样的输出之前，我先写了这个，但如果你想在查找连续字符串之间的字符差异的过程中创建一个不匹配的数据框，它仍然可以使用。

samplestrings <- c("apple", "apple", "banana", "banana", "apple", "apple","aslkd;fa")
library(stringr) 
# use str_pad to make every string equal in number of characters
samplestrings <- str_pad(samplestrings, max(nchar(samplestrings)) , side="right")

findiffs <- rle(samplestrings)

newdf <- data.frame(index = paste0(cumsum(findiffs$length),"-",cumsum(findiffs$length)+1), 
          firststring = samplestrings[cumsum(findiffs$length)],
          secondstring = samplestrings[cumsum(findiffs$length)+1])

newdf <- newdf[-dim(newdf)[1],] 

  index firststring secondstring
1   2-3    apple        banana  
2   4-5    banana       apple   
3   6-7    apple        aslkd;fa

所以，newdf 包含不相同的字符串，那么我们可以使用您使用的方法：

  X0 <- unlist(strsplit(as.character(newdf$firststring),split=""))  ## Nasty but necessary!
  Y0 <- unlist(strsplit(as.character(newdf$secondstring),split=""))  ## ...
  ix <- which(X0 != Y0)
  cbind(ix,X0[ix],Y0[ix]) 

     ix          
 [1,] "1"  "a" "b"
 [2,] "2"  "p" "a"
 [3,] "3"  "p" "n"
 [4,] "4"  "l" "a"
 [5,] "5"  "e" "n"
 [6,] "6"  " " "a"
 [7,] "9"  "b" "a"
 [8,] "10" "a" "p"
 [9,] "11" "n" "p"
[10,] "12" "a" "l"
[11,] "13" "n" "e"
[12,] "14" "a" " "
[13,] "18" "p" "s"
[14,] "19" "p" "l"
[15,] "20" "l" "k"
[16,] "21" "e" "d"
[17,] "22" " " ";"
[18,] "23" " " "f"
[19,] "24" " " "a"

Answer 3

我不完全确定我理解了这个问题。如果是您试图找出列/变量中存在差异的位置？你可以这样做。

• 将您的列转换为字符向量。

我在这里取了你的前 17 个条目并手动将它们放入向量“x”中

x<-c("45CCBC44B", "45CCBC44B", "45CCBC44B", "45CCBC44B", "45CCBC44B", "45CCBC44B",     "45CCBC44B", "45CCBC44B", "45CCBC44B", "<5CCBC:4B", "<5CCBC:4B", "<5CCBC:4B", "<<CCBC::B", "<<GGBG::E", "<<GGBG::E", "55CCBC41B", "55CCBC41B")

然后您可以简单地询问该向量的每个元素是否与进行滞后比较的前一个元素相同：

lagged.x <- c(NA,head(x,-1))
x == lagged.x


[1]    NA  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE  TRUE  TRUE FALSE FALSE   TRUE FALSE  TRUE

这将在存在差异的地方识别为“FALSE”。如果这是你感兴趣的？