R 如何比较一列中的两个相邻字符串并遍历所有字符串？

我用于查找两个字符串之间的差异（）的函数：

错误：

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

---

所以我想知道我该怎么做？非常感谢

我不能完全确定我是否理解这个问题。如果您正试图查找列/变量中存在的差异？你可以这样做

• 将列转换为字符向量

我在这里获取了您的前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")

---

x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）]）

这里有两种稍微不同的方法，它们可以处理不同长度的字符串。如果所有字符串的长度相同，则不需要使用
stru-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"

我想您需要
match
，它返回第一个匹配的索引。移除第一个元素

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

>（m cbind（x[m-1]，x[m]）
[,1]        [,2]       
[1，]“45CCBC44B”Try
diff（grint[-1]，grint[-length（grint）]）
我想可以肯定地说，您正在寻找由生成的输出。哦，我应该添加链接。谢谢您的提醒，我下次会记得的！谢谢您的回答！这几乎是我所需要的。我想它不仅可以找到相邻字符串之间的所有差异，还可以压缩字符串中的每个字符。like:'diff（“苹果”、“香蕉”）ix[1、][1”“a”“b”[2、][2”“p”“a”[3、][3”“p”“n”[4、][4”“l”“a”[5、][5”“e”“n”@Chenlu当然。请随意选择有用的答案，一旦你得到正确的答案或你最喜欢的答案，请随意接受，因为网站就是这样运作的；）另外，在未来，最好能提供所需的输出，以便明确您想要的内容。干杯。谢谢您的建议：）我还没有很好地理解规则，因为我对编程和网站都是一个新手。@Chenlu我用两种稍微不同的方法更新了我的答案。+1我喜欢这种方法，但我想你您缺少最后一对：
“55CCBC11B”“55CCBC41B”
，因为
“55CCBC41B”
出现在向量的两个部分。有时可能需要多于第一个匹配项（请参见
其中（grint==“55CCBC41B”）
。
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")

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

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"

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

  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"

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

> 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"