为矩阵的每一行查找 5 个连续的数字 >= 3

我有以下矩阵

mdat <- matrix(c(6,2,4,4,'*',5,1,6,'*',2,1,5,1,3,3,5,4,'*',5,'*',1,'*',4,'*',2,2,4,3,4,4,4,'*',4,3,3,1,1,3,2,3,3,3,3,3,2,2,'*','*',2,1,2,2,2,2,2,1,1,1,1,1,'*',1,1,1,1,1 ),nrow = 6, ncol = 11, byrow = TRUE)



[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] "6"  "2"  "4"  "4"  "*"  "5"  "1"  "6"  "*"  "2"   "1"  
[2,] "5"  "1"  "3"  "3"  "5"  "4"  "*"  "5"  "*"  "1"   "*"  
[3,] "4"  "*"  "2"  "2"  "4"  "3"  "4"  "4"  "4"  "*"   "4"  
[4,] "3"  "3"  "1"  "1"  "3"  "2"  "3"  "3"  "3"  "3"   "3"  
[5,] "2"  "2"  "*"  "*"  "2"  "1"  "2"  "2"  "2"  "2"   "2"  
[6,] "1"  "1"  "1"  "1"  "1"  "*"  "1"  "1"  "1"  "1"   "1"  

我试图为每行找到 5 个连续数字 >= 3 并将初始数字和最终数字的位置写入 txt 文件，在示例中，第 3 行和第 4 行分别有从第 5 列和第 7 列开始的 5 个数字，应该：

initial = [3,5] final [3,9]
initial = [4,7] final [4,11]

以下解决方案几乎可以工作，但错过了第一个识别行（第 3 行）的 ini 和最终值

element<-0
ini<-1
final<-1
consecutives<-0
zz<-file("C:/consecutives.txt","w")
for (i in 1:6){
for (j in 1:11) {
 if (b[i,j] != "*"){
    element<-as.integer(b[i,j])
       if (element>=3)
     {ini<-j
          consecutives<-consecutives+1
      row<-i 
      if (consecutives>=5){
         final<-j
        writeLines(paste("element",toString(element), "Row", toString(row),  "ini", toString(ini),"final",toString(final)) ,con=zz,sep = "\n")
     }

      }


  }
else consecutives<-0

}

}

close(zz)

如果使用模式 numeric 和NA代替"*":

 mdat <- matrix(c(6,2,4,4,NA,5,1,6,NA,2,1,5,1,3,3,5,4,NA,5,NA,1,NA,4,NA,2,2,4,3,4,4,4,NA,4,3,3,1,1,3,2,3,3,3,3,3,2,2,NA,NA,2,1,2,2,2,2,2,1,1,1,1,1,NA,1,1,1,1,1 ),nrow = 6, ncol = 11, byrow = TRUE)

然后你可以使用rle找到序列TRUE值x >= 3:

apply(mdat, 1, function(x) {
  r <- rle(x >= 3)
  w <- which(!is.na(r$values) & r$values & r$lengths > 4)
  if (length(w) > 0) {
    before <- sum(r$lengths[1:(w[1]-1)])
    c(before+1,before+ r$lengths[w[1]])
  } else
    NULL
})

[[1]]
NULL

[[2]]
NULL

[[3]]
[1] 5 9

[[4]]
[1]  7 11

[[5]]
NULL

[[6]]
NULL