用于在R中构建逻辑的计算-ComputationstobuildalogicinR

2  

Here is an R solution using rleid from data.table:

这是使用data.table中的rleid的R解决方案：

library(data.table)
df[t(apply(df, 1, function(x) shift(rleid(x) == max(rleid(x)))))] <- NA

Result:

结果：

  C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15
1 12  3  3  4  5  6  7 NA NA  NA  NA  NA  NA  NA  NA
2  1  4  5  5  5  5  5  5  6   7   8  NA  NA  NA  NA
3  3  4  4  4  4 45 32 57 23  NA  NA  NA  NA  NA  NA

Note that this works even if the repeating integer appears anywhere in each row not connected to the repeating sequence at the end.

请注意，即使重复整数出现在每行中没有连接到末尾重复序列的任何位置，这仍然有效。

Data:

数据：

df = structure(list(C1 = c(12L, 1L, 3L), C2 = c(3L, 4L, 4L), C3 = c(3L, 
5L, 4L), C4 = c(4L, 5L, 4L), C5 = c(5L, 5L, 4L), C6 = c(6L, 5L, 
45L), C7 = c(7L, 5L, 32L), C8 = c(7L, 5L, 57L), C9 = c(7L, 6L, 
23L), C10 = c(7L, 7L, 23L), C11 = c(7L, 8L, 23L), C12 = c(7L, 
8L, 23L), C13 = c(7L, 8L, 23L), C14 = c(7L, 8L, 23L), C15 = c(7L, 
8L, 23L)), .Names = c("C1", "C2", "C3", "C4", "C5", "C6", "C7", 
"C8", "C9", "C10", "C11", "C12", "C13", "C14", "C15"), class = "data.frame", row.names = c(NA, 
-3L))

1  

With base R you can do it like the following.
This is the third version of the function, thanks to @useR for pointing out the wrong results of the others in some use cases.

使用基础R，您可以像下面这样做。这是该函数的第三个版本，感谢@useR在某些用例中指出其他函数的错误结果。

fun <- function(x, n){
  r <- rle(x)
  n <- length(x)
  x[(n - r$lengths[length(r$lengths)] + 2):n] <- NA
  x
}

x <- c(12,3,3,4,5,6,7,7,7,7,7,7,7,7,7)
fun(x)
#[1] 12  3  3  4  5  6  7 NA NA NA NA NA NA NA NA

y <- c(12,7,7,4,5,6,7,7,7,7,7,7,7,7,7)
fun(y)
#[1] 12  7  7  4  5  6  7 NA NA NA NA NA NA NA NA

Now with a data.frame.

现在有了data.frame。

dat[] <- t(apply(dat, 1, fun))
#  C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15
#1 12  3  3  4  5  6  7 NA NA  NA  NA  NA  NA  NA  NA
#2  1  4  5  5  5  5  5  5  6   7   8  NA  NA  NA  NA
#3  3  4  4  4  4 45 32 57 23  NA  NA  NA  NA  NA  NA

Data.

数据。

dat <- read.csv(text = "
C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C11,C12,C13,C14,C15
12,3,3,4,5,6,7,7,7,7,7,7,7,7,7
1,4,5,5,5,5,5,5,6,7,8,8,8,8,8
3,4,4,4,4,45,32,57,23,23,23,23,23,23,23
")

1  

Data:

数据：

df1 <- read.table(text='C1,C2,C3,C4,C5,C6,C7,C8,C9,C10,C11,C12,C13,C14,C15
12,3,3,4,5,6,7,7,7,7,7,7,7,7,7
                  1,4,5,5,5,5,5,5,6,7,8,8,8,8,8
                  3,4,4,4,4,45,32,57,23,23,23,23,23,23,23', sep = ",", header = TRUE, stringsAsFactors = FALSE)

Code:

码：

 apply(df1, 1, function(x) {
  x <- rle(x)
  len_x <- length(x$lengths)

  if( (x$lengths)[len_x] > 1 ){ # check for end sequence
    x <- list(lengths = c(x$lengths[-len_x], 1, x$lengths[len_x]- 1 ),
              values = c(x$values, NA))
  } 
  inverse.rle(x)
})

Output:

输出：

#       [,1] [,2] [,3]
# [1,]   12    1    3
# [2,]    3    4    4
# [3,]    3    5    4
# [4,]    4    5    4
# [5,]    5    5    4
# [6,]    6    5   45
# [7,]    7    5   32
# [8,]   NA    5   57
# [9,]   NA    6   23
# [10,]   NA    7   NA
# [11,]   NA    8   NA
# [12,]   NA   NA   NA
# [13,]   NA   NA   NA
# [14,]   NA   NA   NA
# [15,]   NA   NA   NA

0  

Before:

之前：

the code:

代码：

Sub marine()
    For i = 2 To 4
        For j = 15 To 2 Step -1
            If Cells(i, j).Value = Cells(i, j - 1).Value Then
                Cells(i, j).Value = "#N/A"
            Else
                Exit For
            End If
        Next j
    Next i
End Sub

and after:

之后：

0  

this only uses one loop:

这只使用一个循环：

Sub nadup()

With Worksheets("Sheet5") 'change to your sheet
    Dim i As Long
    For i = 2 To 4
        Dim t As Long
        t = Application.Match(.Cells(i, 15), Range(i & ":" & i), 0)
        If t <14 Then
            .Range(.Cells(i, t + 1), .Cells(i, 15)).Value = "N/A"
        ElseIf t = 14 Then
            .Cells(i, 15) = "N/A"
        End If
    Next i
End With



End Sub

0  

With arrays

有阵列

Option Explicit

Public Sub ReplaceItems()
    Dim arr(), arr2(), i As Long, j As Long
    With ActiveSheet
        arr() = .Range("A1:O3").Value
        For i = LBound(arr, 1) To UBound(arr, 1)
            ReDim arr2(1 To UBound(arr, 2))
            For j = LBound(arr, 2) To UBound(arr, 2)
                If IsError(Application.Match(arr(i, j), arr2, 0)) Then
                    arr2(j) = arr(i, j)
                Else
                    Select Case arr(i, j)
                    Case 7, 8, 23
                    arr2(j) = "#N/A"
                    Case Else
                    arr2(j) = arr(i, j)
                    End Select
                End If
            Next
            .Range("A" & i).Resize(1, UBound(arr2)).Value = arr2
        Next i
    End With
End Su