Question

假设我有一个

形式的矩阵（或向量）

var outputPDFDocument = PdfDocument.Combine("Output.pdf", "doc1.pdf", "doc2.pdf", "doc3.pdf");
outputPDFDocument.Save("C:\\temp.pdf"); 

PdfFilePrinter.AdobeReaderPath = @"C:\Program Files\Adobe\Reader 8.0\Reader\AcroRd32.exe";  
PdfFilePrinter.DefaultPrinterName = "Take printer";   
PdfFilePrinter printer = new PdfFilePrinter(@"C:\\temp.pdf");  
printer.Print();

如何计算每列中1之间的连续零的数量，并将那些数量小于预定义常数k的零替换为0。或者至少获得每个零序列中的起始索引和元素数量。通常，此数据集中的零比实际值多得多，并且大多数情况下序列的长度大于k

因此，例如，如果k = 1，那么[4,2]; [13,1]和[15,1]将被替换为1.如果k = 2，则除了[4， 1]; [13,1]和[15,1]，[3,1]，[4,1]，[14,2]和[15,2]中的零也将被1替换在这个例子中。

当然，我可以运行一个循环并遍历所有行。我想知道是否有一个包，或一个可以做到的整洁的矢量化技巧。

更新

k = 1

的期望输出示例

>set.seed(1)
>X=ifelse(matrix((runif(30)),ncol = 2)>0.4,0,1)

      [,1] [,2]
 [1,]    1    1
 [2,]    1    1
 [3,]    0    1
 [4,]    0    0
 [5,]    1    1
 [6,]    0    0
 [7,]    0    0
 [8,]    0    0
 [9,]    0    1
[10,]    1    0
[11,]    1    0
[12,]    1    0
[13,]    0    1
[14,]    1    0
[15,]    0    0
    ...
    etc

k = 2的期望输出

      [,1] [,2]
 [1,]    1    1
 [2,]    1    1
 [3,]    0    1
 [4,]    0    1
 [5,]    1    1
 [6,]    0    0
 [7,]    0    0
 [8,]    0    0
 [9,]    0    1
[10,]    1    0
[11,]    1    0
[12,]    1    0
[13,]    1    1
[14,]    1    0
[15,]    1    0

Answer 1

游程工具rle在这里工作：

fill_shortruns <- function(X,k=1,badval=0,newval=1){
    apply(X,2,function(x){
        r <- rle(x)
        r$values[ r$lengths <= k & r$values == badval ] <- newval
        inverse.rle(r)
    })
}

# smaller example

set.seed(1)
X0=ifelse(matrix((runif(10)),ncol = 2)>0.4,0,1)
#      [,1] [,2] [,3] [,4]
# [1,]    1    0    1    0
# [2,]    1    0    1    0
# [3,]    0    0    0    0
# [4,]    0    0    1    1
# [5,]    1    1    0    0

fill_shortruns(X0,2)
#      [,1] [,2] [,3] [,4]
# [1,]    1    0    1    0
# [2,]    1    0    1    0
# [3,]    1    0    1    0
# [4,]    1    0    1    1
# [5,]    1    1    1    1

用期望值替换连续的零

1 个答案: