Question

在R中，我试图在众多数据文件中创建元素的计数矩阵：

rnames <- c("N","A")
mymatrix <- matrix(nrow=2,ncol=0,dimnames=list(rnames))

#loop through hundreds of large files (MB)
#make the vector "names" contain all elements within each file 
for(name in names)
{
#if name is already in the matrix increment by 1 the second row
  if(name %in% colnames(mymatrix))
  {
    mymatrix[2,name] = mymatrix[2,name]+1
  }
#else add a column to the matrix with the specified name
  else
  {
    mymatrix <- transform(mymatrix,name)
    mymatrix[2,name] = 1
  }    
}

我运行了Rprof命令，发现可能嵌入％in％运算符中的match（）函数导致性能问题（执行时间更长）

如果向量中的每个元素存在于我的矩阵中，它是否有更有效的方法来检查它，如果它没有在矩阵中用该向量元素作为列名创建新列？< / p>

如果你想要一个可重现的代码，那么......但请记住，我的原始代码中的名称向量是从包含数千个变量的大文件中读取的，这些变量与mymatrix中不断增加的列号相匹配，最终导致运行时间增加：

rnames <- c("N","A")
mymatrix <- matrix(nrow=2,ncol=0,dimnames=list(rnames))

#suppose this is what the first file contains
names <- c("x","y","z","x","x","y","a")

#suppose this is what the second file contains
names <- c("x","y","z","x","x","x","x","k")


  for(name in names)
  {
    if(name %in% colnames(mymatrix))
    {
      mymatrix[2,name] = mymatrix[2,name] + 1
    }
    else
    {
      mymatrix <- transform(mymatrix,name)
      mymatrix[2,name] = 1
    }

  }


the expected output
> mymatrix
   x  y  z  a  k
N NA NA NA NA NA
A  8  3  2  1  1

Answer 1

我不知道你是如何确定match是瓶颈的。它可能是，但您提供的示例并未显示。

rnames <- c("N","A")
mymatrix <- matrix(nrow=2, ncol=0, dimnames=list(rnames))
set.seed(21)
names <- sample(letters, 1e6, TRUE)
Rprof()
for(name in names) {
  if(name %in% colnames(mymatrix)) {
    mymatrix[2,name] <- mymatrix[2,name] + 1
  } else {
    mymatrix <- transform(mymatrix,name)
    mymatrix[2,name] <- 1
  }
}
Rprof(NULL)

下面的结果显示瓶颈是data.frame方法，由于您使用transform而调用这些方法。 transform.default会将您的矩阵转换为data.frame，然后调用transform.data.frame，其中包括对match的调用。

R> lapply(summaryRprof(), head)
$by.self
                 self.time self.pct total.time total.pct
"[<-.data.frame"     12.02    26.15      25.90     56.35
"[.data.frame"        7.22    15.71      13.32     28.98
"match"               7.20    15.67      11.40     24.80
"%in%"                2.38     5.18      12.34     26.85
"anyDuplicated"       2.22     4.83       3.08      6.70
"names"               2.16     4.70       2.16      4.70

$by.total
                 total.time total.pct self.time self.pct
"[<-"                 27.06     58.88      1.16     2.52
"[<-.data.frame"      25.90     56.35     12.02    26.15
"["                   14.32     31.16      1.00     2.18
"[.data.frame"        13.32     28.98      7.22    15.71
"%in%"                12.34     26.85      2.38     5.18
"match"               11.40     24.80      7.20    15.67

$sample.interval
[1] 0.02

$sampling.time
[1] 45.96

避免transform调用，您的代码会明显加快。 mymatrix2实际上是一个矩阵，而mymatrix是一个data.frame。

rnames <- c("N","A")
mymatrix2 <- matrix(nrow=2, ncol=0, dimnames=list(rnames))
set.seed(21)
names <- sample(letters, 1e6, TRUE)
Rprof()
for(name in names) {
  if(name %in% colnames(mymatrix)) {
    mymatrix2[2,name] <- mymatrix2[2,name] + 1
  } else {
    mymatrix2 <- cbind(mymatrix2, matrix(c(NA,1), 2, 1, dimnames=list(rnames, name)))
  }
}
Rprof(NULL)
lapply(summaryRprof(), head)
$by.self
                self.time self.pct total.time total.pct
"match"              1.28    41.83       2.70     88.24
"colnames"           0.78    25.49       1.42     46.41
"is.data.frame"      0.58    18.95       0.58     18.95
"%in%"               0.34    11.11       3.04     99.35
"dimnames"           0.06     1.96       0.06      1.96
"+"                  0.02     0.65       0.02      0.65

$by.total
                total.time total.pct self.time self.pct
"%in%"                3.04     99.35      0.34    11.11
"match"               2.70     88.24      1.28    41.83
"colnames"            1.42     46.41      0.78    25.49
"is.data.frame"       0.58     18.95      0.58    18.95
"dimnames"            0.06      1.96      0.06     1.96
"+"                   0.02      0.65      0.02     0.65
identical(mymatrix2, as.matrix(mymatrix))
[1] TRUE

R的绩效问题

1 个答案: