R-hist(XX，plot = FALSE)$ count的更快替代方法

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问题描述

我正在寻找一种更快的替代R的hist(x, breaks=XXX, plot=FALSE)$count函数的方法，因为我不需要生成任何其他输出(因为我想在sapply调用中使用它，因此需要进行一百万次迭代其中将调用此函数)，例如

I am on the lookout for a faster alternative to R's hist(x, breaks=XXX, plot=FALSE)$count function as I don't need any of the other output that is produced (as I want to use it in an sapply call, requiring 1 million iterations in which this function would be called), e.g.

x = runif(100000000, 2.5, 2.6)
bincounts = hist(x, breaks=seq(0,3,length.out=100), plot=FALSE)$count

有什么想法吗?

推荐答案

首次尝试使用table和cut:

table(cut(x, breaks=seq(0,3,length.out=100)))

它避免了多余的输出，但是在我的计算机上大约需要34秒:

It avoids the extra output, but takes about 34 seconds on my computer:

system.time(table(cut(x, breaks=seq(0,3,length.out=100))))
   user  system elapsed 
 34.148   0.532  34.696

相比，hist为3.5秒:

system.time(hist(x, breaks=seq(0,3,length.out=100), plot=FALSE)$count)
   user  system elapsed 
  3.448   0.156   3.605

使用tabulate和.bincode的速度比hist快一点:

Using tabulate and .bincode runs a little bit faster than hist:

tabulate(.bincode(x, breaks=seq(0,3,length.out=100)), nbins=100)

system.time(tabulate(.bincode(x, breaks=seq(0,3,length.out=100))), nbins=100)
   user  system elapsed 
  3.084   0.024   3.107

使用tablulate和findInterval相对于table和cut可以显着提高性能，并且相对于hist可以改善:

Using tablulate and findInterval provides a significant performance boost relative to table and cut and has an OK improvement relative to hist:

tabulate(findInterval(x, vec=seq(0,3,length.out=100)), nbins=100)

system.time(tabulate(findInterval(x, vec=seq(0,3,length.out=100))), nbins=100)
   user  system elapsed 
  2.044   0.012   2.055

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