R:给定坐标的快速滑动窗口

Question

我有一个数据表,其中nrow大约是一百万或两个,而ncol大约是200.

一行中的每个条目都有一个与之关联的坐标.

数据的微小部分:

[1,] -2.80331471  -0.8874522 -2.34401863   -3.811584   -2.1292443
[2,]  0.03177716   0.2588624  0.82877467    1.955099    0.6321881
[3,] -1.32954665  -0.5433407 -2.19211837   -2.342554   -2.2142461
[4,] -0.60771429  -0.9758734  0.01558774    1.651459   -0.8137684

前4行的坐标:

9928202 9928251 9928288 9928319

我想要的是一个函数,给定数据和窗口大小将返回相同大小的数据表,并在每列上应用平均滑动窗口.换句话说 - 对于每个行条目i,它将找到坐标之间的坐标[i] -windsize和coords [i] + windsize并用该间隔内的值的平均值替换初始值(对于每列分别) .

速度是这里的主要问题.

这是我第一次接受这样的功能.

doSlidingWindow <- function(intensities, coords, windsize) {
windHalfSize <- ceiling(windsize/2)
### whole range inds
RANGE <- integer(max(coords)+windsize)
RANGE[coords] <- c(1:length(coords)[1])

### get indeces of rows falling in each window
COORDS <- as.list(coords)
WINDOWINDS <- sapply(COORDS, function(crds){ unique(RANGE[(crds-windHalfSize):
    (crds+windHalfSize)]) })

### do windowing

wind_ints <- intensities
wind_ints[] <- 0
for(i in 1:length(coords)) {
    wind_ints[i,] <- apply(as.matrix(intensities[WINDOWINDS[[i]],]), 2, mean)
}
return(wind_ints)
}

最后一个for循环之前的代码非常快,它为我提供了我需要为每个条目使用的索引列表.然而,一切都崩溃了,因为我需要研磨for循环一百万次,获取我的数据表的子集,并确保我有多行可以在apply中同时处理所有列.

我的第二种方法是将实际值粘贴在RANGE列表中,用零填充空白并从zoo包中进行rollmean,对每列重复.但这是多余的,因为rollmean将遍历所有间隙,我将只使用最终原始坐标的值.

任何帮助,使其更快,而不去C,将非常感激.

Answer 1

数据生成:

N <- 1e5 # rows
M <- 200 # columns
W <- 10  # window size

set.seed(1)
intensities <- matrix(rnorm(N*M), nrow=N, ncol=M)
coords <- 8000000 + sort(sample(1:(5*N), N))

原始功能,我用于基准测试的微小修改:

doSlidingWindow <- function(intensities, coords, windsize) {
  windHalfSize <- ceiling(windsize/2)
  ### whole range inds
  RANGE <- integer(max(coords)+windsize)
  RANGE[coords] <- c(1:length(coords)[1])

  ### get indices of rows falling in each window
  ### NOTE: Each elements of WINDOWINDS holds zero. Not a big problem though.
  WINDOWINDS <- sapply(coords, function(crds) ret <- unique(RANGE[(crds-windHalfSize):(crds+windHalfSize)]))

  ### do windowing
  wind_ints <- intensities
  wind_ints[] <- 0
  for(i in 1:length(coords)) {
    # CORRECTION: When it's only one row in window there was a trouble
    wind_ints[i,] <- apply(matrix(intensities[WINDOWINDS[[i]],], ncol=ncol(intensities)), 2, mean)
  }
  return(wind_ints)
}

可能的解决方案:

1)data.table

data.table众所周知,子集化速度很快,但这个页面(以及与滑动窗口相关的其他内容)表明情况并非如此.确实,data.table代码很优雅,但遗憾的是很慢:

require(data.table)
require(plyr)
dt <- data.table(coords, intensities)
setkey(dt, coords)
aaply(1:N, 1, function(i) dt[WINDOWINDS[[i]], sapply(.SD,mean), .SDcols=2:(M+1)])

2)foreach + doSNOW

基本例程易于并行运行,因此,我们可以从中受益:

require(doSNOW)
doSlidingWindow2 <- function(intensities, coords, windsize) {
  NC <- 2 # number of nodes in cluster
  cl <- makeCluster(rep("localhost", NC), type="SOCK")
  registerDoSNOW(cl)

  N <- ncol(intensities) # total number of columns
  chunk <- ceiling(N/NC) # number of columns send to the single node

  result <- foreach(i=1:NC, .combine=cbind, .export=c("doSlidingWindow")) %dopar% {
    start <- (i-1)*chunk+1
    end   <- ifelse(i!=NC, i*chunk, N)
    doSlidingWindow(intensities[,start:end], coords, windsize)    
  }

  stopCluster(cl)
  return (result)
}

Benchmark在我的双核处理器上显示出显着的加速:

system.time(res <- doSlidingWindow(intensities, coords, W))
#    user  system elapsed 
# 306.259   0.204 307.770
system.time(res2 <- doSlidingWindow2(intensities, coords, W))
#  user  system elapsed 
# 1.377   1.364 177.223
all.equal(res, res2, check.attributes=FALSE)
# [1] TRUE

3)Rcpp

是的,我知道你问过" 不去C ".但是,请看一下.此代码内联且相当简单:

require(Rcpp)
require(inline)
doSlidingWindow3 <- cxxfunction(signature(intens="matrix", crds="numeric", wsize="numeric"), plugin="Rcpp", body='
  #include <vector>
  Rcpp::NumericMatrix intensities(intens);
  const int N = intensities.nrow();
  const int M = intensities.ncol();
  Rcpp::NumericMatrix wind_ints(N, M);

  std::vector<int> coords = as< std::vector<int> >(crds);
  int windsize = ceil(as<double>(wsize)/2);  

  for(int i=0; i<N; i++){
    // Simple search for window range (begin:end in coords)
    // Assumed that coords are non-decreasing
    int begin = (i-windsize)<0?0:(i-windsize);
    while(coords[begin]<(coords[i]-windsize)) ++begin;
    int end = (i+windsize)>(N-1)?(N-1):(i+windsize);
    while(coords[end]>(coords[i]+windsize)) --end;

    for(int j=0; j<M; j++){
      double result = 0.0;
      for(int k=begin; k<=end; k++){
        result += intensities(k,j);
      }
      wind_ints(i,j) = result/(end-begin+1);
    }
  }

  return wind_ints;
')

基准测试:

system.time(res <- doSlidingWindow(intensities, coords, W))
#    user  system elapsed 
# 306.259   0.204 307.770
system.time(res3 <- doSlidingWindow3(intensities, coords, W))
#  user  system elapsed 
# 0.328   0.020   0.351
all.equal(res, res3, check.attributes=FALSE)
# [1] TRUE

我希望结果非常激励.虽然数据适合内存Rcpp版本非常快.说,有N <- 1e6,M <-100我得到:

   user  system elapsed 
  2.873   0.076   2.951

当然,在R开始使用交换之后,一切都会变慢.对于不适合内存的非常大的数据,您应该考虑sqldf,ff或bigmemory.