jos*_*ber 11 r moving-average zoo
zoo::rollmean是一个有用的函数,它返回时间序列的滚动均值; 对于x长度n和窗口大小k的向量,它返回向量c(mean(x[1:k]), mean(x[2:(k+1)]), ..., mean(x[(n-k+1):n])).
我注意到我正在开发的一些代码似乎运行缓慢,所以我使用Rcpp包和一个简单的for循环编写了我自己的版本:
library(Rcpp)
cppFunction("NumericVector rmRcpp(NumericVector dat, const int window) {
const int n = dat.size();
NumericVector ret(n-window+1);
double summed = 0.0;
for (int i=0; i < window; ++i) {
summed += dat[i];
}
ret[0] = summed / window;
for (int i=window; i < n; ++i) {
summed += dat[i] - dat[i-window];
ret[i-window+1] = summed / window;
}
return ret;
}")
令我惊讶的是,这个版本的函数比zoo::rollmean函数快得多:
# Time series with 1000 elements
set.seed(144)
y <- rnorm(1000)
x <- 1:1000
library(zoo)
zoo.dat <- zoo(y, x)
# Make sure our function works
all.equal(as.numeric(rollmean(zoo.dat, 3)), rmRcpp(y, 3))
# [1] TRUE
# Benchmark
library(microbenchmark)
microbenchmark(rollmean(zoo.dat, 3), rmRcpp(y, 3))
# Unit: microseconds
# expr min lq mean median uq max neval
# rollmean(zoo.dat, 3) 685.494 904.7525 1776.88666 1229.2475 1744.0720 15724.321 100
# rmRcpp(y, 3) 6.638 12.5865 46.41735 19.7245 27.4715 2418.709 100
即使对于更大的向量,加速也是如此:
# Time series with 5 million elements
set.seed(144)
y <- rnorm(5000000)
x <- 1:5000000
library(zoo)
zoo.dat <- zoo(y, x)
# Make sure our function works
all.equal(as.numeric(rollmean(zoo.dat, 3)), rmRcpp(y, 3))
# [1] TRUE
# Benchmark
library(microbenchmark)
microbenchmark(rollmean(zoo.dat, 3), rmRcpp(y, 3), times=10)
# Unit: milliseconds
# expr min lq mean median uq max
# rollmean(zoo.dat, 3) 2825.01622 3090.84353 3191.87945 3206.00357 3318.98129 3616.14047
# rmRcpp(y, 3) 31.03014 39.13862 42.67216 41.55567 46.35191 53.01875
为什么简单的Rcpp实现运行速度比zoo::rollmean？快100倍？
感谢@DirkEddelbuettel指出在问题中进行的比较并不是最公平的,因为我将C++代码与纯R代码进行比较.以下是一个简单的基本R实现(没有来自zoo包的所有检查); 这非常类似于如何zoo::rollmean实现滚动平均值的核心计算:
baseR.rollmean <- function(dat, window) {
n <- length(dat)
y <- dat[window:n] - dat[c(1, 1:(n-window))]
y[1] <- sum(dat[1:window])
return(cumsum(y) / window)
}
相比之下zoo:rollmean,我们发现这仍然是一个很好的交易:
set.seed(144)
y <- rnorm(1000000)
x <- 1:1000000
library(zoo)
zoo.dat <- zoo(y, x)
all.equal(as.numeric(rollmean(zoo.dat, 3)), baseR.rollmean(y, 3), RcppRoll::roll_mean(y, 3), rmRcpp(y, 3))
# [1] TRUE
library(microbenchmark)
microbenchmark(rollmean(zoo.dat, 3), baseR.rollmean(y, 3), RcppRoll::roll_mean(y, 3), rmRcpp(y, 3), times=10)
# Unit: milliseconds
# expr min lq mean median uq max neval
# rollmean(zoo.dat, 3) 507.124679 516.671897 646.813716 563.897005 593.861499 1220.08272 10
# baseR.rollmean(y, 3) 46.156480 47.804786 53.923974 49.250144 55.061844 76.47908 10
# RcppRoll::roll_mean(y, 3) 7.714032 8.513042 9.014886 8.693255 8.885514 11.32817 10
# rmRcpp(y, 3) 7.729959 8.045270 8.924030 8.388931 8.996384 12.49042 10
为了深入研究为什么我们在使用base R时看到10倍的加速,我使用了Hadley的lineprof工具,zoo在需要时从包源获取源代码:
lineprof(rollmean.zoo(zoo.dat, 3))
# time alloc release dups ref src
# 1 0.001 0.954 0 26 #27 rollmean.zoo/unclass
# 2 0.001 0.954 0 0 #28 rollmean.zoo/:
# 3 0.002 0.954 0 1 #28 rollmean.zoo
# 4 0.001 1.431 0 0 #28 rollmean.zoo/seq_len
# 5 0.001 0.000 0 0 #28 rollmean.zoo/c
# 6 0.006 2.386 0 1 #28 rollmean.zoo
# 7 0.002 0.954 0 2 #31 rollmean.zoo/cumsum
# 8 0.001 0.000 0 0 #31 rollmean.zoo//
# 9 0.005 1.912 0 1 #33 rollmean.zoo
# 10 0.013 2.898 0 14 #33 rollmean.zoo/[<-
# 11 0.299 28.941 0 127 #34 rollmean.zoo/na.fill
显然,几乎所有的时间都花在na.fill函数上,实际上在已经计算了滚动平均值之后调用了函数.
lineprof(na.fill.zoo(zoo.dat, fill=NA, 2:999999))
# time alloc release dups ref src
# 1 0.004 1.913 0 39 #26 na.fill.zoo/seq
# 2 0.002 1.921 0 9 #33 na.fill.zoo/coredata
# 3 0.002 1.921 0 6 #37 na.fill.zoo/[<-
# 4 0.001 0.955 0 10 #46 na.fill.zoo
# 5 0.008 3.838 0 19 #46 na.fill.zoo/[<-
# 6 0.003 0.959 0 2 #52 na.fill.zoo
# 7 0.006 0.972 0 21 #52 na.fill.zoo/[<-
# 8 0.001 0.486 0 0 #57 na.fill.zoo/seq_len
# 9 0.005 0.959 0 6 #66 na.fill.zoo
# 10 0.124 11.573 0 34 #66 na.fill.zoo/[
几乎所有的时间都花在zoo对象的子集上:
lineprof("[.zoo"(zoo.dat, 2:999999))
# time alloc release dups ref src
# 1 0.004 0.004 0 0 character(0)
# 2 0.002 1.922 0 4 #4 [.zoo/coredata
# 3 0.038 11.082 0 29 #19 [.zoo/zoo
# 4 0.004 0.000 0 1 #28 [.zoo
几乎所有的时间子集用于构建具有以下zoo函数的新zoo对象:
lineprof(zoo(y[2:999999], 2:999999))
# time alloc release dups ref src
# 1 0.021 4.395 0 8 c("zoo", "unique") zoo/unique
# 2 0.012 0.477 0 8 c("zoo", "ORDER") zoo/ORDER
# 3 0.001 0.477 0 1 "zoo" zoo
# 4 0.001 0.954 0 0 c("zoo", ":") zoo/:
# 5 0.015 3.341 0 5 "zoo" zoo
设置新的动物园对象需要各种操作(例如,确定唯一的时间点并对它们进行排序).
总之,zoo通过构造一个新的zoo对象而不是使用当前zoo对象的内部,该包似乎为其滚动平均操作增加了大量开销; 与基本R实现相比,这会产生10倍的减速,与Rcpp实现相比,速度减慢100倍.