在 XTS 中滚动列表的次数不等 [英] Rolling list over unequal times in XTS

问题描述

我有分时级别的股票数据，想创建一个包含前 10 秒所有分时的滚动列表.下面的代码有效，但处理大量数据需要很长时间.我想对这个过程进行矢量化或以其他方式使其更快，但我没有想出任何东西.任何朝着正确方向的建议或推动将不胜感激.

I have stock data at the tick level and would like to create a rolling list of all ticks for the previous 10 seconds. The code below works, but takes a very long time for large amounts of data. I'd like to vectorize this process or otherwise make it faster, but I'm not coming up with anything. Any suggestions or nudges in the right direction would be appreciated.

library(quantmod)
set.seed(150)

# Create five minutes of xts example data at .1 second intervals
mins  <- 5
ticks <- mins * 60 * 10 + 1


times <- xts(runif(seq_len(ticks),1,100), order.by=seq(as.POSIXct("1973-03-17 09:00:00"),
                                                       as.POSIXct("1973-03-17 09:05:00"), length = ticks))

# Randomly remove some ticks to create unequal intervals
times <- times[runif(seq_along(times))>.3]

# Number of seconds to look back
lookback  <- 10
dist.list <- list(rep(NA, nrow(times)))

system.time(
  for (i in 1:length(times)) {

    dist.list[[i]] <- times[paste(strptime(index(times[i])-(lookback-1), format = "%Y-%m-%d %H:%M:%S"), "/",
                                  strptime(index(times[i])-1, format = "%Y-%m-%d %H:%M:%S"), sep = "")]
  }
)
>  user  system elapsed 
   6.12    0.00    5.85 

推荐答案

你应该看看 window 功能，它会让你更容易地选择日期.下面的代码使用lapply来完成for循环的工作.

You should check out the window function, it will make your subselection of dates a lot easier. The following code uses lapply to do the work of the for loop.

# Your code
system.time(
  for (i in 1:length(times)) {

    dist.list[[i]] <- times[paste(strptime(index(times[i])-(lookback-1), format = "%Y-%m-%d %H:%M:%S"), "/",
                                  strptime(index(times[i])-1, format = "%Y-%m-%d %H:%M:%S"), sep = "")]
  }
  )

#    user  system elapsed 
#    10.09    0.00   10.11

# My code 
system.time(dist.list<-lapply(index(times),
    function(x) window(times,start=x-lookback-1,end=x))
)
#    user  system elapsed 
#    3.02    0.00    3.03 

所以，大约快了三分之一.

So, about a third faster.

但是，如果你真的想加快速度，并且你愿意放弃毫秒精度(我认为你的原始方法隐含地这样做)，你可以在独特的日期-小时-秒组合上运行循环，因为它们都将返回相同的时间窗口.这应该会加快大约二十或三十倍的速度:

But, if you really want to speed things up, and you are willing to forgo millisecond accuracy (which I think your original method implicitly does), you could just run the loop on unique date-hour-second combinations, because they will all return the same time window. This should speed things up roughly twenty or thirty times:

dat.time=unique(as.POSIXct(as.character(index(times)))) # Cheesy method to drop the ms.
system.time(dist.list.2<-lapply(dat.time,function(x) window(times,start=x-lookback-1,end=x)))

# user  system elapsed 
# 0.37    0.00    0.39 

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