2012-03-19 45 views
3

我有一個股票交易事件的xts序列,我想要處理它以生成1分鐘的OHLC時間序列。比如這套交易:R xts:從第二個事件生成1分鐘的時間序列

Timestamp Price Size 
9:30:00.123 12.32 200 
9:30.00.532 12.21 100 
9:30.32.352 12.22 500 
9:30.45.342 12.35 200 

應導致9:30:00記錄:

Timestamp Open High Low Close 
9:30:00 12.32 12.35 12.21 12.35 

我走近這是分裂按分鐘原始交易系列的方式:

myminseries = do.call(rbind, lapply(split(mytrades, "minutes"), myminprocessing)) 

這產生了我想要的記錄,但存在一個問題:如果股票在給定分鐘內沒有任何交易,我將完全錯過那個分鐘記錄。我想要的是爲缺少的交易分鐘創造一個全0的記錄。例如,如果在9:31:00沒有任何交易,我應該有:

Timestamp Open High Low Close 
9:30:00 12.32 12.35 12.21 12.35 
9:31:00 0  0  0  0 
9:32:00 12.40 12.42 12.38 12.42 

如何回填1分鐘系列?或者我應該使用與split()完全不同的方法嗎?

回答

4

to.period()功能,例如to.minute()在xts這樣做。

德克

+2

看起來他可能會想換align.time周圍也。 – GSee 2012-03-19 22:17:18

5

如果有一個給定分鐘內沒有交易,to.minutes「將錯過分鐘的記錄完全」也。你可以通過合併一個零寬度,嚴格規則的xts系列來解決這個問題。

## Make sample data 
> x <- xts(cumsum(rnorm(600, 0, 0.2)), Sys.time() - 600:1) # 10 minutes of secondly data 
> # remove all data from a couple different minutes 
> x['2012-03-19 17:33'] <- NA 
> x['2012-03-19 17:35'] <- NA 
> x <- na.omit(x) 
> 
> ## Convert to minutes 
> xm <- to.minutes(x) 
> head(xm) 
         x.Open x.High  x.Low x.Close 
2012-03-19 17:31:59 0.1945049 1.661000 -0.35943057 1.6610000 
2012-03-19 17:32:59 1.7283877 1.728388 -0.69288918 1.1398868 
2012-03-19 17:34:59 2.0529582 2.603881 -0.80532315 -0.8053232 
2012-03-19 17:36:59 0.5314270 1.189609 -0.94996548 0.5807342 
2012-03-19 17:37:59 0.3761700 1.943363 0.04046976 0.9101720 
2012-03-19 17:38:59 1.0614807 1.722110 -0.22147145 1.4075637 
> axm <- align.time(xm) #align times to begining of next period 
> 
> # to make strictly regular, create an xts object that has values for each minute 
> tmp <- xts(, seq.POSIXt(start(axm), end(axm), by='min')) 
> out <- cbind(tmp, axm) 
> out 
         x.Open  x.High  x.Low  x.Close 
2012-03-19 17:32:00 0.19450494 1.66100005 -0.35943057 1.66100005 
2012-03-19 17:33:00 1.72838773 1.72838773 -0.69288918 1.13988679 
2012-03-19 17:34:00   NA   NA   NA   NA 
2012-03-19 17:35:00 2.05295818 2.60388093 -0.80532315 -0.80532315 
2012-03-19 17:36:00   NA   NA   NA   NA 
2012-03-19 17:37:00 0.53142696 1.18960858 -0.94996548 0.58073422 
2012-03-19 17:38:00 0.37616997 1.94336348 0.04046976 0.91017202 
2012-03-19 17:39:00 1.06148070 1.72211018 -0.22147145 1.40756366 
2012-03-19 17:40:00 1.28437005 1.28437005 -0.62691689 -0.62691689 
2012-03-19 17:41:00 -0.56820166 0.90339983 -0.77554869 0.26101945 
2012-03-19 17:42:00 -0.07443971 -0.07443971 -0.07443971 -0.07443971 
> na.locf(out) 
         x.Open  x.High  x.Low  x.Close 
2012-03-19 17:32:00 0.19450494 1.66100005 -0.35943057 1.66100005 
2012-03-19 17:33:00 1.72838773 1.72838773 -0.69288918 1.13988679 
2012-03-19 17:34:00 1.72838773 1.72838773 -0.69288918 1.13988679 
2012-03-19 17:35:00 2.05295818 2.60388093 -0.80532315 -0.80532315 
2012-03-19 17:36:00 2.05295818 2.60388093 -0.80532315 -0.80532315 
2012-03-19 17:37:00 0.53142696 1.18960858 -0.94996548 0.58073422 
2012-03-19 17:38:00 0.37616997 1.94336348 0.04046976 0.91017202 
2012-03-19 17:39:00 1.06148070 1.72211018 -0.22147145 1.40756366 
2012-03-19 17:40:00 1.28437005 1.28437005 -0.62691689 -0.62691689 
2012-03-19 17:41:00 -0.56820166 0.90339983 -0.77554869 0.26101945 
2012-03-19 17:42:00 -0.07443971 -0.07443971 -0.07443971 -0.07443971 

或者,如果你真的想零點時,有沒有價值,你可以做out[is.na(out)] <- 0