我正在尝试计算15天均线的指数移动平均值,但希望看到15天均线在每个(结束)天/酒吧的“演变”。所以,这意味着我有15天的酒吧。当每天都有新的数据出现时,我想使用新的信息重新计算EMA。实际上,我有15天的条形图,然后,每天之后,我的新的15天条形图开始增长,每个新的条形图都应该与之前完整的15天条形图一起用于EMA计算。
假设我们从2012-01-01开始(对于这个例子,我们有每个日历日的数据),在2012-01-15结束时,我们有第一个完整的15天条形图。在2012-03-01完成了4个完整的15天条形之后,我们可以开始计算4个条形均线( EMA (x,n=4))。在2012-03-02年末,我们使用到现在为止的信息,计算2012-03-02的EMA,假装2012-03-02的OHLC是进行中的15天吧。因此,我们取4个完整的条形和2012-03-02的条形,并计算均方根均值(x,n=4)。然后我们再等一天,看看新的15天条在进行中发生了什么(详细信息请参见下面的函数to.period.cumulative ),并计算EMA的新值。所以在接下来的15天里...有关详细信息,请参阅下面的函数EMA.cumulative ...
请在下面找到我到目前为止所能想到的。性能对我来说是不可接受的,以我有限的R知识,我不能让它更快。
library(quantmod)
do.call.rbind <- function(lst) {
while(length(lst) > 1) {
idxlst <- seq(from=1, to=length(lst), by=2)
lst <- lapply(idxlst, function(i) {
if(i==length(lst)) { return(lst[[i]]) }
return(rbind(lst[[i]], lst[[i+1]]))
})
}
lst[[1]]
}
to.period.cumulative <- function(x, name=NULL, period="days", numPeriods=15) {
if(is.null(name))
name <- deparse(substitute(x))
cnames <- c("Open", "High", "Low", "Close")
if (has.Vo(x))
cnames <- c(cnames, "Volume")
cnames <- paste(name, cnames, sep=".")
if (quantmod:::is.OHLCV(x)) {
x <- OHLCV(x)
out <- do.call.rbind(
lapply(split(x, f=period, k=numPeriods),
function(x) cbind(rep(first(x[,1]), NROW(x[,1])),
cummax(x[,2]), cummin(x[,3]), x[,4], cumsum(x[,5]))))
} else if (quantmod:::is.OHLC(x)) {
x <- OHLC(x)
out <- do.call.rbind(
lapply(split(x, f=period, k=numPeriods),
function(x) cbind(rep(first(x[,1]), NROW(x[,1])),
cummax(x[,2]), cummin(x[,3]), x[,4])))
} else {
stop("Object does not have OHLC(V).")
}
colnames(out) <- cnames
return(out)
}
EMA.cumulative<-function(cumulativeBars, nEMA = 4, period="days", numPeriods=15) {
barsEndptCl <- Cl(cumulativeBars[endpoints(cumulativeBars, on=period, k=numPeriods)])
# TODO: This is sloooooooooooooooooow...
outEMA <- do.call.rbind(
lapply(split(Cl(cumulativeBars), period),
function(x) {
previousFullBars <- barsEndptCl[index(barsEndptCl) < last(index(x)), ]
if (NROW(previousFullBars) >= (nEMA - 1)) {
last(EMA(last(rbind(previousFullBars, x), n=(nEMA + 1)), n=nEMA))
} else {
xts(NA, order.by=index(x))
}
}))
colnames(outEMA) <- paste("EMA", nEMA, sep="")
return(outEMA)
}
getSymbols("SPY", from="2010-01-01")
SPY.cumulative <- to.period.cumulative(SPY, , name="SPY")
system.time(
SPY.EMA <- EMA.cumulative(SPY.cumulative)
)在我的系统上
user system elapsed
4.708 0.000 4.410 可接受的执行时间将少于一秒...是否有可能使用纯R来实现这一点?
这篇文章链接到了Optimize moving averages calculation - is it possible?,在那里我没有收到任何回复。我现在能够创建一个可重复的示例,并对我想要加速的内容进行更详细的解释。我希望现在这个问题更有意义了。
任何关于如何加速这一过程的想法都将受到高度赞赏。
发布于 2012-01-16 05:36:08
我没有找到一个令人满意的解决方案,我的问题使用R,所以我使用旧的工具,c语言,结果比我预期的要好。感谢你“推”我使用这个伟大的工具的Rcpp,内联等。神奇。我想,当我将来有性能要求,而使用R不能满足时,我会将C添加到R中,性能就在那里。所以,请看下面我的代码和性能问题的解决方案。
# How to speedup cumulative EMA calculation
#
###############################################################################
library(quantmod)
library(Rcpp)
library(inline)
library(rbenchmark)
do.call.rbind <- function(lst) {
while(length(lst) > 1) {
idxlst <- seq(from=1, to=length(lst), by=2)
lst <- lapply(idxlst, function(i) {
if(i==length(lst)) { return(lst[[i]]) }
return(rbind(lst[[i]], lst[[i+1]]))
})
}
lst[[1]]
}
to.period.cumulative <- function(x, name=NULL, period="days", numPeriods=15) {
if(is.null(name))
name <- deparse(substitute(x))
cnames <- c("Open", "High", "Low", "Close")
if (has.Vo(x))
cnames <- c(cnames, "Volume")
cnames <- paste(name, cnames, sep=".")
if (quantmod:::is.OHLCV(x)) {
x <- quantmod:::OHLCV(x)
out <- do.call.rbind(
lapply(split(x, f=period, k=numPeriods),
function(x) cbind(rep(first(x[,1]), NROW(x[,1])),
cummax(x[,2]), cummin(x[,3]), x[,4], cumsum(x[,5]))))
} else if (quantmod:::is.OHLC(x)) {
x <- OHLC(x)
out <- do.call.rbind(
lapply(split(x, f=period, k=numPeriods),
function(x) cbind(rep(first(x[,1]), NROW(x[,1])),
cummax(x[,2]), cummin(x[,3]), x[,4])))
} else {
stop("Object does not have OHLC(V).")
}
colnames(out) <- cnames
return(out)
}
EMA.cumulative<-function(cumulativeBars, nEMA = 4, period="days", numPeriods=15) {
barsEndptCl <- Cl(cumulativeBars[endpoints(cumulativeBars, on=period, k=numPeriods)])
# TODO: This is sloooooooooooooooooow...
outEMA <- do.call.rbind(
lapply(split(Cl(cumulativeBars), period),
function(x) {
previousFullBars <- barsEndptCl[index(barsEndptCl) < last(index(x)), ]
if (NROW(previousFullBars) >= (nEMA - 1)) {
last(EMA(last(rbind(previousFullBars, x), n=(nEMA + 1)), n=nEMA))
} else {
xts(NA, order.by=index(x))
}
}))
colnames(outEMA) <- paste("EMA", nEMA, sep="")
return(outEMA)
}
EMA.c.c.code <- '
/* Initalize loop and PROTECT counters */
int i, P=0;
/* ensure that cumbars and fullbarsrep is double */
if(TYPEOF(cumbars) != REALSXP) {
PROTECT(cumbars = coerceVector(cumbars, REALSXP)); P++;
}
/* Pointers to function arguments */
double *d_cumbars = REAL(cumbars);
int i_nper = asInteger(nperiod);
int i_n = asInteger(n);
double d_ratio = asReal(ratio);
/* Input object length */
int nr = nrows(cumbars);
/* Initalize result R object */
SEXP result;
PROTECT(result = allocVector(REALSXP,nr)); P++;
double *d_result = REAL(result);
/* Find first non-NA input value */
int beg = i_n*i_nper - 1;
d_result[beg] = 0;
for(i = 0; i <= beg; i++) {
/* Account for leading NAs in input */
if(ISNA(d_cumbars[i])) {
d_result[i] = NA_REAL;
beg++;
d_result[beg] = 0;
continue;
}
/* Set leading NAs in output */
if(i < beg) {
d_result[i] = NA_REAL;
}
/* Raw mean to start EMA - but only on full bars*/
if ((i != 0) && (i%i_nper == (i_nper - 1))) {
d_result[beg] += d_cumbars[i] / i_n;
}
}
/* Loop over non-NA input values */
int i_lookback = 0;
for(i = beg+1; i < nr; i++) {
i_lookback = i%i_nper;
if (i_lookback == 0) {
i_lookback = 1;
}
/*Previous result should be based only on full bars*/
d_result[i] = d_cumbars[i] * d_ratio + d_result[i-i_lookback] * (1-d_ratio);
}
/* UNPROTECT R objects and return result */
UNPROTECT(P);
return(result);
'
EMA.c.c <- cfunction(signature(cumbars="numeric", nperiod="numeric", n="numeric", ratio="numeric"), EMA.c.c.code)
EMA.cumulative.c<-function(cumulativeBars, nEMA = 4, period="days", numPeriods=15) {
ratio <- 2/(nEMA+1)
outEMA <- EMA.c.c(cumbars=Cl(cumulativeBars), nperiod=numPeriods, n=nEMA, ratio=ratio)
outEMA <- reclass(outEMA, Cl(cumulativeBars))
colnames(outEMA) <- paste("EMA", nEMA, sep="")
return(outEMA)
}
getSymbols("SPY", from="2010-01-01")
SPY.cumulative <- to.period.cumulative(SPY, name="SPY")
system.time(
SPY.EMA <- EMA.cumulative(SPY.cumulative)
)
system.time(
SPY.EMA.c <- EMA.cumulative.c(SPY.cumulative)
)
res <- benchmark(EMA.cumulative(SPY.cumulative), EMA.cumulative.c(SPY.cumulative),
columns=c("test", "replications", "elapsed", "relative", "user.self", "sys.self"),
order="relative",
replications=10)
print(res)编辑:为了给出我笨拙的性能改进的指示(我相信它可以变得更好,因为实际上我已经创建了double for循环)R这里是一个打印输出:
> print(res)
test replications elapsed relative user.self
2 EMA.cumulative.c(SPY.cumulative) 10 0.026 1.000 0.024
1 EMA.cumulative(SPY.cumulative) 10 57.732 2220.462 56.755所以,根据我的标准,一种科幻类型的改进...
https://stackoverflow.com/questions/8720055
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