如何在R中矢量化函数

Question

我需要一些帮助来矢量化下面的代码，因为我相信它会变得更高效。不过，我不知道如何开始...我创建了一个循环，通过z。 z有3列和112847行，这可能是需要很长时间的原因。 3列包含了在该MACD()功能使用的号码......

library(quantmod) 
library(TTR) 

# get stock data 
getSymbols('LUNA') 

#Choose the Adjusted Close of a Symbol 
stock <- Ad(LUNA) 

#Create matrix for returns only 
y <- stock 

#Create a "MATRIX" by choosing the Adjusted Close 
Nudata3 <- stock 

#Sharpe Ratio Matrix 
SR1<- matrix(NA, nrow=1) 

# I want to create a table with all possible combinations from the ranges below 
i = c(2:50) 
k = c(4:50) 
j = c(2:50) 

# stores possible combinations into z 
z <- expand.grid(i,k,j) 
colnames(z)<- c("one","two","three")    

n = 1 
stretches <- length(z[,1]) 

while (n < stretches){ 

# I am trying to go through all the values in "z" 
Nuw <- MACD((stock), nFast=z[n,1], nSlow=z[n,2], nSig=z[n,3], maType="EMA") 

colnames(Nuw) <- c("MACD","Signal") #change the col names to create signals 
x <- na.omit(merge((stock), Nuw)) 

x$sig <- NA 

# Create trading signals        

sig1 <- Lag(ifelse((x$MACD <= x$Signal),-1, 0)) # short when MACD < SIGNAL 
sig2 <- Lag(ifelse((x$MACD >= x$Signal),1, 0)) # long when MACD > SIGNAL 
x$sig <- sig1 + sig2 



#calculate Returns 
ret <- na.omit(ROC(Ad(x))*x$sig) 
colnames(ret)<- c(paste(z[n,1],z[n,2],z[n,3],sep=",")) 
x <- merge(ret,x) 
y <- merge(y,ret) #This creates a MATRIX with RETURNs ONLY 
Nudata3 <- merge(Nudata3, x) 

((mean(ret)/sd(ret)) * sqrt(252)) -> ANNUAL # Creates a Ratio 
ANNUAL->Shrat        # stores Ratio into ShRat 
SR1 <- cbind(SR1,Shrat)      # binds all ratios as it loops 

n <- (n+1) 

} 

我想知道如何向量化的MACD()功能，以加快这一进程以来的stretches长度约。 112847.我的电脑需要一段时间才能通过循环。

Answer 1

首先和最重要的案例具体优化 - 删除nFast > nSlow，因为它从技术上讲没有意义。其次 - 你正在创建对象并且一遍又一遍地复制它们。这非常昂贵。

第三 - 您可以通过在一个循环中创建一个信号矩阵并以矢量化方式执行其余操作来更好地对其进行编码。

我会编码你在做什么这样的事情。

如果您不明白，请参阅mapply，do.call，merge和sapply的帮助页面。

require(quantmod) 
getSymbols("LUNA") 

#Choose the Adjusted Close of a Symbol 
stock <- Ad(LUNA) 

# I want to create a table with all possible combinations from the ranges below 
i = c(2:50) 
k = c(4:50) 
j = c(2:50) 

# stores possible combinations into z 
z <- expand.grid(i,k,j) 

IMO：这是你第一次优化的地方。除去的情况下，其中i>ķ

z <- z[z[,1]<z[,2], ] 

它减少的情况下的数量从112847到57575

#Calculate only once. No need to calculate this in every iteration. 
stockret <- ROC(stock) 

getStratRet <- function(nFast, nSlow, nSig, stock, stockret) { 
    x <- MACD((stock), nFast=nFast, nSlow=nSlow, nSig=nSig, maType="EMA") 
    x <- na.omit(x) 
    sig <- Lag(ifelse((x$macd <= x$signal),-1, 0)) + Lag(ifelse((x$macd >= x$signal),1, 0)) 
    return(na.omit(stockret * sig)) 
} 

RETURNSLIST <- do.call(merge, mapply(FUN = getStratRet, nFast = z[,1], nSlow = z[,2], nSig = z[,3], MoreArgs = list(stock = stock, stockret = stockret), SIMPLIFY = TRUE)) 

getAnnualSharpe <- function(ret) { 
    ret <- na.omit(ret) 
    return ((mean(ret)/sd(ret)) * sqrt(252)) 
} 


SHARPELIST <- sapply(RETURNSLIST, FUN = getAnnualSharpe) 

结果将是如下。哪一列属于哪个组合i,j,k是微不足道的。

head(RETURNSLIST[, 1:3]) 
##   LUNA.Adjusted LUNA.Adjusted.1 LUNA.Adjusted.2 
## 2007-01-10 0.012739026 -0.012739026    0 
## 2007-01-11 -0.051959739  0.051959739    0 
## 2007-01-12 -0.007968170 -0.007968170    0 
## 2007-01-16 -0.007905180 -0.007905180    0 
## 2007-01-17 -0.005235614 -0.005235614    0 
## 2007-01-18 0.028315920 -0.028315920    0 

SHARPELIST 
## LUNA.Adjusted LUNA.Adjusted.1 LUNA.Adjusted.2 LUNA.Adjusted.3 LUNA.Adjusted.4 LUNA.Adjusted.5 LUNA.Adjusted.6 
##  0.04939150  -0.07428392    NaN  0.02626382  -0.06789803  -0.22584987  -0.07305477 
## LUNA.Adjusted.7 LUNA.Adjusted.8 LUNA.Adjusted.9 
##  -0.05831643  -0.08864845  -0.08221986 



system.time(
+ RETURNSLIST <- do.call(merge, mapply(FUN = getStratRet, nFast = z[1:100,1], nSlow = z[1:100,2], nSig = z[1:100,3], MoreArgs = list(stock = stock, stockret = stockret), SIMPLIFY = TRUE)), 
+ SHARPELIST <- sapply(RETURNSLIST, FUN = getAnnualSharpe) 
+) 
    user system elapsed 
    2.28 0.00 2.29