R向量化数组数据操作 [英] R vectorized array data manipulation
问题描述
我认为将会有更多的人对此主题感兴趣.我有一些最有效的方式要做的特定任务.我的基本数据是:-买入和卖出信号的时间指数-根据时间指标，我可以得出最接近的买卖对之间的ROC(变化率):
I think there will be much more people interested into this subject. I have some specific task to do in the most efficient way. My base data are: - time indices of buy and sell signals - on the diag of time indicies I have ROC (rate of change) between closest buy-sell pairs:
r <- array(data = NA,
dim = c(5, 5),
dimnames = list(buy_idx = c(1,5,9,12,16),
sell_idx = c(3,7,10,14,19)))
diag(r) <- c(1.04,0.97,1.07,1.21,1.1)
任务是在每个可能的窗口(买卖对)上生成移动复合ROC，以及当前我完成任务的方式:
The task is to generate moving compound ROC on every possible window (buy-sell pairs), and the way I'm solving my task currently:
for(i in 2:5){
r[1:(i-1),i] <- r[1:(i-1),i-1] * r[i,i]
}
直到我没有将其循环到更高的位置，我的解决方案的时间是可以接受的.有没有办法将此循环更改为矢量化解决方案?是否有任何记录良好的教程来学习R中的向量化思维类型?-这将比一次性解决方案更有价值！
Until I'm not looping it somewhere upper, the time of my solution is very acceptable. Is there a way to change this loop to vectorized solution? Are there any good well documented tutorials to learn vectorized type of thinking in R? - it would be much more valuable than one time solution!
修改20130709:
与上一个任务/示例高度相关的下一个任务.在每笔交易中应用税值(以％值计的税).当前解决方案:
Next task highly related to previous task/example. Apply tax value on each transaction (tax in % values). Current solution:
diag(r[,]) <- diag(r[,]) * ((1-(tax/100))^2)
for(i in 2:dim(r)[2]){
r[1:(i-1),i] <- r[1:(i-1),i] * ((1-(tax/100))^(2*(i:2)))
}
您知道更有效的方法吗?或更正确(如果不能解决所有问题).
Do you know any more efficient way? or more correct if this doesn't handle everything.
推荐答案
如果 d 是您的对角线元素，那么到处 j> = i ， r[i，j] 是 prod(d [i:j])，也可以写成 prod(d [1:j])/prod(d [1:(i-1)]).因此，此技巧使用累积乘积的外部比率:
If d are your diagonal elements, then everywhere j >= i, r[i,j] is prod(d[i:j]), which can also be written prod(d[1:j]) / prod(d[1:(i-1)]). Hence this trick using the outer ratio of the cumulative product:
d <- c(1.04,0.97,1.07,1.21,1.1)
n <- length(d)
p <- cumprod(c(1, d))
r <- t(outer(p, 1/p, "*"))[-n-1, -1]
r[lower.tri(r)] <- NA
一些基准测试表明，对于某些(不是全部)输入大小，它比OP更好:
Some benchmarks showing that it does better than OP for some (not all) input sizes:
OP <- function(d) {
r <- diag(d)
for(i in 2:length(d)){
r[1:(i-1),i] <- r[1:(i-1),i-1] * r[i,i]
}
r
}
flodel <- function(d) {
n <- length(d)
p <- cumprod(c(1, d))
r <- t(outer(p, 1/p, "*"))[-n-1, -1]
r[lower.tri(r)] <- NA
r
}
d <- runif(10)
microbenchmark(OP(d), flodel(d))
# Unit: microseconds
# expr min lq median uq max
# 1 flodel(d) 83.028 85.6135 88.4575 90.153 144.111
# 2 OP(d) 115.993 122.0075 123.4730 126.826 206.892
d <- runif(100)
microbenchmark(OP(d), flodel(d))
# Unit: microseconds
# expr min lq median uq max
# 1 flodel(d) 490.819 545.528 549.6095 566.108 684.043
# 2 OP(d) 1227.235 1260.823 1282.9880 1313.264 3913.322
d <- runif(1000)
microbenchmark(OP(d), flodel(d))
# Unit: milliseconds
# expr min lq median uq max
# 1 flodel(d) 97.78687 106.39425 121.13807 133.99502 154.67168
# 2 OP(d) 53.49014 60.10124 72.56427 85.17864 91.89011
编辑以回答20130709的新增内容:
我假设 tax 是一个标量，然后让 z<--(1- tax/100)^ 2 .您的最终结果是 r 乘以以不同次幂得出的 z 矩阵.您要避免的是一遍又一遍地计算这些能力.这就是我要做的:
I'll assume tax is a scalar and let z <- (1- tax/100)^2. Your final result is r multiplied by a matrix of z raised at different powers. What you want to avoid is compute these powers over and over. Here is what I would do:
pow <- 1L + col(r) - row(r)
pow[lower.tri(pow)] <- NA
tax.mult <- (z^(1:n))[pow]
r <- r * tax.mult
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