将嵌套的for循环转换为R中的foreach? [英] Converting nested for loops to foreach in R?
问题描述
我写了一个函数，其中有2个for循环嵌套在一个foreach循环中.我要尝试遍历以下结构的数据帧列表: [[5]] [10,30] .
I have written a function with 2 for loops nested within a foreach loop. I have a list of dataframes that I am trying to loop through in the structure of: [[5]][10,30].
但是，我试图用1,000,000条路径(即 for(i在1:1000000中))运行此功能，显然，性能太差了.
However, I am trying to run this function with 1,000,000 paths (i.e. for (i in 1:1000000)), and obviously, the performance is terrible.
我真的很想与 foreach 循环并行运行.我发现将任何形式的 apply 函数与foreach结合使用也无法正常工作.当然，如果有更好的方法可以做到这一点，我也很乐意看到这些:
I'd really like to run this in parallel with foreach loops. I have found that using any form of apply functions in conjunction with foreach does not work properly as well. Of course, if there are even better ways to do this, I'd love to see those too:
library(foreach)
library(doParallel)
# input: matr is a list of 5 matrices
cum_returns <- function(matr) {
time_horizon <- 30
paths <- 10
asset <- 5
foreach (x = matr) %dopar% {
for (i in 1:paths) {
x[i,] <- append(x[i,],100,0)
for (m in 2:(time_horizon + 1)) {
# loop through each row of matrix to apply function
x[i,m] <- x[i,m-1] + x[i,m]
}
}
return(x)
}
}
该函数的目标是转换以下格式的数据帧:
The goal of the function is to convert dataframes in this format:
V1 V2 V3 V4 V5 V6
result.4 -0.3937681 0.42309970 -0.2283395 -0.8331735437 0.7874238 -0.1453797
result.9 -1.5680301 0.41994580 -2.1580822 1.6118210199 -1.1626008 1.7275690
result.4.1 -0.5495332 -0.82372187 0.3571042 1.0774779108 -0.7305624 0.6109353
result.9.1 -0.6323561 1.70637893 0.6652303 0.7848319013 -1.0563251 0.8036310
result.4.2 -0.3242765 -0.75415454 0.7407225 -1.7877216475 1.5852460 0.1917951
result.9.2 -0.5348290 -0.05270434 1.5113037 0.8491153876 -2.0715359 -2.0216315
result.4.3 -0.7013342 -0.89451784 -0.2683157 -0.2759993796 0.2709717 1.3437261
result.9.3 1.6187813 -1.53112097 0.6938031 -1.4157996794 -0.6058584 0.4324761
result.4.4 -0.6069532 0.07735158 0.7632158 1.0759685528 -0.3157746 -1.1726851
result.9.4 -0.4945204 1.20070722 -0.1619356 -0.0009728659 -2.0367133 1.4713883
对于这种格式，通过在每行的开头添加100，然后在每行中添加以下每个值，以在每行的字段中创建累加和:
To this format, through appending 100 at the start of each row then adding each of the following values in each row to create a cumulative sum in each row's fields:
V1 V2 V3 V4 V5 V6
result.4 100 99.60623 100.02933 99.80099 98.96782 99.75524
result.9 100 98.43197 98.85192 96.69383 98.30565 97.14305
result.4.1 100 99.45047 98.62674 98.98385 100.06133 99.33076
result.9.1 100 99.36764 101.07402 101.73925 102.52408 101.46776
result.4.2 100 99.67572 98.92157 99.66229 97.87457 99.45982
result.9.2 100 99.46517 99.41247 100.92377 101.77289 99.70135
result.4.3 100 99.29867 98.40415 98.13583 97.85983 98.13080
result.9.3 100 101.61878 100.08766 100.78146 99.36566 98.75981
result.4.4 100 99.39305 99.47040 100.23361 101.30958 100.99381
result.9.4 100 99.50548 100.70619 100.54425 100.54328 98.50657
推荐答案
无需遍历行和列.您可以使用R的能力进行矢量化计算，以将整个列加在一起，然后用对cbind()的单个调用来替换对append()的重复调用.
There's no need to loop over the rows and columns. You can use R's ability to do vectorized calculations to add whole columns together, and replace the repeated calls to append() with a single call to cbind().
foreach (x = matr) %dopar% {
x <- cbind(100,x)
for (m in 2:(time_horizon + 1)) {
# loop through each row of matrix to apply function
x[,m] <- x[,m-1] + x[,m]
}
x
}
即使不使用我的计算机上的多个内核，这也非常快，每个矩阵中有1,000,000行.
Even without using multiple cores on my computer this is pretty quick with 1,000,000 rows in each matrix.
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