Question

我有工作代码来计算运行的drawdown.duration，其中drawdown.duration定义为当前月份与上一个peak之间的月份数。但是，我将代码实现为for循环，并且运行速度非常慢。

在R中是否有更有效/更快的方法来实现此目的？

该代码使用名为data.frame的{{1}}（特别是tibble，因为我一直在使用dplyr）。

returnsWithValues

我已经使用> structure(list(date = structure(c(789, 820, 850, 881, 911, 942 ), class = "Date"), value = c(0.94031052, 0.930751624153046, 0.926756311376762, 0.874209664097166, 0.843026010916249, 2.1), peak = c(1, 1, 1, 1, 1, 2.1), drawdown = c(-0.05968948, -0.0692483758469535, -0.0732436886232377, -0.125790335902834, -0.156973989083751, 0)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -6L)) # A tibble: 6 x 4 date value peak drawdown <date> <dbl> <dbl> <dbl> 1 1972-02-29 0.940 1 -0.0597 2 1972-03-31 0.931 1 -0.0692 3 1972-04-30 0.927 1 -0.0732 4 1972-05-31 0.874 1 -0.126 5 1972-06-30 0.843 1 -0.157 6 1972-07-31 2.1 2.1 0循环实现了drawdown.duration：

for

哪个给出正确的答案为：

returnsWithValues <- returnsWithValues %>% mutate(drawdown.duration = NA)

    # add drawdown.duration col
    for (row in 1:nrow(returnsWithValues)) {
        if(returnsWithValues[row,"value"] == returnsWithValues[row,"peak"]) {
            returnsWithValues[row,"drawdown.duration"] = 0
        } else {
            if(row == 1){
                returnsWithValues[row,"drawdown.duration"] = 1
            } else {
                returnsWithValues[row,"drawdown.duration"] = returnsWithValues[row - 1,"drawdown.duration"] + 1
            }
        }
    }

Answer 1

我将根据需要删除for循环，并使用索引的想法。

indices <- function(returnsWithValues){
    indices_logical<-(returnsWithValues[["value"]] == returnsWithValues[["peak"]]) #return a logical vector where true values are for equal and false for not.
    indices_to_zero<-which(indices_logical) # which values are true
    indices_drawdpwn<-which(!indices_logical) # which values are false
    returnsWithValues[indices_to_zero,"drawdown.duration"] <- 0
    returnsWithValues[indices_drawdpwn,"drawdown.duration"] <- 1:length(indices_drawdpwn) #basically you compute this if I understand correctly
    returnsWithValues

这是包装在函数中的for循环。

for_loop<-function(returnsWithValues){
    # add drawdown.duration col
    for (row in 1:nrow(returnsWithValues)) {
        if(returnsWithValues[row,"value"] == returnsWithValues[row,"peak"]) {
            returnsWithValues[row,"drawdown.duration"] = 0
        } else {
            if(row == 1){
                returnsWithValues[row,"drawdown.duration"] = 1
            } else {
                returnsWithValues[row,"drawdown.duration"] = returnsWithValues[row - 1,"drawdown.duration"] + 1
            }
        }
    }
    returnsWithValues
}

与for循环相比，这是一个基准。

microbenchmark::microbenchmark(
      "for loop" = flp<-for_loop(returnsWithValues),
      indices = ind<-indices(returnsWithValues),
      times = 10
)

Unit: microseconds
        expr      min       lq     mean    median       uq      max neval
    for loop 8671.228 8699.555 8857.198 8826.8185 8967.631 9196.708    10
     indices   92.781   99.349  106.328  102.8385  115.360  122.749    10
all.equal(ind,flp)
[1] TRUE

Answer 2

我认为只要每个peak值都是唯一的并且以后不会在另一个组中重复，就可以做到：

returnsWithValues %>%
    group_by(peak) %>%
    mutate(drawdown.duration = cumsum(value != peak))

如果确实有重复的峰值，则可能需要一种方法来将连续的peak值之内分组，例如

returns %>%
    # Start counting the number of groups at 1, and every time
    #   peak changes compared to the previous row, add 1
    mutate(peak_group = cumsum(c(1, peak[-1] != head(peak, -1)))) %>%
    group_by(peak_group) %>%
    mutate(drawdown.duration = cumsum(value != peak))

提高延期的速度

2 个答案: