提高drawdown.duration实施的速度

Question

我有工作代码计算正在运行的drawdown.duration ，其中drawdown.duration定义为当前月和前一个peak之间的月数。 然而，我实现了代码作为for循环，它运行得很慢。

是否有更有效/更快的方式在R实现这一点？

该代码需要一个data.frame （具体为tibble ，因为我一直在努力dplyr ）命名returnsWithValues 。

> 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   

我使用for循环实现了drawdown.duration ：

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
            }
        }
    }

这给出了正确的答案：

> returnsWithValues
# A tibble: 6 x 5
  date       value  peak drawdown drawdown.duration
  <date>     <dbl> <dbl>    <dbl>             <dbl>
1 1972-02-29 0.940   1    -0.0597                 1
2 1972-03-31 0.931   1    -0.0692                 2
3 1972-04-30 0.927   1    -0.0732                 3
4 1972-05-31 0.874   1    -0.126                  4
5 1972-06-30 0.843   1    -0.157                  5
6 1972-07-31 2.1     2.1   0                      0

Answer 1

我将根据需要删除for循环，我将使用索引的想法。

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

这是你在函数中包含的循环。

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
}

这是与你的for循环相比的基准。

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

我想这会做到这一点，只要每个peak都是唯一的，以后不会在另一个组中重复：

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

如果确实有重复的峰值，则可能需要一种在连续peak内进行分组的方法，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))