Odtnhj

I have working code that calculates a running drawdown.duration where drawdown.duration is defined as the number of months between the current month and the previous peak. I implemented the code, however, as a for loop and it runs quite slow.

Is there a more efficient/faster way to implement this in R?

The code takes a data.frame (specifically a tibble since I have been working with dplyr) named 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 

I have implemented drawdown.duration using a for loop:

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
 
 
 

Which gives the correct answer as:

> 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

I think this will do it, as long as each peak value is unique and not repeated in another group later on:

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

If you do have repeated peak values, you might need a way to group just within consecutive peak values, e.g.

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))

I will remove the for loop as you want and I will use the idea of indexing.

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

Here is you for loop wrapped in a function.

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

Here is a benchmark compared to your for loop.

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