How can I write this -working- code to improve performance?

Question

I have this sample set out of a bigger dataset (25K records). I noticed that the code of my app slows down on this piece and I want to check for performance improvement.

context : my financial years starts July and ends in June. Therefore my records have a Financial Period and Financial Year that is different than a month and a calendar Year. I want to add extra columns indicating the calendar year and month.

FinancialPeriod-FinancialYear : 01 2018 is 7 2017 (Jul 2017),
FinancialPeriod-FinancialYear : 07 2018 is 1 2018 (Jan 2018), etc..

Reproducible example :

dt<-data.table(FinancialPeriod =c(3,4,4,5,1,2,8,8,11,12,2,3,10,1,6), FinancialYear=c(2018), Amount=c(12,14,16,18,12))
dt$Month<-dt$FinancialPeriod + 6
dt$Year<-dt$FinancialYear
t1<-proc.time()
for(row in 1:nrow(dt)){
  if (dt[row,"Month"] > 12){
    dt[row,"Month"]<- dt[row,"Month"] -12
  } 
  else {
    dt[row,"Year"]<- dt[row,"Year"] -1
  }
}
proc.time()-t1
dt

The code above works, but works slow. I would like to have suggestions on how to improve.

Answer 1

My suggestion is to use a full date, e.g., the begin of a financial period. So, we can use a FinancialDate instead of a FinancialPeriod-FinancialYear combo, e.g., OP's example

FinancialPeriod-FinancialYear : 01 2018

becomes

FinancialDate: 2018-01-01

This approach has several benefits:

1. There is only one date object describing the period instead of two objects which is easier to handle, e.g., for aggregations. If required, month and year are easily extracted from a date object.
2. We can take advantage of the usual date arithmetic, e.g., of the lubridate package, to convert a FinancialDate into a CalendarDate.
3. Also, plotting works better for date objects.

Here is the code:

library(lubridate)
dt[, FinDate := make_date(FinancialYear, FinancialPeriod)]
dt[, CalDate := FinDate %m+% months(6)]   # convert to calendar date by adding offset
dt[, c("CalMonth", "CalYear") := list(month(CalDate), year(CalDate))]
dt

    FinancialPeriod FinancialYear Amount Month Year    FinDate    CalDate CalMonth CalYear
 1:               3          2018     12     9 2018 2018-03-01 2018-09-01        9    2018
 2:               4          2018     14    10 2018 2018-04-01 2018-10-01       10    2018
 3:               4          2018     16    10 2018 2018-04-01 2018-10-01       10    2018
 4:               5          2018     18    11 2018 2018-05-01 2018-11-01       11    2018
 5:               1          2018     12     7 2018 2018-01-01 2018-07-01        7    2018
 6:               2          2018     12     8 2018 2018-02-01 2018-08-01        8    2018
 7:               8          2018     14    14 2018 2018-08-01 2019-02-01        2    2019
 8:               8          2018     16    14 2018 2018-08-01 2019-02-01        2    2019
 9:              11          2018     18    17 2018 2018-11-01 2019-05-01        5    2019
10:              12          2018     12    18 2018 2018-12-01 2019-06-01        6    2019
11:               2          2018     12     8 2018 2018-02-01 2018-08-01        8    2018
12:               3          2018     14     9 2018 2018-03-01 2018-09-01        9    2018
13:              10          2018     16    16 2018 2018-10-01 2019-04-01        4    2019
14:               1          2018     18     7 2018 2018-01-01 2018-07-01        7    2018
15:               6          2018     12    12 2018 2018-06-01 2018-12-01       12    2018

Answer 2

Another option:

i <- DT1[Month > 12L, which=TRUE]
DT1[i, Month := Month - 12L][-i, Year := Year - 1L]

Note that there is a fifelse under development at the rdatatable github site.

Timing using actual dim:

# A tibble: 4 x 14
  expression      min     mean   median      max `itr/sec` mem_alloc  n_gc n_itr total_time result                  memory                 time    gc              
  <chr>      <bch:tm> <bch:tm> <bch:tm> <bch:tm>     <dbl> <bch:byt> <dbl> <int>   <bch:tm> <list>                  <list>                 <list>  <list>          
1 for_loop      1.92s    1.92s    1.92s    1.92s     0.521    4.66GB   110     1      1.92s <NULL>                  <Rprofmem [25,175 x 3~ <bch:t~ <tibble [1 x 3]>
2 mtd0         1.45ms   2.26ms    1.7ms   7.49ms   442.       6.55MB    31   221   500.02ms <data.table [25,000 x ~ <Rprofmem [67 x 3]>    <bch:t~ <tibble [221 x ~
3 mtd1       498.46us 650.79us  566.1us   2.98ms  1537.     699.56KB     6   768   499.81ms <data.table [25,000 x ~ <Rprofmem [24 x 3]>    <bch:t~ <tibble [768 x ~
4 basemtd    950.97us   1.41ms   1.02ms  50.65ms   709.       5.06MB    41   355   500.45ms <dbl [25,000]>          <Rprofmem [52 x 3]>    <bch:t~ <tibble [355 x ~

Base R is already very fast for 25k rows of data.

data:

set.seed(0L)
DT <- data.table(FinancialPeriod=sample(12L, 25e3L, TRUE), 
    FinancialYear=c(2018))
DT[, c("Month","Year") := .(FinancialPeriod + 6L, FinancialYear)]
DT0 <- copy(DT)
DT1 <- copy(DT)
DF <- setDF(copy(DT))

timing code:

bench::mark(
    for_loop = {
        for(row in 1:nrow(DF)){
            if (DF[row,"Month"] > 12){
                DF[row,"Month"]<- DF[row,"Month"] -12
            } 
            else {
                DF[row,"Year"]<- DF[row,"Year"] -1
            }
        }
    },
    mtd0=DT0[, c("Month", "Year") := list(ifelse(Month > 12, Month-12, Month), 
        ifelse(Month <= 12, Year-1, Year))],
    mtd1={
        i <- DT1[Month > 12L, which=TRUE]
        DT1[i, Month := Month - 12L][-i, Year := Year - 1L]
    },
    basemtd={
        DF0$Month <- ifelse(DF0$Month > 12L, DF0$Month - 12L, DF0$Month)  
        DF0$Year <- ifelse(DF0$Month <= 12L, DF0$Year - 1L, DF0$Year)  
    },
    check=FALSE
)

Answer 3

It looks like you are using data.table so I used data.table syntax. There are several solutions but one option is to use ifelse:

dt[, c("Month", "Year") := list(ifelse(Month > 12, Month-12, Month), ifelse(Month <= 12, Year-1, Year))]

Or, if you want to skip creating the initial variables Month and Year:

dt[, c("Month", "Year") := list(ifelse(FinancialPeriod + 6 > 12, FinancialPeriod -6, FinancialPeriod + 6), 
                                ifelse(FinancialPeriod + 6 <= 12, FinancialYear-1, FinancialYear))]

Benchmarks:

Original Data:

Unit: microseconds
     expr     min       lq      mean   median       uq     max neval
 for_loop 21183.8 21765.90 23600.586 22169.80 22981.15 73243.2   100
   ifelse   298.7   332.05   350.509   356.35   369.10   495.8   100

Larger dataset:

data.table(sapply(dt, rep, 100))
Unit: microseconds
     expr       min      lq       mean     median        uq       max neval
 for_loop 1645857.5 1655792 1690718.78 1696271.35 1704510.1 1746731.2    10
       o1     395.4     398     425.29     425.05     446.6     462.3    10

Code:

microbenchmark::microbenchmark(
  for_loop = {
    for(row in 1:nrow(dt)){
      if (dt[row,"Month"] > 12){
        dt[row,"Month"]<- dt[row,"Month"] -12
      } 
      else {
        dt[row,"Year"]<- dt[row,"Year"] -1
      }
    }
  },
  ifelse = {
    dt[, c("Month", "Year") := list(ifelse(Month > 12, Month-12, Month), ifelse(Month <= 12, Year-1, Year))]
  },
  times = 100
)