Why is method 1 faster than method 2? - r

Question

My question: I was debugging some code at work, running it block-by-block, when I realized a small block was taking an unusual amount of time. I killed it and made a minor (but logically equivalent) tweak, and it ran almost instantly. I would like to understand why. The following code is in R, however, I imagine the answer may not be specific to R, and may apply to most programming languages of a similar paradigm or 'method-of-compiling'?
The code & information:
Using R version 3.6.1
Libraries loaded: dplyr, knitr, DataExplorer, glue, zoo
old_df is data frame of 5653380 obs. of 91 variables.
field1 is a col of policy numbers with class "character". Not unique, each occurs many times.
date_col1 and date_col2 are columns with class "date".
Method 1:
new_df <- old_df %>%
group_by(field1) %>%
mutate(checkfield = date_col1 - date_col2) %>%
filter(checkfield < 0) %>%
filter(row_number() == 1)
old_df$filter <- ifelse(old_df$field1 %in% new_df$field1,1,0)
Method 2:
new_df <- old_df %>%
group_by(field1) %>%
filter(date_col1 < date_col2) %>%
filter(row_number() == 1)
old_df$filter <- ifelse(old_df$field1 %in% new_df$field1,1, 0)
As you can probably see, the intended output of both methods is to add a flag, "1", in the column "filter" for policy numbers where date_col1 < date_col2. I did not write method 1, and my goal in writing method 2 was to change it as little as possible while also making it faster, so please avoid spending too much time talking about problems with method 1 that are not related to why it is unbearably slower than method 2. Feel free to mention such things, but I would like the crux to be why method 1 was taking 20, 30 minutes etc. For example, I believe in method 1, the first filter call could be above the group_by call. This might increase speed by an unnoticeable amount. I am not too concerned about this.
My thoughts: Clearly method 2 might be a little faster because it avoids making the column "checkfield", but I dont think this is the issue, as I ran method 1 line by line, and it appears to be the line 'filter(checkfield < 0)' where things went awry. For testing, I defined two dates x,y and checked class(x-y) which returned "difftime". So in this filter call, we are comparing "difftime" to a "numeric". Perhaps this requires some type of type-juggling to make the comparison, where as method 2 compares a date object to a date object?
Let me know what you think! I am very curious about this.

Answer 1

My explorations so far, with a simplified example and a slightly smaller data set (only a million rows and a minimal subset of columns) have the individual tests (test_cf for filtering on the checkfield variable, test_lt for filtering on the date comparison) taking about the same time, which both take about the same time as creating the checkfield column. Doing both at once (comb, creating and comparing) takes 2.5 x longer, not sure why.
Perhaps you can use this as a starting point for bisecting/benchmarking to find the culprit.
test elapsed relative
2 comb 5.557 2.860
1 make_cf 1.943 1.000
4 test_cf 2.122 1.092
3 test_lt 2.109 1.085
I used rbenchmark::benchmark() because I prefer the output format: microbenchmark::microbenchmark() might be more accurate (but I would be surprised if it made a big difference).
code
library(dplyr)
n <- 1e6 ## 5653380 in orig; reduce size for laziness
set.seed(101)
## sample random dates, following
## https://stackoverflow.com/questions/21502332/generating-random-dates
f <- function(n)
sample(seq(as.Date('1999/01/01'), as.Date('2000/01/01'), by="day"),
replace=TRUE,
size=n)
dd <- tibble(
date_col1=f(n),
date_col2=f(n)
## set up checkfield so we can use it without creating it
) %>% mutate(cf=date_col1-date_col2)
Benchmark:
library(rbenchmark)
benchmark(
make_cf=dd %>% mutate(checkfield=date_col1-date_col2),
comb=dd %>% mutate(checkfield=date_col1-date_col2) %>% filter(checkfield<0),
test_lt=dd %>% filter(date_col1<date_col2),
test_cf=dd %>% filter(cf<0),
columns=c("test","elapsed","relative")
)

Answer 2

I believe most of the increased time is due to creating the new column. As you can see, M1 and M3 have similar times. Of course that difference of ~2 milliseconds between M1 and M3 will multiply based on the data size
library(tidyverse)
library(microbenchmark)
set.seed(42)
n = 1e5
d = seq.Date(Sys.Date() - 10000, Sys.Date(), 1)
x = sample(d, n, TRUE)
y = sample(d, n, TRUE)
df1 = data.frame(x, y, id = sample(LETTERS, n, TRUE))
microbenchmark(M1 = {
df1 %>%
group_by(id) %>%
mutate(chk = x - y) %>%
filter(chk < 0) %>%
filter(row_number() == 1)
},
M2 = {
df1 %>%
group_by(id) %>%
filter(x < y) %>%
filter(row_number() == 1)
},
M3 = {
df1 %>%
group_by(id) %>%
mutate(chk = x - y) %>%
filter(x < y) %>%
filter(row_number() == 1)
})
#Unit: milliseconds
# expr min lq mean median uq max neval
# M1 13.130673 13.405151 15.088266 14.096772 15.56080 22.636533 100
# M2 5.931192 6.208457 6.564363 6.402879 6.71973 9.354252 100
# M3 11.360640 11.607993 12.449220 12.001383 12.57732 18.260131 100
For the point about comparing difftime to numeric, there doesn't seem to be a lot of difference
library(microbenchmark)
set.seed(42)
n = 1e7
x = sample(d, n, TRUE)
y = sample(d, n, TRUE)
df1 = data.frame(x, y)
df1$difference = df1$x - df1$y
class(df1$difference)
#[1] "difftime"
microbenchmark(date_vs_date = {
df1 %>% filter(x < y)
},
date_vs_numeric ={
df1 %>% filter(difference < 0)
})
#Unit: milliseconds
# expr min lq mean median uq max neval
# date_vs_date 177.1789 222.4112 243.6617 233.7221 244.2765 430.4273 100
# date_vs_numeric 181.6222 217.1121 251.6127 232.7213 249.8218 455.8285 100