Day 1
Problem
Given a txt file consisting of one column of integers with missing rows indicating the start of a new individual, find:
- The individual with the highest sum
- The three individuals with the highest sum
Thoughts
This was a relatively straightforward problem in R. Using cumsum on missing values to create an index is an idiom that comes up a lot in Advent problems.
Translating to Julia, I had a couple of minor issues.
- Forgot that
ismissingwould need to be broadcast, which ruined my groups in Julia for a bit. - I tried a couple of suggestions that didn’t work correctly for me before finding the
* -1strategy for a descending sort in@orderby.
On the plus side,
- The
@asidemacro is a really nice feature when you want two outputs. - As expected, the Julia version is way faster!
Code
R
Load the data into a tibble using the tidyverse packages.
library(tidyverse)
df <- tibble(
num = as.numeric(
read_lines("data/1.txt")
)
)
Julia
Load the data into a DataFrame using CSV.
using DelimitedFiles, CSV
using DataFrames, DataFramesMeta
df = CSV.read(
"data/1.txt",
DataFrame,
ignoreemptyrows = false,
header = ["x1"]
)
Group the individuals by adding a cumulative sum of missing values, then sum within groups.
solve_day <- function(df){
prep <- df |>
mutate(id = cumsum(is.na(num))) |>
group_by(id) |>
summarize(total_cal = sum(num, na.rm = TRUE)) |>
arrange(desc(total_cal))
p1 <- slice(prep, 1) |> pull(total_cal)
p2 <- slice(prep, 1:3) |> pull(total_cal) |> sum()
return(c(p1, p2))
}
Group the individuals by adding a cumulative sum of missing values, then sum within groups.
function solve_day(df)
p2 = @chain df begin
@transform :id = cumsum(ismissing.(:x1))
groupby(:id)
@combine :total_cal = sum(skipmissing(:x1))
@aside p1 = maximum(_.total_cal)
@orderby(:total_cal * -1)
sum(_.total_cal[1:3])
end
return([p1, p2])
end
Run our solve_day function to get our solution:
solve_day(df)
[1] 74711 209481
Run our solve_day function to get our solution:
solve_day(df)
2-element Vector{Int64}:
74711
209481
Run benchmark using bench::mark():
select(bench::mark(solve_day(df)), median, mem_alloc)
# A tibble: 1 × 2
median mem_alloc
<bch:tm> <bch:byt>
1 11.6ms 192KB
Run benchmark using BenchmarkTools:
using BenchmarkTools
median(@benchmark solve_day(df))
BenchmarkTools.TrialEstimate:
time: 197.050 μs
gctime: 0.000 ns (0.00%)
memory: 107.63 KiB
allocs: 613
