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Parsing numbers from a text file
This is my solution to the first task of the Advent of Code 2023.
The task description is:
On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.
For example:
1abc2 pqr3stu8vwx a1b2c3d4e5f treb7uchetIn this example, the calibration values of these four lines are
12,38,15, and77. Adding these together produces142.Consider your entire calibration document. What is the sum of all of the calibration values?
Here is my solution, which provides the correct sum (at least I passed the task):
use std::fs::read_to_string;
use regex::Regex;
fn main() {
let re = Regex::new(r"^[\D]*(\d)(|.*(\d))\D*$").unwrap();
let mut sum: u32 = 0;
for line in read_to_string("calibration_data.txt").unwrap().lines() {
println!("{line}");
let Some(caps) = re.captures(line) else {
panic!("no match! in line {line}");
};
let first: &str = &caps.get(1).map_or("0", |m| m.as_str());
let second: &str = &caps.get(3).map_or(first, |m| m.as_str());
let mut compound: String = String::new();
compound.push_str(first);
compound.push_str(second);
println!("{first} + {second} = {compound}");
let compound: u32 = compound.parse().unwrap();
sum += compound;
}
println!("***********************");
println!("Calibration sum: {sum}");
println!("***********************");
}
Is there any way I could have done this better, especially regarding the extraction of the numbers from the text lines?
2 answers
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The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| GeraldS | (no comment) | Jul 19, 2024 at 06:35 |
You don't need a regex for this. To find first you can simply iterate through the line until you find a digit. To find second you can do the same but in reverse. This is more efficient than running a regex.
For a problem as small as the example, it doesn't matter and arguably the regex adds readability. But I assume AoC makes you run this on a much bigger input as well.
You also don't need to construct the calibration number as a string. first*10 + second gives the calibration number.
When finding the digits, you can do a time-space tradeoff by creating a hashmap that maps each char 0-9 to the corresponding digit. Plugging the chars into this map is probably faster than parsing the string to an integer.
using rfind and find as str.chars().find(|c| "0123456789".contains(c)) and str.chars().find(|c| "0123456789".contains(c)), then a manual match block is the simplest solution I think:
edit str.find returns index, so better to call chars() first for functional-style.
fn char_to_int(c: char) -> u32 {
match c {
'0' => 0,
'1' => 1,
'2' => 2,
'3' => 3,
'4' => 4,
'5' => 5,
'6' => 6,
'7' => 7,
'8' => 8,
'9' => 9,
_ => u32::MAX, // make summing logic panic! for pleasure on invalid
}
}
#[test]
fn test_advent_2023() {
let data = [
"1abc2",
"pqr3stu8vwx",
"a1b2c3d4e5f",
"treb7uchet",
];
let mut sum: u32 = 0;
for line in data {
let first: u32 = line.chars().find(|c| "0123456789".contains(*c)).map(char_to_int).unwrap();
let second: u32 = line.chars().rfind(|c| "0123456789".contains(*c)).map(char_to_int).unwrap();
sum += first*10 + second;
}
println!("{sum:?}");
assert_eq!(sum, 42) // make the test fail for output
}
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