The module is designed to speed up the solution of certain coding challenges. The module is inspired by Advent of Code, I have not used it for anything else, but it is probably useful for other things.
The use of the module is different from the traditional ones, my goal
was to be able to create a new solution easily. The Solution
base
class provided by the module contains some frequently used or useful
functions for debugging and only the solutions need to be added.
Each time I start from the code below.
#!/usr/bin/python3 -u
import aochallenge
class Solution(aochallenge.Solution):
def __init__(self):
super().__init__()
data = self.load(True,',',int)
# def part1(self):
# def part2(self):
# def solve_more(self):
solution = Solution()
solution.main()
To be more precise, I also use type annotation, which I have taken from here.
The part1
and part2
methods are called and the returned value is
displayed by the original class. If the two parts of the challenge build
on each other, you can also use solve_more
(generator), in which the
solutions are returned with the yield
keyword, so that the computation
can continue without saving previous results. Check out also the
example at the end of this document.
In the constructor, it makes sense to load and, if necessary, preprocess the input data, which is well done by the load function of the Solution class. Note that this function examines the program's input arguments and decides which input file to load (test or main) based on them.
For each challenge there are one or more test inputs and there is your challenge one. The class expects the input files to be named appropriately to be able to load automatically. For example, if the file name is "aoc-2201.py", then the input files should be named "aoc-2201__.input". E.g.
aoc/2022/01/
|---- aoc-2201.py source code
|---- aoc-2201-t.input test input
\---- aoc-2201.input challenge input
In this case, you can run your code with the test data as follows:
$ ./aoc-2201.py -t
And with the challenge data, simply:
$ ./aoc-2201.py
In some special cases the input is a single line of data or some other
simple constructs. In this case it is unnecessary to create files for
each, you can simply pass a look-up-table to the load
function. E.g.
INPUT = {
None: 'My challenge input data',
't': 'My test 1 input data',
't2': 'My test 2 input data',
}
...
def __init__(self):
super().__init__()
data = self.load(lut=INPUT)
The load
method is used to prepare the data for further processing.
The input can come from a file or from a predefined look-up-table. If
the latter is not specified, file handling is automatic (see above).
load
function does not only import data, but does some preprocessing
on them:
def load(self,
splitlines: bool = False,
splitrecords: str|None = None,
recordtype: list|tuple|type|None = None,
*,
lut: dict[str|None, Any]|None = None,
) -> list[str|int|list]
Parameters:
splitlines
: boolean value whether the input data lines have to be splitted into a list.splitrecords
: string value used to separate the records in each line. For example, if there are comma-separated values, this field is ",". If set toNone
, items within the row are not split.recordtype
: type of records. For example, if the values are numbers, it can beint
or evenfloat
. Theload
function does the conversion. If the parameter type islist
ortuple
, the various fields may have different types. E.g.(str, list)
means, that the first record should be astr
, but all further ones have to be casted toint
.lut
(keyword only parameter): if this parameter is specified, the input data will be read from it instead of from an input file.
Note, that if splitlines
is False
but splitrecords
is defined,
only the first row will be processed. This means that if you have a
one-row data set, the return element is not a two-dimensional list with
a single nested list, but a simple list of values from the first row.
The class contains some debugging solutions to display temporary results.
-
print_condensed
: Prints content of a 2-dimensional container of characters "condensed". E.g. if data is[['#', '#', '.'], ['.', '#', '.'], ['.', '#', '#']]`
the following will be printed:
##. .#. .##
Note that the 2-dimensional container can be also a list of strings.
-
def print_csv
: Prints content of a 2-dimensional container in a comma separated way -
def print_arranged
: Prints content of a 2-dimensional container arranged into columns
PART 1: Add up all the numbers in each row separated by commas and print the maximum of these sums.
PART 2: Find the 3 largest sums, add them up and determine the final result.
Using part1
and part2
:
#!/usr/bin/python3 -u
import aochallenge
class Solution(aochallenge.Solution):
def __init__(self):
super().__init__()
self.data = self.load(True,',',int)
def part1(self):
return max(sum(row) for row in self.data)
def part2(self):
return sum(sorted(sum(row) for row in self.data)[-3:])
solution = Solution()
solution.main()
Using solve_more
:
#!/usr/bin/python3 -u
import aochallenge
class Solution(aochallenge.Solution):
def __init__(self):
super().__init__()
self.data = self.load(True,',',int)
def solve_more(self):
sums = sorted(sum(row) for row in self.data)
yield sums[-1]
yield sum(sums[-3:])
solution = Solution()
solution.main()