I noticed I kept reimplementing the same 2d-grid structure in many of my personal projects, so I decided to make it into a library. This data structure does not attempt to be the fastest or best implementation of a 2d-grid, but it's simple to use and has zero dependencies.
```rust use simple_grid::Grid;
let grid = Grid::new(10, 10, (1..=100).collect::
println!("{}", grid.toprettystring()); // prints: // 1 2 3 4 5 6 7 8 9 10 // 11 12 13 14 15 16 17 18 19 20 // 21 22 23 24 25 26 27 28 29 30 // 31 32 33 34 35 36 37 38 39 40 // 41 42 43 44 45 46 47 48 49 50 // 51 52 53 54 55 56 57 58 59 60 // 61 62 63 64 65 66 67 68 69 70 // 71 72 73 74 75 76 77 78 79 80 // 81 82 83 84 85 86 87 88 89 90 // 91 92 93 94 95 96 97 98 99 100 ```
```rust
let grid = Grid::new(10, 10, (1..=100).collect::
let elementsinrow3: Vec
let elementsincolumn7: Vec
```rust
let mut grid = Grid::new(10, 10, (1..=100).collect::
// get a mutable reference to a cell *grid.getmut((8, 2)).unwrap() = 1000; asserteq!(grid.get((8, 2)).unwrap(), &1000);
// can also access directly via the index operator grid[(5,5)] = 1001; assert_eq!(grid.get((5, 5)).unwrap(), &1001); ```
This is only available if the serde
feature is enabled.
The linalg
feature includes some methods that are useful for linear algebra:
rust
let grid1 = Grid::new(2, 2, vec![1, 2, 3, 4]);
let grid2 = Grid::new(2, 2, vec![1, 0, 1, 0]);
let sum = grid1 + grid2;
assert_eq!(sum, Grid::new(2, 2, vec![2, 2, 4, 4]));
rust
let grid = Grid::new(3, 3, vec![3., 0., 2., 2., 0., -2., 0., 1., 1.]);
let inverse = grid.inverse().unwrap();
for (actual, expected) in inverse
.cell_iter()
.zip(Grid::new(3, 3, vec![0.2, 0.2, 0., -0.2, 0.3, 1.0, 0.2, -0.3, 0.]).cell_iter())
{
let diff = actual - expected;
assert!(diff < 0.000001);
}
To solve the following system:
2x + y - z = 8
-3x - y + 2z = -11
-2x + y + 2z = -3
rust
// the equation system represented as a Grid where the rightmost column is the right side of the equal signs
let mut grid = Grid::new(
4,
3,
vec![2., 1., -1., 8., -3., -1., 2., -11., -2., 1., 2., -3.],
);
let solution = grid.gaussian_elimination();
assert_eq!(solution.unwrap_single_solution(), vec![2., 3., -1.]))
Giving the solution
x = 2, y = 3, z = -1