Simple Grid

我注意到我在许多个人项目中反复实现了相同的2D网格结构，所以我决定将其做成一个库。此数据结构并不试图成为最快的或最好的2D网格实现，但它使用简单且没有依赖。

示例用法

创建网格并访问其单元格

use simple_grid::Grid;

let grid = Grid::new(10, 10, (1..=100).collect::<Vec<u32>>());
assert_eq!(grid.get((5, 2)).unwrap(), &26);

println!("{}", grid.to_pretty_string());
// 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

遍历单元格

let grid = Grid::new(10, 10, (1..=100).collect::<Vec<u32>>());

let elements_in_row_3: Vec<u32> = grid.row_iter(3).copied().collect();
assert_eq!(
    elements_in_row_3,
    vec![31, 32, 33, 34, 35, 36, 37, 38, 39, 40]
);

let elements_in_column_7: Vec<u32> = grid.column_iter(7).copied().collect();
assert_eq!(
    elements_in_column_7,
    vec![8, 18, 28, 38, 48, 58, 68, 78, 88, 98]
);

修改内容

let mut grid = Grid::new(10, 10, (1..=100).collect::<Vec<u32>>());

// get a mutable reference to a cell
*grid.get_mut((8, 2)).unwrap() = 1000;
assert_eq!(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);

序列化/反序列化

此功能仅在启用serde功能时可用。

线性代数

linalg功能包括一些对线性代数有用的方法

算术运算

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]));

逆、转置等。

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);
}

高斯消元法

以解决以下系统

2x+y-z= 8

-3x-y+2z= -11

-2x+y+2z= -3

// 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.]))

给出解 x = 2, y = 3, z = -1