我有一个二维矢量:
let vec2d = vec![
vec![1, 1, 1],
vec![1, 1, 1],
];
我可以通过这种方式生成存储的元素总数:
let mut n_vec_element: i32 = 0;
for i in vec2d.iter() {
n_vec_element += i.len() as i32;
}
println!("2D vector elements :{}", n_vec_element); // prints 6
当我增加尺寸时,循环变长:
let mut n_vec_element: i32 = 0;
let vec3d = vec![
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
],
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
]
];
for i in vec3d.iter() {
// I must add another iter everytime I increment the dimension by 1.
// Else, it returns the number of stored vector instead of the vector
// elements.
for j in i.iter() {
n_vec_size += j.len() as i32;
}
};
println!("3D vector elements :{}", n_vec_element); // prints 18
必须有一个更简洁的方法来做到这一点,但我还没有想出来。最初,我尝试使用vector的len()
函数,但正如我上面所说,它返回存储的向量数而不是其元素。
您不需要显式循环来执行此操作:
let vec2d = vec![
vec![1, 1, 1],
vec![1, 1, 1],
];
let n_vec_element: usize = vec2d.iter().map(Vec::len).sum();
assert_eq!(n_vec_element, 6);
对于3d矢量,您可以这样做:
let vec3d = vec![
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
],
vec![
vec![1, 3, 5 as i32],
vec![2, 4, 6 as i32],
vec![3, 5, 7 as i32],
]
];
let n_vec_element: usize = vec3d.iter().flatten().map(Vec::len).sum();
assert_eq!(n_vec_element, 18);
使用4D矢量,你可以放2个flatten
等。
使用专门化功能(即使用夜间编译器),您可以使用一种独特的方法来概括它:
#![feature(specialization)]
trait Count {
fn count(self) -> usize;
}
impl<T> Count for T {
default fn count(self) -> usize {
1
}
}
impl<T> Count for T
where
T: IntoIterator,
T::Item: Count,
{
fn count(self) -> usize {
self.into_iter().map(|x| x.count()).sum()
}
}
fn main() {
let v = vec![1, 2, 3];
assert_eq!(v.count(), 3);
let v = vec![vec![1, 2, 3], vec![4, 5, 6]];
assert_eq!(v.count(), 6);
}