image - 如何将 DynamicImage 转换为 ImageBuffer？

Question

当我运行以下代码时，最后一行会导致此错误:

error[E0308]: mismatched types
  --> src\main.rs:25:13
   |
25 |     box_filter(&img, 5, 5);
   |                ^^^^ expected struct `ImageBuffer`, found enum `DynamicImage`
   |
   = note: expected reference `&ImageBuffer<Luma<u8>, Vec<u8>>`
              found reference `&DynamicImage`

extern crate image;
extern crate imageproc;

use image::GenericImageView;
use imageproc::filter::box_filter;

fn main() {
    println!("Hello, world!");

    // Use the open function to load an image from a Path.
    // `open` returns a `DynamicImage` on success.
    // `unwrap` returns a `panic` when it receives a `None`.
    let img = image::open(
        "C:/Users/hp/Desktop/Multiprocessor real-time scheduling/Project2/data/aLIEz.jpg",
    )
    .unwrap();

    // The dimensions method returns the images width and height.
    println!("dimensions {:?}", img.dimensions());

    // The color method returns the image's `ColorType`.
    println!("{:?}", img.color());

    // Write the contents of this image to the Writer in PNG format.
    img.save(
        "C:/Users/hp/Desktop/Multiprocessor real-time scheduling/Project2/data_output/test.png",
    )
    .unwrap();

    box_filter(&img, 5, 5);
}

我不知道如何转换 DynamicImage进入 ImageBuffer这样我就可以应用过滤器。我看过 image 0.23.12 的文档依赖，但我不明白如何做到这一点。

Answer 1

image::open().unwrap() 结果为 image::DynamicImage

imageproc::filter::box_filter() 需要 image::GrayImage

转换 image::DynamicImage变成 image::GrayImage使用
to_luma8()
方法

这个更新的示例应该可以工作:

extern crate image;
extern crate imageproc;
use image::GenericImageView;
use imageproc::filter::box_filter;

fn main() {
    let dynamic_image = image::open("path/here").unwrap();
    let gray_image = dynamic_image.to_luma8();
    let result = box_filter(&gray_image, 5, 5);
}

如果您需要 image::RgbImage 或 image::RgbaImage 然后调用 into_rgb8() 或 into_rgba8() .然而 box_filter()仅支持 image::GrayImage .