Question

我试图编写自己的使用线性插值的图像旋转功能（参见下面的代码）。在示例256x256图像上运行我的代码大约需要8秒，或每像素约0.12ms。在同一图像上使用双线性插值运行Matlab的imrotate函数大约需要0.2秒，或每像素约0.003ms - 大约提高一百倍。

我猜测有一些我错过的矢量化优化，但我无法弄清楚在哪里。非常感谢任何建议。

以下代码;

function [ output ] = rot_input_img_by_angle( input_img, angle )
%rot_input_img_by_angle Rotates the given image by angle about position
%   Given an image in the format [y, x, c], rotates it by the given angle
%   around the centre of the image

    if(nargin < 2)
        error('input_img and angle parameters are both required');
    end

    if(angle == 0)
        output = input_img;
        return;
    end

    position = [0 0];

    [height, width, channels] = size(input_img);
    num_pixels = height * width;
    half_width = width/2 - 0.5;
    half_height = height/2 - 0.5;

    % Compute the translation vector to move from a top-left origin to a
    % centred-origin
    T = [-half_width half_height]';

    % A lambda function for creating a 2D rotation matrix
    rotmat = @(th) [cos(th) -sin(th); sin(th) cos(th)];

    % Convert angle to radians and generate rotation matrix R for CR
    % rotation
    R = rotmat(deg2rad(angle));

    output = zeros(height, width, channels);

    for y=1:height

        for x=1:width

            loc = [x-1 y-1]';

            % Transform the current pixel location into the
            % origin-at-centre coordinate frame
            loc = loc .* [1; -1] + T;

            % Apply the inverse rotation mapping to this ouput pixel to
            % determine the location in the original input_img that this pixel
            % corresponds to
            loc = R * loc;

            % Transform back from the origin-at-centre coordinate frame to
            % the original input_img's origin-at-top-left frame
            loc = (loc - T) .* [1; -1] + [1; 1];


            if((loc(1) < 1) || (loc(1) > width) || (loc(2) < 1) || (loc(2) > height))
                % This pixel falls outside the input_img - leave it at 0
                continue;
            end

            % Linearly interpolate the nearest 4 pixels
            left_x = floor(loc(1));
            right_x = ceil(loc(1));
            top_y = floor(loc(2));
            bot_y = ceil(loc(2));

            if((left_x == right_x) & (top_y == bot_y))

                % The sample pixel lies directly on an original input_img pixel
                output(y, x, :) = input_img(y, x, :);

            else

                % The sample pixel lies inbetween several pixels

                % Location of the nearest 4 pixels
                px_locs = [left_x right_x left_x right_x; top_y top_y bot_y bot_y];

                px_dists = distance(loc, px_locs);
                px_dists = px_dists ./ sum(px_dists);

                % Take the linearly interpolated average of each color
                % channel's value
                for c=1:channels
                    output(y, x, c) = ...
                        px_dists(1) * input_img(px_locs(1, 1), px_locs(2, 1), c) + ...
                        px_dists(2) * input_img(px_locs(1, 2), px_locs(2, 2), c) + ...
                        px_dists(3) * input_img(px_locs(1, 3), px_locs(2, 3), c) + ...
                        px_dists(4) * input_img(px_locs(1, 4), px_locs(2, 4), c);
                end
            end

        end

    end

    output = cast(output, class(input_img));


end

Answer 1

您可以通过使用

键入来查看matlab使用的功能

edit imrotate

此外，文档说：

%   Performance Note
%   ----------------
%   This function may take advantage of hardware optimization for datatypes
%   uint8, uint16, and single to run faster.

在这种情况下，Matlab调用imrotatemex，即编译为从Matlab调用的C代码，通常更快。我不知道你的形象和系统，所以如果发生这种情况我就不能说了。

您仍然可以通过矢量化来显着加速代码。不是循环遍历图像中的每个x和y值，而是使用meshgrid构建包含x和y的所有组合的数组，并将操作应用于数组。这个SO问题包含matlab中最近邻插值旋转的实现，它是矢量化的：

Image rotation by Matlab without using imrotate

Answer 2

我认为当MATLAB使用英特尔的IPP库时会发生神奇的事情：

http://software.intel.com/en-us/intel-ipp

Matlab如何快速地转动？

2 个答案: