感谢您的所有答案和有用的建议.我接受了pseudoDust,Hugues,Dima和High Performance Mark的建议并编写了我自己的代码.我的代码如下：

clc;clear all;close all;

x=32;

I = imread('coins.png');

I = imresize(I,[x x]);

I = double(I(:,:,1));

tic

[dx dy] = gradient(I);

toc

tic

[m,n]=size(I);

A = [I(:,2:end) zeros(m,1)];

B = [zeros(m,1) I(:,1:end-1)];

dx1 = [I(:,2)-I(:,1) (A(:,2:end-1)-B(:,2:end-1))./2 I(:,end)-I(:,end-1)];

A = [I(2:end,:) ; zeros(1,n)];

B = [zeros(1,n) ; I(1:end-1,:)];

dy1 = [I(2,:)-I(1,:) ; (A(2:end-1,:)-B(2:end-1,:))./2 ; I(end,:)-I(end-1,:)];

toc

nnz(dx-dx1)

nnz(dy-dy1)

我的基本想法是：渐变平均2个相邻位置(左和右或顶部和底部),除了它取值和相邻位置之间的差异的边缘.然后,我用matlab梯度函数(dx,dy)生成的矩阵检查了我生成的矩阵(dx1,dy1).

Elapsed time is 0.010232 seconds.

Elapsed time is 0.000066 seconds.

ans =

0

ans =

0

所以我相信我的代码是正确的.至少可以说,时间结果令人惊讶.然后我用matlab计算我的代码,用于不同大小的图像.

我得到了这个结果：

%x=16

Elapsed time is 0.010790 seconds.

Elapsed time is 0.000057 seconds.

%x=32

Elapsed time is 0.010564 seconds.

Elapsed time is 0.000069 seconds.

%x=64

Elapsed time is 0.010627 seconds.

Elapsed time is 0.000152 seconds.

%x=128

Elapsed time is 0.011346 seconds.

Elapsed time is 0.000669 seconds.

%x=256

Elapsed time is 0.017311 seconds.

Elapsed time is 0.004468 seconds.

%x=512

Elapsed time is 0.044148 seconds.

Elapsed time is 0.030435 seconds.

%x=1024

Elapsed time is 0.093386 seconds.

Elapsed time is 0.093029 seconds.

%x=2048

Elapsed time is 0.345423 seconds.

Elapsed time is 0.387762 seconds.

所以我的结论是这样的：对于高达1024X1024的图像大小,我的代码比matlab中内置的渐变命令更快.

编辑：我更新了我的答案并添加了此图表：

它清楚地表明,对于较小的数组大小,我的代码明显快于matlab梯度函数.

我的代码是否正确？伙计们请仔细检查并检查.请提供反馈.我实际上是matlab的新手,我对这个结果非常惊讶.请检查我在做什么是否正确？