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计算机器人 transformation matrix 相关内容时，对于关节角度进行离散，循环计算很慢，随着角度划分越来越细，怎么提高速度是一个问题。

最优解决方法

fun_handle = matlabFunction(T_t2b_RPY_tmp);

T_t2b_RPY_tmp是 transformation matrix， 其中使用 符号类型 syms 关节角度，直接把它转换成函数。注意查看函数变量顺序

fun_handle =function_handle with value:@(theta_P,theta_R,theta_Y)reshape([cos(theta_Y).*sin(theta_R)+cos(theta_R).*sin...

然后直接使用循环进行计算

tic
fun_handle = matlabFunction(T_t2b_RPY_tmp);
AAA = zeros(4,4,prod(num_point));
p = 0;
for i = Rollfor j = Pitchfor k = Yawp = p + 1;AAA(:,:,p) = fun_handle(j,i,k);endend
end
toc

使用符号计算 时间差别 上千倍

tic
BBB = zeros(4,4,prod(num_point));
p = 0;
for i = Rollfor j = Pitchfor k = Yawp = p + 1;BBB(:,:,p) = double(subs(T_t2b_RPY_tmp, {theta_R theta_P theta_Y}, {i j k}));endend
end
toc

问题：两个计算结果有差别

把两者结果作差，D1 = AAA-BBB;， 可以看到结果不一样，不过都小于 10^-4.

val(:,:,1) =1.0e-15 *0    0.0000         0         00   -0.0612         0         00.0612         0    0.0000    0.61230         0         0         0val(:,:,2) =1.0e-04 *-0.2846    0.0000    0.2190    0.1895-0.2190   -0.0000   -0.2846    0.15370.0000         0    0.0000    0.00000         0         0         0val(:,:,3) =1.0e-04 *-0.1169    0.0000    0.4760   -0.2399-0.4760   -0.0000   -0.1169   -0.16910.0000         0    0.0000    0.00000         0         0         0

isequal(round(AAA,5), round(BBB,5)) 对比精度

这个是由于两者计算精度不一样，具体原理没有搞清楚，详情参考

• Why is there a significant difference in the assignment results of functions using “subs” and “feval”
• Problem of creating function handle from the result of symbolic differentiation
• Getting different results from function handle & syms for a same equation. How to avoid it?
• Why I got different result by using matlabFunction and subs to do symbolic substitution?