How can I calculate fzero from values of matrices in cell array?

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Szabó-Takács Beáta
评论： Szabó-Takács Beáta 2015-4-22
Dear All, I have three cell arrays: cv_o, std_m and mean_m. Both cells contains 359 2D matrices. The size of matrices is same. I would like to calculate the root of an equation using the values of these matrices. I tried to calculate fzero by the following way:
for jj=1:359;
for i=1:464;
for j=1:201;
mat(i,j)=cv_o{jj};
mstd(i,j)=std_m{jj};
mmean(i,j)=mean_m{jj};
if mat(i,j) == 0;
b(i,j) = 0;
else
b(i,j)=fzero(@(x)mat(i,j)-(mstd(i,j)./mmean(i,j)).^x,(0));
end
end
end
end
But I got the following error message: Subscripted assignment dimension mismatch. Could someone write me how I should solve calculate the root?

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Michael Haderlein
Michael Haderlein 2015-4-20
编辑：Michael Haderlein 2015-4-20
Dear Beáta,
Check this line (same holds for the next two lines):
mat(i,j)=cv_o{jj};
As you said, cv_o{jj} will return a matrix; but mat(i,j) is one scalar value and this mismatch is causing the error. Given that varname(i,j) returns a scalar, you'll see that also your equation can be solved analytically: x=log(mat(i,j))/log(mstd(i,j)/mmean(i,j)) I guess you want to have matrix operations here, but right now it's all scalar. Possibly you just need to remove all those (i,j), but I cannot say that for sure. Also, you'll not get b(i,j) but rather b(jj).
  2 个评论
Michael Haderlein
Michael Haderlein 2015-4-21
To keep the thread readable, I copy-paste your post here. Please also answer as comment here. The sequence of answers might change and the thread will be confusing then.
Answer by Szabó-Takács Beáta:
Dear Michael, Thank you for your help! I changed the calculation according to your suggestion by the following way:
for jj=1:360;
for i=1:464;
for j=1:201;
mat=cv_o{jj};
mstd=std_m{jj};
mmean=mean_m{jj};
if mat(i,j) == 0;
b(i,j)=0;
elseif mat(i,j) > 0;
b(i,j)=fzero(@(x)mat(i,j)-(mstd(i,j)./mmean(i,j)).^x,(0.1));
BB{jj}=b;
end
end
end
end
It calculates the power b until jj=29. After that I got the following message: Error using fzero (line 309) Function value at starting guess must be finite and real. I do not understand why? I checked the values of cv_o. It does not contain NaN or Inf values. And I am not sure that values of b are right. I had tried calculate the function without using i, j but as you suggested but it did not create b(jj). Could you write me what the matter is?
=========== my response ==============
I had another look on your function, why do you actually use fzero at all? You are using element-wise operators (with dot), so my analytical solution from the above answer holds, too. You can simplify your code to
for jj=1:360;
mat=cv_o{jj};
mstd=std_m{jj};
mmean=mean_m{jj};
BB{jj}=log(mat)./log(mstd./mmean);
BB{jj}(mat==0)=0;
end
In principle, by using cellfun you could even cut it down to a one-liner without any loop (I was using rand values for cv_o, std_m, mean_m; also please note the transpose operator needed to show equality):
>>PP=cellfun(@(x,y,z) log(x)./log(y./z), cv_o, std_m, mean_m,'uniform',false);
>> isequal(PP',BB)
ans =
1
With respect to the error message, maybe mmean is zero somewhere. Then your function is inf (or NaN if the same mstd is also zero) and causes the error.

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更多回答（1 个）

Szabó-Takács Beáta
Dear Michael, Thank you for your help! I changed the calculation according to your suggestion by the following way:
for jj=1:360;
for i=1:464;
for j=1:201;
mat=cv_o{jj};
mstd=std_m{jj};
mmean=mean_m{jj};
if mat(i,j) == 0;
b(i,j)=0;
elseif mat(i,j) > 0;
b(i,j)=fzero(@(x)mat(i,j)-(mstd(i,j)./mmean(i,j)).^x,(0.1));
BB{jj}=b;
end
end
end
end
It calculates the power b until jj=29. After that I got the following message: Error using fzero (line 309) Function value at starting guess must be finite and real. I do not understand why? I checked the values of cv_o. It does not contain NaN or Inf values. And I am not sure that values of b are right. I had tried calculate the function without using i, j but as you suggested but it did not create b(jj). Could you write me what the matter is?

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