add number to cell

Question

NA 2019-1-31

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/442465-add-number-to-cell

编辑： Jan 2019-2-6

I have A={[1,2,4],[2,3,4],[2,4,5],[4,5,6,9,10,11],[4,7,9],[7,8],[6,12,13]} and ref={[1],[1],[1],[3],[2],[1]} (ref means that which index in A is chosen as reference). So according to each referance ang is calculated in each array. ang={[0; -0.08;-0.17],[0;-0.14;0.09],[0;-0.09;-0.05],[0.05; 0.09; 0; -0.02;-0.03;-0.03],[0;-0.02;-0.01],[0.004;0],[0;-0.02;-0.01]}

[0; -0.08;-0.17] means that first elemnt is chosen as referance so -0.08 is calculated according to 1 as reference.

As I want to change all referance according to 1.

[1,2,4]=[0; -0.08;-0.17] this one is not changing

[2,3,4] as 2 is reference here so I need scale up all element to -0.08 (add all element to -0.08)

[2,3,4]=[0+(-0.08);-0.14+(-0.08);-0.09+(-0.08)] =[-0.08;-0.22;-0.17]

for [2,4,5] as reference is not changing only need to add all to -0.08

[2,4,5]=[0;-0.09+(-0.08);-0.05+(-0.08)]=[0;-0.17;-0.13]

for [4,5,6,9,10,11] as 6 is reference, I need to scale up. we know that ang(4)should be-0.17 and ang(4) here is 0.05. need to change 0.05 to -0.17 .

(0.05+(X)=-0.17, X=-0.22) need to add all to -0.22

[4,5,6,9,10,11]=[0.05+(X); 0.09-0.22; 0+(-0.22); -0.02+(-0.22);-0.03+(-0.22);-0.03+(-0.22)] =[-0.17;-0.13;-0.22; -0.24;-0.25;-0.25]

for [4,7,9] ang(4) should be -0.17. so add all to -0.17

[4,7,9]=[0+(-0.17);-0.02+(-0.17);-0.01+(-0.17)]=[-0.17;-0.19;-0.18]

for [7,8] from the above we know ang(7) should be -0.19 so add all to -0.19

[7,8]=[0.004+(X);0+(-0.186)]=[-0.19;-0.186] 0.004+(X)=-0.19 X=-0.186

[6,12,13]=[0+(-0.22);-0.02+(-0.22);-0.01+(-0.22)]

result should be [1,2,3,4,5,6,7,8,9,10,11,12,13]=[0,-0.08,-0.22;-0.17;-0.13;-0.22;-0.19;-0.186; -0.18;-0.25;-0.25]]

I also try this

tmp = cellfun(@(c,p,m)c+c(p(m)),ang,A,ref,'uni',0);

but it is not working

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Jan 2019-2-1

编辑：Jan 2019-2-1

@Naime: Whenever you mention in the forum, that you get an error, post a copy of the complete error message. Provide all information, which is required to help you.

NA 2019-2-1

编辑：NA 2019-2-1

but I really don't get the "I need scale up all element to -0.08 (add all element to -0.08)". Where does that -0.08 come from?

I mean for this array [2,3,4], I need to add all ang [0;-0.14;0.09] to -0.08.

As in the first one [1,2,4]=[0; -0.08;-0.17] ang(2) is -0.08.

So, if I want to compare all ang I need to chage reference. as reference is different in each array of A.

Same for ''we know that ang(4)should be-0.17''Do we? Why?

After adding -0.08 to [0;-0.14;0.09] I have [-0.08;-0.22;-0.17]. so [-0.08;-0.22;-0.17] correspond to [2,3,4]. ang(4)=-0.17

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Answer 1

Jan 2019-2-6

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/442465-add-number-to-cell#answer_359752

编辑：Jan 2019-2-6

在 MATLAB Online 中打开

I haven't seen a demand for such an indirekt algorithm yet, but if this really satisfies your needs, there is no reason to change the code. With the above mentioned simplifications:

A   = {[1,2,4],[2,3,4],[2,4,5],[4,5,6,9,10,11],[4,7,9],[7,8],[6,12,13]};
ref = {[1],[1],[1],[3],[1],[2],[1]};
ang = {[0; -0.08;-0.17],[0;-0.14;-0.09],[0;-0.09;-0.05],[0.05; 0.09; 0; -0.02;-0.03;-0.03],[0;-0.02;-0.07],[0.004;0],[0;-0.02;-0.01]};
new_ang = zeros(1, max([A{:}]));
for k = 1:numel(A)
    index = A{k};
    first = index(1);
    if ang{k}(1) == 0
        % if any(A{k} == 1)
        %    new_ang(index) = new_ang(first) + ang{k};
        % elseif any(A{k} == first)
        %    new_ang(index) = new_ang(first) + ang{k};
        % end
        
        % Because both assigments are equal, this is more elegant:
        if any(A{k} == 1) || any(A{k} == first)
           new_ang(index) = new_ang(first) + ang{k};
        end
    else
        new_ang(index) = new_ang(first) - ang{k}(1) + ang{k};       
    end
end

Or

new_ang = zeros(1, max([A{:}]));
for k = 1:numel(A)
    index = A{k};
    first = index(1);
    if ang{k}(1) ~= 0 || (any(A{k} == 1) || any(A{k} == first))
        new_ang(index) = new_ang(first) - ang{k}(1) + ang{k};       
    end
end

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Answer 2

Jan 2019-2-1

1
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https://ww2.mathworks.cn/matlabcentral/answers/442465-add-number-to-cell#answer_359032

在 MATLAB Online 中打开

I strongly recommend to avoid the smart and neat cellfun, if its compact notation confuses the programmer. A simple loop might look less professional, but it can save hours for debugging.

tmp = cell(size(ang));
for k = 1:numel(A)
    tmp{k} = ang{k} + ang{A{ref{k}}};
end

In your cellfun approach, A is treated as 2nd input, such that the anoynmous function

@(c,p,m)c+c(p(m))

defines p=A{k}, but this is not wanted. I assume you want:

tmp = cellfun(@(c,m) c+c(A(m)), ang, ref, 'uni', 0);

Anonymous functions cannot be debugged directly, but the suggested loop can. So do not make the programming harder than needed.

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NA 2019-2-4

编辑：NA 2019-2-4

在 MATLAB Online 中打开

I have

A={[1,2,4],[2,3,4],[2,4,5],[4,5,6,9,10,11],[4,7,9],[7,8],[6,12,13]} 
ang={[0; -0.08;-0.17],[0;-0.14;-0.09],[0;-0.09;-0.05],
    [0.05; 0.09; 0; -0.02;-0.03;-0.03],[0;-0.02;-0.07],
    [0.004;0],[0;-0.02;-0.01]}

for [1,2,4] ang(1)=0, ang(2)=-0.08, ang(4)=-0.17.

for [2,3,4] we know from above that ang(2)=-0.08, so add all [0;-0.14;-0.09] to -0.08, ang(2)=-0.08, ang(3)=-0.14-0.08=-0.22, ang(4)=-0.09-0.08=-0.17.

[2,4,5] we know from above that ang(2)=-0.08, so add all [0;-0.09;-0.05] to -0.08, ang(2)=-0.08, ang(4)=-0.09-0.08=-0.17, ang(5)=-0.05-0.08=-0.13.

[4,5,6,9,10,11] we know from above that ang(4)=-0.17, in [0.05; 0.09; 0; -0.02;-0.03;-0.03] ang(4) is not zero so 0.05+X= -0.17 , X=-0.22

So add all to -0.22, ang(4)= 0.05+-0.22= -0.17, ang(5)= 0.09+-0.22=-0.13,

ang(6)= 0+-0.22=-0.22, ang(9)= -0.02+-0.22=-0.24, ang(10)= -0.03+-0.22=-0.25, ang(11)= -0.03+-0.22=-0.25,

[4,7,9] we know from above that ang(4)=-0.17 so add all [0;-0.02;-0.07] to -0.17, ang(4)= -0.17, ang(7)= -0.02-0.17=-0.19 ,ang(9)= -0.07-0.17=-0.24

for [7,8] we know from above that ang(7)=-0.19, in [0.004;0] ang(7)=0.004

ang(7) is not zero so 0.004+X= -0.19 , X=-0.186, So add all to -0.186, ang(7)= -0.19, ang(8)= 0+-0.186=-0.186

for [6,12,13] we know from above that ang(6)=-0.22 so add all [0;-0.02;-0.01] to -0.22, ang(6)= -0.22, ang(12)= -0.02-0.22=-0.24 ,ang(13)= -0.01-0.22=-0.23

result should be [1,2,3,4,5,6,7,8,9,10,11,12,13]=[0,-0.08,-0.22;-0.17;-0.13;-0.22;-0.19;-0.186; -0.24;-0.25;-0.25;-0.24,-0.23]]

NA 2019-2-4

在 MATLAB Online 中打开

 the variable A is used only to determine the size, but the elements are not used.

in this system I have a 13 measurements (1 to 13) and measuremnts is grouped according to A={[1,2,4],[2,3,4],[2,4,5],[4,5,6,9,10,11],[4,7,9],[7,8],[6,12,13]};

So, A is important as illustrate which measurement has which value (ang).

ang={[0; -0.08;-0.17],[0;-0.14;-0.09],[0;-0.09;-0.05],[0.05; 0.09; 0; -0.02;-0.03;-0.03],[0;-0.02;-0.07],[0.004;0],[0;-0.02;-0.01]};

In each group base measuremnet (measurement that is zero) is different. As I want to compare all 13 measurements, I need to have one base and other values changing according to this base (measurement 1)

this base measurement is understandable according to ref={[1],[1],[1],[3],[1],[2],[1]} ;

 Correct one
 [7,8]=[0.004+(X);0+(-0.194)]=[-0.19;-0.194]   
0.004+(X)=-0.19   X=-0.194

As measurement 8 is located only in one group [7,8], I could not compare value of this measurement according to other. But 7 is in group [4,7,9] and calculate ang(7) = -0.02-0.17=-0.19 , ang (7) should be -0.19 in all groups

[7,8] without changing to one base is [0.004;0]. if I change to one base ang(7) = -0.19

for this reason 0.004 should be change to -0.19 so, need to add 0.004 to -0.194 to get -0.19.

So ang(8) become-0.194

NA 2019-2-5

在 MATLAB Online 中打开

A={[1,2,4],[2,3,4],[2,4,5],[4,5,6,9,10,11],[4,7,9],[7,8],[6,12,13]};
ref={[1],[1],[1],[3],[1],[2],[1]};
ang={[0; -0.08;-0.17],[0;-0.14;-0.09],[0;-0.09;-0.05],[0.05; 0.09; 0; -0.02;-0.03;-0.03],[0;-0.02;-0.07],[0.004;0],[0;-0.02;-0.01]};
idx = [A{:}];
new_ang = zeros(1,max(idx));
for k = 1:numel(A)
    index=A{k};
    first_element=index(1);
    if ang{k}(1)==0
        if any(ismember(A{k},1))==1
            
            new_ang(index)=new_ang(first_element)+ang{k};
            
        else
            n=find(new_ang~=0);
            
            if any(ismember(A{k},first_element))==1
                
                new_ang(index)=new_ang(first_element)+ang{k};
            end
            
        end
    else 
        
        X=new_ang(first_element)-ang{k}(1);
        new_ang(index)=X+ang{k};
       
    end
    
end