Save values in a loop in a vector

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Daniel 2016-9-8

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评论： mbonus 2016-9-12

Hi !

I would need some help for my home assignment cause im stucked. I need to iterate thru a vector and look for some values that are equal to 1.00, 0.80, 0.60, 0.40, 0.20 and 0.10. Then i need to store those values in another vector, how do i do this? Below u see the code for plotting those but i need to save them all in a vector so that i can make a nice table!

for i=1:size(d)

    if d(i)==1.00000
        disp(h(i))
    elseif d(i)==0.80000
        disp(h(i))
    elseif d(i)==0.60000
        disp(h(i))
    elseif d(i)==0.40000
        disp(h(i))
    elseif d(i)==0.20000
        disp(h(i))
    elseif d(i)==0.10000
        disp(h(i))
    end
end

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回答（2 个）

mbonus 2016-9-8

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编辑：mbonus 2016-9-8

在 MATLAB Online 中打开

before the loop

v = zeros(size(d));

then just insert this line for each part of the if structure

v(i) = d(i);

If you need to get rid of the indexes with no values you can do this after the loop

v(find(0)) = [];

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Daniel 2016-9-9

and result i get is all i positions in vector d which is == to 1.00, 0.80 and so on. But what i need is to use those i`s and save the values in z(i) in a new vector Z

mbonus 2016-9-12

Could post an example of what you get and what it should look like?

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Thorsten 2016-9-9

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I found it a bit hard to understand what your are looking for. As far as I understood, this can solve your problem:

p =  [1.00, 0.80, 0.60, 0.40, 0.20, 0.10];
for i = 1:numel(p) % for all values in p
  Z{i} = z(h == p(i)); % find all positions in h that equal p(i), and assign the corresponding positions in z to a new variable Z
end

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Daniel 2016-9-9

For exampel: I have d=[1 22 31 41 51 6 7 8 9 10]; z=[1 2 3 4 5 6 7 8 9 10];

Now, i want to search in d for a value of 41. Then, give me that position, which in this case is on fourth column, hence i = 4

Now, go into the z vector and find me the value of column 4 and take that value and place it into Z vector :)

Thorsten 2016-9-9

编辑：Thorsten 2016-9-9

在 MATLAB Online 中打开

Z = z(d==41)

In my code above d == h and p(i) == 41. Because there can be, depending on your data, in principle one, two, or even more matches for d== 41, you have to store one, or two, etc values in Z, i.e., a different number of values for each i. That's why I use Z{i}. If you can guarantee that there is always one and only one match for each i, you can use Z(i).

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