Finding the values right after and right before some values in a matrix

Hi. Suppose I have the matrix A of size (m*n).

 A = [12  44
      93  43
      128 44
      145 41
      180 41
      220 40
      280 40];

also I have the vector V e.g.

 V = [13 20 70 90 95 100 102 110 129 130 145 158 170 185 190 200 207 220 270 280 285 290];

I want to find which one of the rows of matrix A based on its first column values is right after and right before the values of vector V, and if one of the values of the first column of matrix A is equal to the values of vector V, I want to find that row.

For example :

 A(2,1) is right before V(5) and V(6) and V(7) and V(8)
 A(3,1) is right after  V(5) and V(6) and V(7) and V(8)

and in case of equality I want to find this

 A(4,1) = V(11)
 A(5,1) = V(18)
 A(7,1) = V(20)

In other words, I want to find exactly which members of vector V are between which members of first column of matrix A with their indices like this :

 V(1) and V(2) and V(3) and V(4) is between A(1,1) and A(2,1)

Thanks for your help.

[KL]

what do you mean by "right before"? If I want to explain it myself I would say 93 is right before 94 (and 94 only!). It is not right before 95!

and right after is the immediate next number, not every number after it.

[mr mo]

I am looking for this. For example

 V(1) and V(2) and V(3) and V(4) is between A(1,1) and A(2,1)

Answer 1

[KL]

 res = find(V>A(1,1) & V<A(2,1))

for all elements,

 res = arrayfun(@(x,y) find(V>x & V<y), A(1:end-1,1), A(2:end,1),'uni',0)

[mr mo]

I need to do this process for A(:,1) not only the first and second row. also I want to show which members of V are between which members of A with their index. Thanks a lot.

[KL]

that's easy too, check the edited answer. The result you will get as a cell array and to access it,

 res{1,1} or res{2,1}

[mr mo]

I checked it, but the result is not clear enough. This code is only show the indices of members of vector V and doesn't tell these indices of V are between what values of first column of matrix A. Thanks a lot.

[mr mo]

Also this code doesn't show the equality in members of vector V and members of first column of matrix A.

[KL]

"I checked it, but the result is not clear enough"

It is clear enough and you just don't read it properly to understand it.

 res = arrayfun(@(x,y) find(V>x & V<y), A(1:end-1,1), A(2:end,1),'uni',0)

In this arrayfun, I pass A(1:end-1,1) and A(2:end,1), which means, during every iteration, V is checked if it's greater than one element of A and less than the next element. The result is,

 res =

6×1 cell array

    [1×4 double]   %between A(1,1) and A(2,1)
    [1×4 double]   %between A(2,1) and A(3,1)
    [1×2 double]   %between A(3,1) and A(4,1)
    [1×2 double]   %between A(4,1) and A(5,1)
    [1×4 double]   %between A(5,1) and A(6,1)
    [        19]   %between A(6,1) and A(7,1)

And yes, I forgot the give you the code to find equality, it's much simpler,

 res_eq = arrayfun(@(x) find(V==x), A(:,1),'uni',0)

Hope this time you understand the result.

[mr mo]

Thanks again for your help. I understand the code but I want the output of this code show the indices of V in the form of numbers not in cell arrays and show the values of A in numbers not in the form of

 %between A(3,1) and A(4,1)

for example I want the output of this code is a matrix named out, which its first and second members in the first row are the A(1,1) and A(2,1) and the other members in this row are the members of vector V that are lying between the A(1,1) and A(2,1)

 out(1,:) = [ 12 93 13 20 70 90];

Thanks a lot.

[KL]

You see, we can make a matrix if the number of elements is same for each row, if you take res, first two rows have 4 elements and then followed by 2 element rows and again 4 element row, finally just one element.

For this reason only we use cell array.

https://de.mathworks.com/help/matlab/cell-arrays.html

They are very easy to use. You just need to use {} braces. with the current "res", if you type on the command line,

 res{1,1}

you will get the indices.

Anyway, if you need the values of A and V, we can create a cell array called out like,

 out = arrayfun(@(x,y) [x, y, V(find(V>x & V<y))], A(1:end-1,1), A(2:end,1),'uni',0)

now if you want to access the contents of first row,

 >> out{1,1}
 ans =

    12   93   13   20   70   90

[mr mo]

Thanks a lot. You are right about the "we can make a matrix if the number of elements is same for each row". I just forgot that. This is what I need.

[mr mo]

Suppose I want to find that only V(10)=130 is lying between which two members of A(:,1). How can I do this process to have the out cell array like your last code?

And in case of equality how can I have the out cell array like the case of inequality? Thanks a lot.

Answer 2

[Guillaume]

An alternative method that does not involve loops (or arrayfun) but is not necessary faster than KL's:

comp = abs(A(:, 1) - V);
comp(A(:, 1) > V) = inf;
[~, beforeVidx] = min(comp, [], 2);
isequaltoV = A(:, 1) == V(beforeVidx)';
%for display only:
table(A, beforeVidx, isequaltoV)

[mr mo]

Thanks for your help. I run the first line and this error was happened :

 comp = abs(A(:, 1) - V);
 Error using  - 
 Matrix dimensions must agree.

[Guillaume]

You're using an old version of matlab (<R2016b). It's always a good idea to mention such things. In pre-R2016b,

comp = bsxfun(@minus, A(:, 1), V);

[mr mo]

Also this error was happened.

 comp(A(:, 1) > V) = inf;
Error using  > 
Matrix dimensions must agree.

[Guillaume]

Same issue, same solution, use bsxfun:

comp(bsxfun(@gt, A(:, 1), V)) = inf;