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

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mr mo on 6 Nov 2017
Commented: Guillaume on 8 Nov 2017
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)

KL on 6 Nov 2017
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 on 6 Nov 2017
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)

KL on 6 Nov 2017
Edited: KL on 6 Nov 2017
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)

KL on 7 Nov 2017
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.
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 on 7 Nov 2017
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 on 7 Nov 2017
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.

Guillaume on 7 Nov 2017
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)

Show 1 older comment
Guillaume on 7 Nov 2017
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 on 7 Nov 2017
Also this error was happened.
comp(A(:, 1) > V) = inf;
Error using >
Matrix dimensions must agree.
Guillaume on 8 Nov 2017
Same issue, same solution, use bsxfun:
comp(bsxfun(@gt, A(:, 1), V)) = inf;