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

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mr mo
mr mo 2017-11-6
评论： Guillaume 2017-11-8
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.
  2 个评论
KL
KL 2017-11-6
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
mr mo 2017-11-6
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)

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回答（2 个）

KL
KL 2017-11-6
编辑：KL 2017-11-6
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)
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mr mo
mr mo 2017-11-7
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
mr mo 2017-11-7
编辑：mr mo 2017-11-7
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.

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Guillaume
Guillaume 2017-11-7
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)
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mr mo
mr mo 2017-11-7
Also this error was happened.
comp(A(:, 1) > V) = inf;
Error using >
Matrix dimensions must agree.

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