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How do I find the nearest value to a number that is bigger than that number?

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Janna Hinchliff
Janna Hinchliff 2019-1-29
回答： achillakos 2022-10-27
I'm using the command
[~,indexToClosestMatch] = min(arrayfun(@(x)min(abs(x-val)),matrixcellarray{j}(:,1)));
to find the value in matrixcellarray{j}(:,1) that is closest in value to the number val. I want to additionally specify that I only want to consider values in matrixcellarray{j}(:,1) that are higher than val - i.e. if matrixcellarray{j}(:,1) was [1 2 5 6] and val was 3, I would want it to match to 5, not 2, as although 2 is closer, it is lower than 3. How can I achieve this?

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Guillaume
Guillaume 2019-1-29
编辑：Guillaume 2019-1-29
First, your command is overly complicated. arrayfun always pass scalar values to its function, so the x is your anonymous function is always be to be scalar (in turn each element of matrixcellarray{j}(:,1)). So abs(x-val) is scalar, and the minimum of a scalar value is always that value. The arrayfun is completely unnecessary, your code is the same as:
[~, indexToClosestMatch] = min(abs(matrixcellarray{j}(:,1) - val));
There are many ways you could achieve what you want. One way:
distancefromval = matrixcellarray{j}(:,1) - val;
distancefromval(distancefromval < 0) = +Inf;
[~, indexToClosestMatch] = min(distancefromval);
Another way:
greaterindices = find(matrixcellarray{j}(:,1) > val);
[~, intermediateindex] = min(matrixcellarray{j}(greaterindices, 1) - val));
indexToClosestMatch = greaterindices(intermediateindex);

更多回答（1 个）

achillakos
achillakos 2022-10-27
Ran in:
An easy way to quickly find the closest, either higher or lower, entries of an array to a specific value is to use the interp1 function with the 'next' option.
In the case that you have an array arr = [1 2 5 6] and the value is val = 3, and you want the closest higher entry, that would be:
arr = [1 2 5 6];
val = 3;
[~,indexToClosestMatch] = find(arr==interp1(arr,arr,val,'next'),1)
indexToClosestMatch = 3
In your case you just need to substitue arr for matrixcellarray{j}(:,1), and that would look like:
[~,indexToClosestMatch] = find(matrixcellarray{j}(:,1)==interp1(matrixcellarray{j}(:,1),matrixcellarray{j}(:,1),val,'next'),1)

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