check whether an interval instant belongs to an array of intervals
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Muhammad Atif
am 14 Apr. 2020
Kommentiert: Walter Roberson
am 14 Apr. 2020
Hi everbod!
have an array of intervals like this C = {[1.54,2.49], [4.97,5.88], [6.69,7.98], [8.05,8.25],[8.58,9.09], [10.48,10.86], [11.16,12.97], [14.51,15.24],[15.38,15.60], [17.20,17.74], [19.16,22.38], [23.69,23.73],[24.59,25.47], [25.59,26.22], [26.65,29.27]};
I want to check if current intervals instant belongs to this array. Where interval divided into different parts e.g. 1.54 to 2.49 is the first part and so on..How can I make program for this?
for example
A=2;
whether or not belongs to 'C'? Help me please
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Kevin Hellemans
am 14 Apr. 2020
Hi,
I would use cellfun to loop through the intervals and check in each cell. The function belows takes the cell array C and value A, checks if A is part of C and, returns true/false (tf) and if true, provides the position as optional output.
function [tf, position] = ValueInArray(C,A)
result = cellfun(@(x) inInterval(x,A),C);
tf = sum(result) > 0;
if tf
position = find(result);
else
position = [];
end
end
function tf = inInterval(interval,val)
lowerLimit = val >= min(interval);
upperLimit = val <= max(interval);
tf = lowerLimit && upperLimit;
end
With the values provided, this is my output:
[tf, position] = ValueInArray(C,A)
tf =
logical
1
position =
1
1 Kommentar
Walter Roberson
am 14 Apr. 2020
tf = sum(result) > 0;
if tf
position = find(result);
else
position = [];
end
can be simplified to
position = find(result);
provided that C is scalar. (If it is not scalar then your code would fail.)
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