If-condition: if number is +-5% of another

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Lukas Netzer
Lukas Netzer am 16 Mai 2021
Bearbeitet: Jan am 16 Mai 2021
I'm trying to set an if-condition like that:
for x = 1:1:size(t{n})
if t{n}.sh(x) = +-0.05 t{n}.DiffMileagekm(x) %here is where I am trying to get the condition
somecalculation
end
end
E.g. sh(1) = 50 and DiffMileagekm(1) = 49, sh is in the range of +-5% of Diffmileagekm and therefor the calculation will be performed.
Is there a way to do that?
  1 Kommentar
Lukas Netzer
Lukas Netzer am 16 Mai 2021
I now set the if-condition like that:
if (t{n}.sh(x) >= 0.95 * t{n}.DiffMileagekm(x)) && (t{n}.sh(x) <= 1.05 * t{n}.DiffMileagekm(x))
Is there a better way to do that - or is it just fine like that?

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Jan
Jan am 16 Mai 2021
Bearbeitet: Jan am 16 Mai 2021
Are you sure, that this is correct:
for x = 1:1:size(t{n})
if t{n}.sh(x) = ...
Does t{n}.sh have as many elements as the size of t{n} is?
Note: 1:size(X) might not do, what you expect. size(X) replies a vector. 1:[a,b] replies 1:a and ignores b. This is at least confusing. Prefer numel(X) or specify the dimension with size(X, 1).
If you code is really correct, this would do what you are asking for:
sh = t{n}.sh; % Easier to read and faster
DiffM = t{n}.DiffMileagekm;
for x = 1:size(t{n}, 1)
if abs((sh(x) - DiffM(x))) < 0.05 * DiffM(x) % [EDITED]
...
end
end

Weitere Antworten (1)

Image Analyst
Image Analyst am 16 Mai 2021
Your if statement is not proper. For one thing you don't have a * in front of the t. Secondly you're using = instead of ==. Third, +- doesn't mean plus or minus -- it means minus (if it even works). Fixed code:
for x = 1:1:size(t{n})
difference = abs(t{n}.sh(x) - t{n}.DiffMileagekm(x)); % abs() takes care of plus or minus (above or below)
if difference < (0.05 * t{n}.DiffMileagekm(x))
% Difference is less than your threshold so do something.
somecalculation
end
end
You might also look at the ismembertol() function.

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