Issue with find() function

9 次查看（过去 30 天）
Mitchell Frankel
Mitchell Frankel 2014-6-5
回答： Image Analyst 2014-6-5
If I create an array x = 0.1:0.05:0.6; and then ask find(x==0.15), I get an empty matrix. It also fails for x==0.40? I don't understand this at all.
  1 个评论
Cedric
Cedric 2014-6-5
If you understand the answers below, read about EPS. In the end, you will probably end up doing something like:
pos = find( abs(x - 0.15) < eps(1) ) ;

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2014-6-5
It’s called ‘floating point approximation error’. The concept is similar to the decimal approximation of 1/3 = 0.33333.... To get around it, you need to state the conditions in your find calls to allow for some imprecision.
Run these to get an idea of how the concept applies to your problem:
x = 0.1:0.05:0.6;
ix11 = find(x <= 0.15, 1, 'last')
ix12 = find(x >= 0.15, 1, 'first')
x1 = x([ix11 ix12])
ix21 = find(x <= 0.40, 1, 'last')
ix22 = find(x >= 0.40, 1, 'first')
x2 = x([ix21 ix22])

更多回答（2 个）

Azzi Abdelmalek
Azzi Abdelmalek 2014-6-5
Image Analyst
Image Analyst 2014-6-5
Here's some pretty general code based on the FAQ Azzi referred you to:
x = 0.1 : 0.05 : 0.6 % Sample data
targetValue = 0.15 % Whatever value you want to find
tolerance = .005 % We want to find x within this value of the target.
% Find values in range.
indexesInRange = abs(x - targetValue) <= tolerance
xThatAreInRange = x(indexesInRange)
% Or, in an if statement, if you want to use it in an if statement instead...
if abs(x(2) - targetValue) <= tolerance
% x(2) is close enough to the targetValue.
else
% x(2) is far away from the targetValue.
end

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by