Finding array index for a given array element does not work all the time

4 次查看（过去 30 天）
Anders Tagmose Lundsfryd
评论： Stephen23 2024-3-1
Ran in:
Hello,
I've been using the find(ismemeber()) function to find the index of an array element, however it does not work all the time for some reason. As you can see from the picture, it returns and empty vector at some given search values. As it can be seen from the array, it should return 393 for the last line of code. I am using the latest installation of matlab at the time of writting (2023b update 7).
To me this looks like a bug with the ismemeber() function, as it returns a false for the second entry(0.3920) but not the first entry(0.3910), se last two lines of code below. However, it could just be that i am doing something wrong? As I construct the array with a start, end, and stepsize, it shouldn't be a round of error?
Best regards Anders
mf = 0:0.001:0.6;
find(ismember(mf,0.3910))
ans = 392
find(ismember(mf,0.3920))
ans = 1×0 empty double row vector
any(ismember(mf,0.3910))
ans = logical
1
any(ismember(mf,0.3920))
ans = logical
0
  1 个评论

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Dyuman Joshi
Dyuman Joshi 2024-3-1
Ran in:
Welcome to the world of floating point numbers, where not all numbers can be represented exactly in binary form.
See this thread for more information - https://in.mathworks.com/matlabcentral/answers/57444-why-is-0-3-0-2-0-1-not-equal-to-zero?s_tid=faqs_link
When comparing floating point numbers, the best practice is to use a tolerance -
mf = 0:0.001:0.6;
val = 0.3910;
tol = 1e-6;
idx1 = find(abs(mf - val) < tol)
idx1 = 392
You can also use ismembertol here -
out = ismembertol(mf,val);
idx2 = find(out)
idx2 = 392

更多回答（1 个）

Chunru
Chunru 2024-3-1
Ran in:
Mind the round off error in searching double array
mf = 0:0.001:0.6;
i = find(abs(mf - 0.3910) < 1e-6)
i = 392

类别

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

标签

产品


版本

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by