Why find cannot handle this very simple task?

X = -0.1:.001:.25;
find(X == .077)

I get the following:

ans = 1×0 empty double row vector

However X(178) = 0.077. How to get back index 178?

Answer 1

[Stephan]

You are dealing with doubles, they are not precisly 0.077 - use round:

X = -0.1:.001:.25;
find(round(X,3) == .077)

[Mr M.]

ok, i see, but what about the right hand side, .077 is ok? round(.077,3) is not better?

[Mr M.]

I can not undersatnd: if .077 is good, than -0.1:.001:.25 is why not good? since elements of this vector is -0.100+0.001+0.001+etc., and each term is rounded to 3th order

[Steven Lord]

I wouldn't use round here. Decide how close is "close enough" and use that as the tolerance.

X = -0.1:0.001:0.25;
closeEnough = 1e-8;
X(find(abs(X-0.077) < closeEnough))

ans = 0.0770

Or work with integer values and convert to non-integer values as needed:

X2 = -100:250;
X2(X2 == 77)/1000

ans = 0.0770

Or use ismembertol which tries to choose a good "close enough" tolerance.

X(ismembertol(X, 0.077))

ans = 0.0770

Answer 2

[Jan]

Welcome to the world of calculations with numbers represented with limited precision.

See: FAQ: Why is 0.3-0.2 ~= 0.1