problems with function "find". matlab doesn't match two numbere which are built equal

M cannot be found because Matlab says they are different, but i built them in order to be equal. length(data)=7155

clear all;clc;close all;

data0 = load('data0000.txt');
data1 = load('data0001.txt');
data2 = load('data0002.txt');

Rg=17.5;

fs = 100; % Hz
dt = 1/fs;
data= [data0; data1; data2];
time= 0:dt:(length(data)-1)*dt;

N=find(time==12.01);
M=find(time==46.5);
data=data(N:M,:);
time=time(N:M);

[Luna]

Could you please attach data0001.txt and data0002.txt ?

Answer 1

[dpb]

dt = 1/fs;
data= [data0; data1; data2];
time= 0:dt:(length(data)-1)*dt;

is the root of the problem; can be demonstrated without the other data files...

>> time= 0:dt:(7155-1)*dt;   % build the time vector based on length and dt
>> find(time==46.5)          % and look for exact match...
ans =
1×0 empty double row vector
>>

which fails. Huh!???!!! Whassup w/ that--

>> find(time>46.5,1)  % ok where is value close to it?
ans =
      4651
>> time(4650:4652)    % see what look like, to four decimal points matches
ans =
 46.4900   46.5000   46.5100
>> time(4651)==46.5     % is it really so or did something go wrong internally
ans =
logical
 0
>> time(4651)-46.5      % what's the difference between what "looks like" match and target?
ans =
 7.1054e-15
>>

It is close, but not exact.

This is floating point roundoff wherein the multiple application of the FP approximation to 1/100 is accrued in the computations and sometimes the calculated number is not quite the same as a value entered as text or computed by another technique not subject to the same rounding.

>> time=linspace(0,7154/100,7155);
>> t1= 0:dt:(7155-1)*dt;
>> all(t1==time)
ans =
logical
 0
>> sum(t1==time)
ans =
      3941
>> max(abs(t1-time))
ans =
 1.4211e-14

>> find(time==46.5)  % with |linspace|, don't see same rounding error
ans =
      4651
>>

However, still is much more robust to use ismembertol when searching/matching for floating point values unless can explicitly control rounding.

[Marco Lorello]

now it's clear! thank you very much, it seemed to me like black magic till now ahah

Answer 2

[jonas]

It is better to compare floats using a tolerance

tol=0.01;
M = find(abs(time-46.5)<tol);

I was however able to find the first number, N, using your code.

N =

        1202

the second number, M, is probably in another textfile because there is no value near 46.5 in the one you attached

[dpb]

ismembertol can be useful for such purposes.