# Using "find" for finding decimal values

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MiauMiau il 13 Dic 2012
Hi
I use the matlab command importdata:
X = importdata('filename.csv');
to read in a csv file with three columns.
Now, for finding a specific values in the matrix X, I simple use the find command as follows:
idx = find(X(:,1 ) == 17)
But, the same seems not to be possible for decimal numbers. This for instance would not work:
idx = find(X(:,1 ) == 17.9203)
even though 17.9203 is to be found in the original csv file. What is the problem and what can I do?
Thanks
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### Risposta accettata

Matt Fig il 13 Dic 2012
Modificato: Matt Fig il 13 Dic 2012
MiauMiau, the numbers you are looking for are obviously different from the short format you see. Do this:
[~,idx] = min(abs(X(:,1) - 6.0018));
Y = X(idx,1);
[~,idx] = min(abs(X(:,1) - 17.9203));
fprintf('%15.15f %15.15f\n',Y,X(idx,1))
I have a feeling you will need to set your tolerance much higher, like 10^-4.
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MiauMiau il 13 Dic 2012
Thanks - it works perfectly, wow! I thought there is a number like 17.9203 because I used two input files which were supposed to be identical - obviously there were not. May I ask how you found out the tolerance has to be set to 10^-3? I don't understand that yet. Thank you very much!
Matt Fig il 14 Dic 2012
I found out by looking at how close your guess value was to the actual value. You may have to do some tweaking but you get the idea now.

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### Più risposte (4)

Azzi Abdelmalek il 13 Dic 2012
Try
idx = find(abs(X(:,1 )-17.9203)<eps)
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Jan il 13 Dic 2012
Modificato: Jan il 13 Dic 2012
Substracting two numbers in the magnitude of 20 can have a roundoff error of eps(20). Then this is better:
idx = find(abs(X(:,1 ) - 17.9203) <= eps(17.9203)) % EDITED: was "<"
Even 10*eps(17.9203) would be a reliable limit.
[EDITED] Thanks, Kye. I cannot test this currently. But I assume that even "<=" can fail, when the values are at the limits of 2^n. Does eps(16 + eps(16)) reply eps(32)? Then a factor of 2 would be obligatory.
Kye Taylor il 13 Dic 2012
should be <=

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Jan il 13 Dic 2012
Modificato: Jan il 13 Dic 2012
There is no exact representation of decimal floating point numbers in binary format for all values. You find a lot of corresponding discussion in this forum:
0.1 + 0.2 - 0.3 == 0
>> false
This is no bug, but the expected behaviour, when floating point numbers are represented with a limited precision.
A consequence is, that you cannot compare numbers like 17.9203 and 17.92029999999999999999 sufficiently and even the display in the command window can be rather confusing.
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MiauMiau il 13 Dic 2012
I was not aware of that, that is pretty horrible! Anyway, both of your suggestions did not work, so also for:
idx = find(abs(X(:,1 ) - 17.9203) < eps(17.9203))
I did get an empty matrix returned - although the value is to be found in X. Also, when I changed the number a bit, I then got too many answers returned. But what strikes me most, is that in the first case, I just did not get any answer at all!..?!
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Azzi Abdelmalek il 13 Dic 2012
Yes, I said Matlab because we are working with Matlab. I think, representing real numbers for numerical programming is almost similar to what an ADC (analogic-digital converter) do. There is a quantification of a real number, which means there is an error of quantification which depends on the number of used bits and method.
Jan il 13 Dic 2012
@Azzi: I've stressed this detail, because the OP seems to be confused about this topic already. I did not assume, that you struggle with floating point arithmetics.

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MiauMiau il 13 Dic 2012
I am really sorry for my use of the word "horrible".
If you meant the following command though ( I have used another nummerical value):
idx = find(abs(X(:,1 ) - 6.0018) <= eps(6.0018))
I - again - got an empty 0,1-matrix.
Also, I was not criticising the obviously necessary restriction on measurement and computation precision but I was amazed that there is not a simpler command to handle that. Anyway. What shall I try next?
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Jan il 13 Dic 2012
Why do you assume that 6.0018 is an element of X?

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