Using "find" for finding decimal values

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MiauMiau . 2012년 12월 13일
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?

채택된 답변

Matt Fig
Matt Fig 2012년 12월 13일
편집: Matt Fig 님. 2012년 12월 13일
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))
And show us the ouput in a comment on this answer, don't add another answer!
I have a feeling you will need to set your tolerance much higher, like 10^-4.
  댓글 수: 4
Matt Fig
Matt Fig 2012년 12월 14일
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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추가 답변 (4개)

Azzi Abdelmalek
Azzi Abdelmalek 2012년 12월 13일
idx = find(abs(X(:,1 )-17.9203)<eps)
  댓글 수: 2
Kye Taylor
Kye Taylor 2012년 12월 13일
should be <=

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Jan 2012년 12월 13일
편집: Jan 님. 2012년 12월 13일
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.

MiauMiau 2012년 12월 13일
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!..?!
  댓글 수: 7
Jan 2012년 12월 13일
@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 2012년 12월 13일
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?
  댓글 수: 1
Jan 2012년 12월 13일
Why do you assume that 6.0018 is an element of X?

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