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dlmread adds low precision digits

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Hello all,
I'm using dlmread to import datas and when the datas contains integers and float numbers, dlmread sometimes makes an error on the last digit of the output precision.
I'm not sure I am clear, so here is an example :
If my text file ('test.rpt') contains only the row :
1 10.8000 31. 10.5001
the command
A = dlmread('test.rpt')
gives back :
A =
1.000000000000000 10.800000000000001 31.000000000000000 10.500100000000000
I don't understand why this 10e-15 is added on the second number. Ok, it's not a big error, but it prevent me from easly comparing two values...
Does anyone know how to prevent this behaviour ?

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Adam Danz
Adam Danz on 27 Aug 2019
I know 0 about rpt files so I can't recommend alternative methods to read in the data. You're currently reading in the data usling dlmread (which has been replaced by readmatrix() starting in r2019a).
Some methods allow you to read in data as text in which case 10.8000 would be read as a string (or char array) '10.8000'. Then you could convert that to numeric by using str2double() or num2str() (the prior is better).
dpb
dpb on 27 Aug 2019
> x=10.8
x =
10.8000
>> fprintf('%.15f\n',x)
10.800000000000001
>>
so it doesn't matter how you enter it, 10.8 is not exactly representable.
"I don't remember seeing this when using fscanf and specifing a number format."
Probably because you just didn't notice until you did try to do an exact comparison on floating point values.
>> fprintf('%.15f\n',round(x,1))
10.800000000000001
>>
It's doing the best it can within the constraints of IEEE floating point representation.

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Accepted Answer

James Tursa
James Tursa on 27 Aug 2019
Edited: James Tursa on 27 Aug 2019
Any function (dlmread, fscanf, etc) in any language (MATLAB, C, etc.) reading text numbers into floating point will have this issue. You may not have noticed it before, but it has always been there. With some exceptions (integers, etc), in general you simply can't represent many decimal numbers exactly in IEEE binary floating point format. For your numbers, here are the exact conversions:
>> sprintf('%.50f',1)
ans =
'1.00000000000000000000000000000000000000000000000000'
>> sprintf('%.50f',10.8)
ans =
'10.80000000000000071054273576010018587112426757812500'
>> sprintf('%.50f',31)
ans =
'31.00000000000000000000000000000000000000000000000000'
>> sprintf('%.50f',10.5001)
ans =
'10.50009999999999976694198267068713903427124023437500'
Integers (that are not too big) it does fine with, but your other decimal numbers cannot be represented internally exactly. Read the links that dpb has provided.

  1 Comment

Arnaud WILHELM
Arnaud WILHELM on 28 Aug 2019
All right, I understand now.
Thanks everyone for these anwsers !

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