MATLAB: Extracting a specific digit from a float w/o floating-point implementation rounding issues

binaryfloating pointMATLABrounding

I have a set of base-10 nonnegative floats with either three or four digits to the right of the decimal point. Can anyone suggest a reliable way to extract the very rightmost digit (that is, the 1E-4 place) without running into occasional floating-point issues?
My first attempt, lastDigit = round(10*rem(num*1000,1)), works most of the time:
>> num = 130404809.288; lastDigit = round(10*rem(num*1000,1)) % fourth digit right of the decimal point is zero
lastDigit =
     0
>> num = 130404809.2882; lastDigit = round(10*rem(num*1000,1))
lastDigit =
     2
But every once in a while, the finite nature of internal binary representation produces undesired results:
>> num = 136147309.434; lastDigit = round(10*rem(num*1000,1)) % should return zero
lastDigit =
    10
I realize this is a common subtlety, and that my challenge is really not so much math as string parsing—but this is the data I have to work with.
Naively, I tried various combinations of floor, round, rem, etc. — but I haven't been able to find a clean way to extract that fourth digit that doesn't run into cases like the one above every so often. Can anyone set me straight?

Best Answer

Some people will incorrectly tell you that this is not possible, but in reality there is a simple and efficient solution:
>> N = 130404809.288;
>> mod(round(10000*N),10)
ans =  0
>> N = 130404809.2882;
>> mod(round(10000*N),10)
ans =  2
>> N = 136147309.434;
>> mod(round(10000*N),10)
ans =  0
This relies on your statement that the values have "with either three or four digits to the right of the decimal point", i.e. that the fifth digit is zero (the sixth and any further digits, including any floating point error, are totally irrelevant). This means that after multiplying by 10000 (NOT by 1000 as you tried) a simple round will remove all floating point error from the number:
xyz.abcd0err % input
xyzabcd.0err % *10000
xyzabcd      % round
      d      % mod 10
You were almost there, just off by a factor of ten.
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