Thanks, that helps. Jorge Godoy wrote: Johan Geldenhuys <[EMAIL PROTECTED]> writes:Now that I have the answer (16.800000000000001), How do I round it of to 16.80 ? I only need it to be to the closest 1/100.print "%.2f" % 16.80000000000000116.80a = 16.8000000000001 b = "%.2f" % a b'16.80'float(b)16.800000000000001str(float(b))'16.8'(The reason for the '000000000001' is because of the numeric base used to represent data -- binary -- and the numeric base I'm requesting the answer to be in -- decimal.) Be seeing you, |
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