floating point rounding

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  • hg
    #1

    floating point rounding

    Hi,

    Here is my issue:

    f = 1.5 * 0.01
    f
    >0.014999999999 999999
    '%f' % f
    >>'0.015000'

    But I really want to get 0.02 as a result ... is there a way out ?

    Thanks,

    hg

    Tags: None
    • hg
      #2
      Re: floating point rounding

      hg wrote:
      Hi,
      >
      Here is my issue:
      >
      f = 1.5 * 0.01
      f
      >>0.01499999999 9999999
      '%f' % f
      >>>'0.015000'
      >
      >
      But I really want to get 0.02 as a result ... is there a way out ?
      >
      Thanks,
      >
      hg
      round

        Comment

        • John Henry
          #3
          Re: floating point rounding

          On Mar 9, 5:45 am, hg <h...@nospam.or gwrote:
          hg wrote:
          Hi,
          >
          Here is my issue:
          >
          f = 1.5 * 0.01
          f
          >0.014999999999 999999
          '%f' % f
          >>'0.015000'
          >
          But I really want to get 0.02 as a result ... is there a way out ?
          >
          Thanks,
          >
          hg
          >
          round

          Or more precisely:

          round(0.0149999 99999999999,2)

          if that's what you wish to do.

            Comment

            • hg
              #4
              Re: floating point rounding

              John Henry wrote:
              On Mar 9, 5:45 am, hg <h...@nospam.or gwrote:
              >hg wrote:
              Hi,
              >>
              Here is my issue:
              >>
              f = 1.5 * 0.01
              f
              >>0.01499999999 9999999
              '%f' % f
              >>>'0.015000'
              >>
              But I really want to get 0.02 as a result ... is there a way out ?
              >>
              Thanks,
              >>
              hg
              >>
              >round
              >
              >
              Or more precisely:
              >
              round(0.0149999 99999999999,2)
              >
              if that's what you wish to do.
              Indeed.

              hg

                Comment

                • greg
                  #5
                  Re: floating point rounding

                  John Henry wrote:
                  Or more precisely:
                  >
                  round(0.0149999 99999999999,2)
                  No, that *won't* solve the problem. Using a slightly
                  different example,
                  >>x = 1.5 * 0.1
                  >>x
                  0.1500000000000 0002
                  >>round(x, 2)
                  0.1499999999999 9999

                  The problem is that floats are stored internally in
                  binary, not decimal, and numbers like 0.1 and 0.01
                  have no exact representation as a binary float.
                  Using round() doesn't help, because the result is
                  still a binary float, and the result you're after
                  still can't be represented.

                  The best you can do is to *display* it rounded
                  to the number of digits you want using formatting,
                  e.g.
                  >>"%.2f" % x
                  '0.15'

                  Alternatively, use the Decimal module, which stores
                  numbers as decimal and does arithmetic in ways that
                  will match your calculator. It's slower, though.

                  --
                  Greg

                    Comment

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