python blogging software

Discussion in 'Python' started by Harald Massa, Jul 6, 2003.

  1. Harald Massa

    Harald Massa Guest

    which one should I use? Recommendations?

    Harald Massa, Jul 6, 2003
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  2. Harald Massa

    Van Gale Guest

    Van Gale, Jul 7, 2003
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  3. Harald Massa

    Carl Banks Guest

    It's a nice little iterator. Three problems, though:

    1. It can accumulate rounding errors
    2. It can return an integer on the first iteration (minor)
    3. The last iteration might not happen because of floating point
    uncertainties, even if rounding errors are minimized

    A more robust version would look something like this (but it still has
    problem #3):

    def frange(start, stop, step=1.0):
    sign = cmp(0,step)
    for i in xrange(sys.maxint):
    v = start+i*step
    if cmp(v,stop) != sign:
    yield v

    The most robust way to handle this is to iterpolate, i.e., instead of
    passing start, stop, and step, pass start, stop, and n_intervals:

    def interiter(start, stop, n_intervals):
    diff = stop - start
    for i in xrange(n_intervals+1):
    yield start + (i*diff)/n_intervals

    An IEEE 754 (whatever) geek can probably point out even more accurate
    ways to do these.
    Carl Banks, Jul 28, 2003
  4. Harald Massa

    Carl Banks Guest

    Just to be sure: what I have should have said was there might be more
    extra iteration. The Python convention, of course, it not to include
    the final point. However, with floating point, it could happen to
    slightly undershoot the stop value, which would cause an evident extra
    Carl Banks, Jul 28, 2003
  5. Harald Massa

    Neal Holtz Guest

    You can easily invent examples where the number of elements generated
    may (or may not) suprise you. Here is an implementation that attempts
    to account for some of the problems of floating point arithmetic:

    # (See

    def epsilon():
    """Compute machine epsilon: smallest number, e, s.t. 1 + e > 1"""
    one = 1.0
    eps = 1.0
    while (one + eps) > one:
    eps = eps / 2.0
    return eps*2.0

    _Epsilon = epsilon()

    def frange2( start, stop, step=1.0, fuzz=_Epsilon ):
    sign = cmp( step, 0.0 )
    _fuzz = sign*fuzz*abs(stop)
    for i in xrange(sys.maxint):
    value = start + i*step
    if cmp( stop, value + _fuzz ) != sign:
    yield value

    And we can generate a bunch of test cases to compare this
    implementation with the first one above:

    for step in frange( 0.0, 1.0, 0.01 ):
    for stop in frange( 0.01, 10.0, 0.01 ):
    l1 = [x for x in frange( 0.0, stop, step )]
    l2 = [x for x in frange2( 0.0, stop, step )]
    if l1 != l2:
    print stop, step, len(l1), len(l2)
    print repr(stop), repr(step)
    print l1
    print l2

    The first of many the differences noted is this:

    0.06 0.01 7 6
    0.060000000000000005 0.01
    [0.0, 0.01, 0.02, 0.029999999999999999, 0.040000000000000001,
    0.050000000000000003, 0.059999999999999998]
    [0.0, 0.01, 0.02, 0.029999999999999999, 0.040000000000000001,

    Under most normal output conditions, you would think that the stop
    value was 0.06 and the step was 0.01, so you would expect to get 6
    elements in the sequence. Yet the first implementation gives you 7.
    This might be surprising, as under most output conditions the last
    value would appear to be equal to the stop value. Of course, the
    actual stop value is just slightly greater than 0.06. The second
    implementation gives you 6 elements in the sequence; though it can be
    argued which is "correct", the second implementation is probably less
    surprising in most cases.

    This is just another example, of course, of the idea that if it
    really, really matters how many elements are in the sequence, don't
    use floating point arithmetic to control the # of iterations.
    Neal Holtz, Jul 28, 2003
  6. To guarantee the exact end points, maybe:

    def interiter(start, stop, n_intervals):
    for i in xrange(n_intervals+1):
    yield ((n_intervals-i)/fn)*start + (i/fn)*stop

    but shouldn't that be xrange(n_intervals) and leave out
    the final point (thus exact but invisible ;-).

    Bengt Richter
    Bengt Richter, Jul 28, 2003
  7. Harald Massa

    Carl Banks Guest

    Well, I thought about that.

    In the interests of practicality beats purity, when you're
    interpolating (instead of stepping as in range) you should include the
    final endpoint. In my experience, when when you divide a segment into
    intervals, you almost always want to include both endpoints.

    If you use the Python convention of not including the final point in
    the iteration, and someone decides they do want the final endpoint,
    then it's kind of a pain to calculate what point you should use to get
    the final "endpoint", and you lose accuracy on top of it.

    So, my suggestion when iterating over an interval is this: If you pass
    in a stepsize, then don't include the final point. If you pass in a
    number of intervals, then do include the final point. I think the
    practicality of it justifies the different conventions.
    Carl Banks, Jul 28, 2003
  8. Harald Massa

    Paul Rubin Guest

    I think you're right about this, but it goes against the Python notion
    of a "range" function, so an frange that includes both endpoints
    should be called something different, maybe "fstep" or something like
    that. For what I was doing, the frange that I posted did the job.

    I see now that that the problem is subtle enough that IMO, a fully
    worked out solution should be written and included in the Python
    library or at least the Python cookbook.
    Paul Rubin, Jul 28, 2003
  9. How about if we indicate the kind of interval, e.g.,
    ... fn=float(n_intervals)
    ... for i in xrange(ends[0]=='(', n_intervals+(ends[1]==']')):
    ... yield ((n_intervals-i)/fn)*start + (i/fn)*stop
    ... ...
    0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
    1.0 2.0 3.0 4.0 5.0 6.0 7.0
    1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 ...
    1.0 2.0 3.0 4.0 5.0 6.0 7.0
    0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0

    Bengt Richter
    Bengt Richter, Jul 28, 2003
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