[Numpy-discussion] Properties of ma.masked singleton
Paul F. Dubois
paul at pfdubois.com
Fri Jan 6 23:19:08 CST 2006
I'll look into your complaints. Your proposed solution does not work:
if x is masked:
if x is None:
become indistinguishable. Likewise for assignment.
Your example if x[2:4] is masked is a red herring. Obviously this
statement is false no matter how you define 'masked'.
I redid ma in a bit of a hurry and didn't pay any attention to the
issues you raise as to len(masked), which I agree should be 1.
Alexander Belopolsky wrote:
> In the current numpy version ma.masked singleton has the following
> >>> len(ma.masked)
> which is a change from old MA, where
> >>> len(MA.masked)
> >>> ma.masked.shape
> >>> MA.masked.shape
> It changes shape when filled:
> >>> ma.masked.filled()
> and so on.
> The code contains numerous "if x is masked" statements to support all
> this logic.
> I would like to suggest a somewhat radical change for your review.
> There are two main uses of ma.masked:
> 1. To set mask:
> >>> x = ma.zeros(5)
> >>> x[2:4] = ma.masked
> >>> print x
> [0 0 -- -- 0]
> 2. To check if an element is masked:
> >>> x is ma.masked
> The second property allows a cute looking idiom "x[i] is masked", but
> only works for single element access:
> >>> x[2:4] is ma.masked
> It also leads to strange results:
> >>> x.shape
> My proposal is to eliminate the property #2 from ma and redefine masked
> as None. Single element getitem will return a rank zero MaskedArray. We
> can also add "is_masked" property, which will allow a convenient check
> in the form "x[i].is_masked".
> The main benefit of this proposal is that x[i] will be duck typing
> compatible with numpy scalars.
> -- sasha
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