[Numpy-discussion] Question on numpy.ma.masked_values
Tue Mar 20 04:53:49 CDT 2012
By default, the mask of a MaskedArray is set to the special value
`np.ma.nomask`. In other terms::
np.ma.array(...) <=> np.ma.array(..., mask=np.ma.nomask)
In practice, np.ma.nomask lets us quickly check whether a MaskedArray
has a masked value : if its .mask is np.ma.nomask, then no masked
value, otherwise it's a full boolean array and we can use any.
If you want to create a MaskedArray w/ a full boolean mask, just use::
In that case, the mask is automatically created as a boolean array
with the same shape as the data, with False everywhere. If you used
True, the mask would be full of True...
Now, just to be clear, you'd want
'np.ma.masked_values(...,shrink=False) to create a maked array w/ a
full boolean mask by default, right ?
On 3/15/12, Gökhan Sever <email@example.com> wrote:
> Submitted the ticket at http://projects.scipy.org/numpy/ticket/2082
> On Thu, Mar 15, 2012 at 1:24 PM, Gökhan Sever <firstname.lastname@example.org> wrote:
>> On Thu, Mar 15, 2012 at 1:12 PM, Pierre GM <email@example.com> wrote:
>>> Ciao Gökhan,
>>> AFAIR, shrink is used only to force a collapse of a mask full of False,
>>> not to force the creation of such a mask.
>>> Now, it should work as you expected, meaning that it needs to be fixed.
>>> Could you open a ticket? And put me in copy, just in case.
>>> Your trick is a tad dangerous, as it erases the previous mask. I'd prefer
>>> to create x w/ a full mask, then use masked_values w/ shrink=False...
>>> if you're sure there's x= no masked values, go for it.
>>> This condition checking should make it stronger:
>> I7 x = np.array([1, 1.1, 2, 1.1, 3])
>> I8 y = np.ma.masked_values(x, 1.5)
>> I9 if y.mask == False:
>> y.mask = np.zeros(len(x), dtype=np.bool)*True
>> I10 y.mask
>> O10 array([False, False, False, False, False], dtype=bool)
>> I11 y
>> masked_array(data = [1.0 1.1 2.0 1.1 3.0],
>> mask = [False False False False False],
>> fill_value = 1.5)
>> How do you create "x w/ a full mask"?
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