Python numpy masked array initialization

Question

I used masked arrays all the time in my work, but one problem I have is that the initialization of masked arrays is a bit clunky. Specifically, the ma.zeros() and ma.empty() return masked arrays with a mask that doesn't match the array dimension. The reason I want this is so that if I don't assign to a particular element of my array, it is masked by default.

In [4]: A=ma.zeros((3,))
...
masked_array(data = [ 0.  0.  0.],
             mask = False,
       fill_value = 1e+20)

I can subsequently assign the mask:

In [6]: A.mask=ones((3,))
...
masked_array(data = [-- -- --],
             mask = [ True  True  True],
       fill_value = 1e+20)

But why should I have to use two lines to initialize and array? Alternatively, I can ignore the ma.zeros() functionality and specify the mask and data in one line:

In [8]: A=ma.masked_array(zeros((3,)),mask=ones((3,)))
...
masked_array(data = [-- -- --],
             mask = [ True  True  True],
       fill_value = 1e+20)

But I think this is also clunky. I have trawled through the numpy.ma documentation but I can't find a neat way of dealing with this. Have I missed something obvious?

Pierre GM · Accepted Answer · 2012-11-13 01:51:07Z

5

Well, the mask in ma.zeros is actually a special constant, ma.nomask, that corresponds to np.bool_(False). It's just a placeholder telling NumPy that the mask hasn't been set. Using nomask actually speeds up np.ma significantly: no need to keep track of where the masked values are if we know beforehand that there are none.

The best approach is not to set your mask explicitly if you don't need it and leave np.ma set it when needed (ie, when you end up trying to take the log of a negative number).

Side note #1: to set the mask to an array of False with the same shape as your input, use

np.ma.array(..., mask=False)

That's easier to type. Note that it's really the Python False, not np.ma.nomask... Similarly, use mask=True to force all your inputs to be masked (ie, mask will be a bool ndarray full of True, with the same shape as the data).

Side note #2: If you need to set the mask after initialization, you shouldn't use an assignment to .mask but assign to the special value np.ma.masked, it's safer:

a[:] = np.ma.masked

edited Nov 13, 2012 at 1:51

answered Nov 13, 2012 at 1:33

Pierre GM

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1 Comment

Thom Chubb Over a year ago

Hi Pierre, thanks for your reply. In my applications I always want a mask so that if I don't end up assigning to an index it remains masked by default (I have edited my question to reflect this better). Your "side note" seems to give me exactly what I have been looking for!

pkeller · Accepted Answer · 2016-10-27 21:54:14Z

0

Unfortunately your Side note#2 recommendation breaks for an array with more than one dimension:

a = ma.zeros( (2,2) )
a[0][0] = ma.masked
a
masked_array(data =
 [[ 0.  0.]
 [ 0.  0.]],
         mask =
 False,
   fill_value = 1e+20)

Like the OP, I haven't found a neat way around this. Masking a whole row will initialise the mask properly:

 a[0] = ma.masked
 a
 masked_array(data =
 [[-- --]
 [0.0 0.0]],
             mask =
 [[ True  True]
 [False False]],
       fill_value = 1e+20)

but if this isn't what you want to do you then have to do a[0] = ma.nomask to undo it. Doing a[0] = ma.nomask immediately after a = ma.zeros( (2,2) ) has no effect.

edited Oct 27, 2016 at 21:54

user6613600

answered Jun 9, 2015 at 15:18

pkeller

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1 Comment

Thom Chubb Over a year ago

I'm not exactly sure what's going on here, but the answer may lie in the difference between "basic" and "advanced" indexing: docs.scipy.org/doc/numpy/reference/arrays.indexing.html 1) If you assign a[0,0]=ma.masked, you'll find that it works as you expected. 2) If you assign a value in the way that you have done (e.g. a[0][0]=1.), it behaves as expected. I'm not sure why the behaviour would be different between assigning a special value and a float though.

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