Numpy masking 3D array using np.where() index

Question

I created an index based on several conditions

transition = np.where((rain>0) & (snow>0) & (graup>0) & (xlat<53.) & (xlat>49.) & (xlon<-114.) & (xlon>-127.)) #indexes the grids where there are transitions

with the shape of (3,259711) that looks like the following:

array([[  0,   0,   0, ...,  47,  47,  47], #hour
       [847, 847, 848, ..., 950, 950, 951], #lat gridpoint
       [231, 237, 231, ..., 200, 201, 198]]) #lon gridpoint

I have several other variables (e.g. temp) with the shape of (48, 1015, 1359) corresponding to hour, lat, lon.

Seeing as the index are my valid gridpoints, how do I mask all the variables, like temp so that it retains the (48,1015,1359) shape, but masks the values outside the index.

What exactly do you mean by mask? For what kind of operation? For somethings a masked array (np.ma) subclass might be useful. — hpaulj
– hpaulj, Commented Feb 1, 2017 at 17:47
Thanks for the suggestion. I have already tried using numpy.ma.masked_where(idx, temp), but I have issues with my shape IndexError: Inconsistant shape between the condition and the input (got (3, 259711) and (48, 1015, 1359)) @hpaulj — jdiction
– jdiction, Commented Feb 1, 2017 at 18:08
masked_where wants the boolean mask of values you want hidden, not the index tuple produced by where. What's the condition matrix that produced this where tuple? — hpaulj
– hpaulj, Commented Feb 1, 2017 at 19:10
@hpaulj Hmm, ah okay, is there a way to use the index (using np.where), so that those indices are not masked, while the rest are masked? — jdiction
– jdiction, Commented Feb 1, 2017 at 20:49

hpaulj · Accepted Answer · 2017-02-01 19:59:07Z

In [90]: arr = np.arange(24).reshape(6,4)
In [91]: keep = (arr % 3)==1
In [92]: keep
Out[92]: 
array([[False,  True, False, False],
       [ True, False, False,  True],
       [False, False,  True, False],
       [False,  True, False, False],
       [ True, False, False,  True],
       [False, False,  True, False]], dtype=bool)
In [93]: np.where(keep)
Out[93]: 
(array([0, 1, 1, 2, 3, 4, 4, 5], dtype=int32),
 array([1, 0, 3, 2, 1, 0, 3, 2], dtype=int32))

Simple application of the keep mask gives a 1d array of the desired values. I could also index with the where tuple.

In [94]: arr[keep]
Out[94]: array([ 1,  4,  7, 10, 13, 16, 19, 22])

With keep, or rather it's boolean inverse, I can make a masked array:

In [95]: np.ma.masked_array(arr,mask=~keep)
Out[95]: 
masked_array(data =
 [[-- 1 -- --]
 [4 -- -- 7]
 [-- -- 10 --]
 [-- 13 -- --]
 [16 -- -- 19]
 [-- -- 22 --]],
             mask =
 [[ True False  True  True]
 [False  True  True False]
 [ True  True False  True]
 [ True False  True  True]
 [False  True  True False]
 [ True  True False  True]],
       fill_value = 999999)

np.ma.masked_where(~keep, arr) does the same thing - just a different argument order. It still expects the boolean mask array.

I can do the same starting with the where tuple:

In [105]: idx = np.where(keep)
In [106]: mask = np.ones_like(arr, dtype=bool)
In [107]: mask[idx] = False
In [108]: np.ma.masked_array(arr, mask=mask)

There may be something in the np.ma class that does this with one call, but it will have to do the same sort of construction.

This also works:

x = np.ma.masked_all_like(arr)
x[idx] = arr[idx]

Collectives™ on Stack Overflow

Numpy masking 3D array using np.where() index

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related