Numpy masked array initialization from another array - numpy

Question

Is there a cleaner way to initialize a numpy masked array from a non-ma, with all masked values False, than this?
masked_array = np.ma.masked_array(array, mask=np.zeros_like(array, dtype='bool'))
The duplicate reference to array seems unnecessary and clunky. If you do not give the mask= parameter, the mask defaults to a scalar boolean, which prevents sliced access to the mask.

Answer 1

You should be able to just set the mask to False:
>>> array = np.array([1,2,3])
>>> masked_array = np.ma.masked_array(array, mask=False)
>>> masked_array
masked_array(data = [1 2 3],
mask = [False False False],
fill_value = 999999)
I saw hpaulj’s comment and played around with different ways of solving this issue and comparing performance. I can’t explain the difference, but #hpaulj seems to have a much deeper understanding of how numpy works. Any input on why m3() executes so much faster would be most appreciated.
def origM():
array = np.array([1,2,3])
return np.ma.masked_array(array, mask=np.zeros_like(array, dtype='bool'))
def m():
array = np.array([1,2,3])
return np.ma.masked_array(array, mask=False)
def m2():
array = np.array([1,2,3])
m = np.ma.masked_array(array)
m.mask = False
return m
def m3():
array = np.array([1,2,3])
m = array.view(np.ma.masked_array)
m.mask = False
return m
>>> origM()
masked_array(data = [1 2 3],
mask = [False False False],
fill_value = 999999)
All four return the same result:
>>> m()
masked_array(data = [1 2 3],
mask = [False False False],
fill_value = 999999)
>>> m2()
masked_array(data = [1 2 3],
mask = [False False False],
fill_value = 999999)
>>> m3()
masked_array(data = [1 2 3],
mask = [False False False],
fill_value = 999999)
m3() executes the fastest:
>>> timeit.timeit(origM, number=1000)
0.024451958015561104
>>> timeit.timeit(m, number=1000)
0.0393978749634698
>>> timeit.timeit(m2, number=1000)
0.024049583997111768
>>> timeit.timeit(m3, number=1000)
0.018082750029861927