Is there an efficient Numpy mechanism to retrieve the integer indexes of locations in an array based on a condition is true as opposed to the Boolean mask array?
For example:
x=np.array([range(100,1,-1)])
#generate a mask to find all values that are a power of 2
mask=x&(x-1)==0
#This will tell me those values
print x[mask]
In this case, I'd like to know the indexes i of mask where mask[i]==True. Is it possible to generate these without looping?
Another option:
In [13]: numpy.where(mask)
Out[13]: (array([36, 68, 84, 92, 96, 98]),)
which is the same thing as numpy.where(mask==True).
numpy.flatnonzero(mask)You should be able to use numpy.nonzero() to find this information.
numpy.nonzero does not give a tuple with an array of values as the second component. The tuple is used in case the Boolean array is multidimensional. From the docs: “Returns a tuple of arrays, one for each dimension […]”.np.arange(100,1,-1)
array([100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89, 88,
87, 86, 85, 84, 83, 82, 81, 80, 79, 78, 77, 76, 75,
74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62,
61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49,
48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36,
35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23,
22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10,
9, 8, 7, 6, 5, 4, 3, 2])
x=np.arange(100,1,-1)
np.where(x&(x-1) == 0)
(array([36, 68, 84, 92, 96, 98]),)
Now rephrase this like :
x[x&(x-1) == 0]
If you prefer the indexer way, you can convert your boolean list to numpy array:
print x[nd.array(mask)]
