Menu
For a given NumPy array, it is easy to perform a "normal" sum along one dimension. For example:
X = np.array([[1, 0, 0], [0, 2, 2], [0, 0, 3]])
X.sum(0)
=array([1, 2, 5])
X.sum(1)
=array([1, 4, 3])
Instead, is there an "efficient" way of computing the bitwise OR along one dimension of an array similarly? Something like the following, except without requiring for-loops or nested function calls.
Example: bitwise OR along zeroeth dimension as I currently am doing it:
np.bitwise_or(np.bitwise_or(X[:,0],X[:,1]),X[:,2])
=array([1, 2, 3])
What I would like:
X.bitwise_sum(0)
=array([1, 2, 3])
492
Axis 1 is the direction along the columns In a multi-dimensional NumPy array, axis 1 is the second axis. When we're talking about 2-d and multi-dimensional arrays, axis 1 is the axis that runs horizontally across the columns.
The [:, :] stands for everything from the beginning to the end just like for lists. The difference is that the first : stands for first and the second : for the second dimension. a = numpy. zeros((3, 3)) In [132]: a Out[132]: array([[ 0., 0., 0.], [ 0., 0., 0.], [ 0., 0., 0.]])
Array indexing is the same as accessing an array element. You can access an array element by referring to its index number. The indexes in NumPy arrays start with 0, meaning that the first element has index 0, and the second has index 1 etc.
c_ = <numpy.lib.index_tricks.CClass object> Translates slice objects to concatenation along the second axis. This is short-hand for np. r_['-1,2,0', index expression] , which is useful because of its common occurrence.
numpy.bitwise_or.reduce(X, axis=whichever_one_you_wanted)
Use the reduce method of the numpy.bitwise_or ufunc.
97
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
