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I have two numpy array (2 dimensional) e.g.
a1 = array([["a","b"],["a","c"],["b","b"],["a","b"]])
a2 = array([["a","b"],["b","b"],["c","a"],["a","c"]])
What is the most elegant way of getting a matrix like this:
array([[1,0,0,0],
[0,0,0,1],
[0,1,0,0],
[1,0,0,0]])
Where element (i,j) is 1 if all(a1[i,:] == a2[j,:]) and otherwise 0
(everything involving two for loops I don't consider elegant)
516
The key to making it fast is to use vectorized operations, generally implemented through NumPy's universal functions (ufuncs). This section motivates the need for NumPy's ufuncs, which can be used to make repeated calculations on array elements much more efficient.
NumPy arrays are faster and more compact than Python lists. An array consumes less memory and is convenient to use. NumPy uses much less memory to store data and it provides a mechanism of specifying the data types.
NumPy Arrays Are NOT Always Faster Than Lists " append() " adds values to the end of both lists and NumPy arrays. It is a common and very often used function. The script below demonstrates a comparison between the lists' append() and NumPy's append() .
Method 1: We generally use the == operator to compare two NumPy arrays to generate a new array object. Call ndarray. all() with the new array object as ndarray to return True if the two NumPy arrays are equivalent.
>>> (a1[:,numpy.newaxis] == a2).all(axis=2)
array([[ True, False, False, False],
[False, False, False, True],
[False, True, False, False],
[ True, False, False, False]], dtype=bool)
If you really need integers, convert to int as last step:
>>> (a1[:,numpy.newaxis] == a2).all(axis=2).astype(int)
array([[1, 0, 0, 0],
[0, 0, 0, 1],
[0, 1, 0, 0],
[1, 0, 0, 0]])
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