Counting the number of times the string occurs in the matrix (numpy)

Question

Is there a better way to count how many times a given string appears in the two-dimensional array numpy than

def get_count(array_2d, row): count = 0 # iterate over rows, compare for r in array_2d[:,]: if np.equal(r, row).all(): count += 1 return count # let make sure it works array_2d = np.array([[1,2], [3,4]]) row = np.array([1,2]) count = get_count(array_2d, row) assert(count == 1) 

Answer 1

One simple way: broadcasting -

 (array_2d == row).all(-1).sum() 

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Considering memory efficiency, there is seen one approach, examining each line of array_2d as the indexing set in a grid n-dimensional , and assuming the inputs are positive numbers -

 dims = np.maximum(array_2d.max(0),row) + 1 array_1d = np.ravel_multi_index(array_2d.T,dims) row_scalar = np.ravel_multi_index(row,dims) count = (array_1d==row_scalar).sum() 

Here, a message dedicated to the various aspects associated with it.

Note. Using np.count_nonzero may be much faster to calculate Boolean summation elements instead via .sum() . So, think about using it for both of the above aprobashey.

Here's a quick test execution time -

 In [74]: arr = np.random.rand(10000)>0.5 In [75]: %timeit arr.sum() 10000 loops, best of 3: 29.6 µs per loop In [76]: %timeit np.count_nonzero(arr) 1000000 loops, best of 3: 1.21 µs per loop