Count all values in a matrix greater than a value

This is very straightforward with boolean arrays:

p31 = numpy.asarray(o31)za = (p31 < 200).sum() # p31<200 is a boolean array, so `sum` counts the number of True elements

The numpy.where function is your friend. Because it's implemented to take full advantage of the array datatype, for large images you should notice a speed improvement over the pure python solution you provide.

Using numpy.where directly will yield a boolean mask indicating whether certain values match your conditions:

>>> dataarray([[1, 8],       [3, 4]])>>> numpy.where( data > 3 )(array([0, 1]), array([1, 1]))

And the mask can be used to index the array directly to get the actual values:

>>> data[ numpy.where( data > 3 ) ]array([8, 4])

Exactly where you take it from there will depend on what form you'd like the results in.

There are many ways to achieve this, like flatten-and-filter or simply enumerate, but I think using Boolean/mask array is the easiest one (and iirc a much faster one):

>>> y = np.array([[123,24123,32432], [234,24,23]])array([[  123, 24123, 32432],       [  234,    24,    23]])>>> b = y > 200>>> barray([[False,  True,  True],       [ True, False, False]], dtype=bool)>>> y[b]array([24123, 32432,   234])>>> len(y[b])3>>>> y[b].sum()56789

Update:

As nneonneo has answered, if all you want is the number of elements that passes threshold, you can simply do:

>>>> (y>200).sum()3

which is a simpler solution.

Speed comparison with filter:

### use boolean/mask array ###b = y > 200%timeit y[b]100000 loops, best of 3: 3.31 us per loop%timeit y[y>200]100000 loops, best of 3: 7.57 us per loop### use filter ###x = y.ravel()%timeit filter(lambda x:x>200, x)100000 loops, best of 3: 9.33 us per loop%timeit np.array(filter(lambda x:x>200, x))10000 loops, best of 3: 21.7 us per loop%timeit filter(lambda x:x>200, y.ravel())100000 loops, best of 3: 11.2 us per loop%timeit np.array(filter(lambda x:x>200, y.ravel()))10000 loops, best of 3: 22.9 us per loop*** use numpy.where ***nb = np.where(y>200)%timeit y[nb]100000 loops, best of 3: 2.42 us per loop%timeit y[np.where(y>200)]100000 loops, best of 3: 10.3 us per loop