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:

>>> data
array([[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
>>> b
array([[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