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