Count all values in a matrix greater than a value

Question

I have to count all the values in a matrix (2-d array) that are greater than 200.

The code I wrote down for this is:

za=0   
p31 = numpy.asarray(o31)         


        

Answer 1

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