计算矩阵中小于一个值的所有值

我必须计算矩阵(2-d 数组)中小于200的所有值。

我为此写的代码是:

za=0
p31 = numpy.asarray(o31)
for i in range(o31.size[0]):
for j in range(o32.size[1]):
if p31[i,j]<200:
za=za+1
print za

o31是一个图像，我把它转换成一个矩阵，然后找到值。

还有更简单的方法吗？

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最佳答案

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.

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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
小开

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
小开

Here's a variant that uses fancy indexing and has the actual values as an intermediate:

p31 = numpy.asarray(o31)
values = p31[p31<200]
za = len(values)
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To count the number of values larger than x in any numpy array you can use:

n = len(matrix[matrix > x])

The boolean indexing returns an array that contains only the elements where the condition (matrix > x) is met. Then len() counts these values.

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You can use numpy.count_nonzero, converting the whole into a one-liner:

za = numpy.count_nonzero(numpy.asarray(o31)<200)  #as written in the code

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