Python 计算矩阵中大于某个值的所有值
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Count all values in a matrix greater than a value
提问by gran_profaci
I have to count all the values in a matrix (2-d array) that are greater than 200.
我必须计算矩阵（二维数组）中大于 200 的所有值。
The code I wrote down for this is:
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
o31is an image and I am converting it into a matrix and then finding the values.
o31是一个图像，我将它转换成一个矩阵，然后找到值。
My question is, is there a simpler way to do this?
我的问题是，有没有更简单的方法来做到这一点？
采纳答案by abought
The numpy.wherefunction 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.
该numpy.where功能是你的朋友。因为它的实现是为了充分利用数组数据类型，所以对于大图像，您应该注意到与您提供的纯 python 解决方案相比速度有所提高。
Using numpy.where directly will yield a boolean mask indicating whether certain values match your conditions:
直接使用 numpy.where 将产生一个布尔掩码，指示某些值是否符合您的条件：
>>> 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:
并且掩码可用于直接索引数组以获取实际值：
>>> 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.
您从那里获取它的确切位置将取决于您希望结果的形式。
回答by K Z
There are many ways to achieve this, like flatten-and-filter or simply enumerate, but I think using Boolean/mask arrayis the easiest one (and iirc a much faster one):
有很多方法可以实现这一点，比如展平和过滤或简单地枚举，但我认为使用布尔/掩码数组是最简单的一种（而 iirc 更快）：
>>> 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:
更新：
As nneonneo has answered, if all you want is the number of elements that passes threshold, you can simply do:
正如 nneonneo 所回答的那样，如果您想要的只是通过阈值的元素数量，您可以简单地执行以下操作：
>>>> (y>200).sum()
3
which is a simpler solution.
这是一个更简单的解决方案。
Speed comparison with filter:
速度比较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
回答by nneonneo
This is very straightforward with boolean arrays:
这对于布尔数组非常简单：
p31 = numpy.asarray(o31)
za = (p31 < 200).sum() # p31<200 is a boolean array, so `sum` counts the number of True elements
回答by Karol
Here's a variant that uses fancy indexing and has the actual values as an intermediate:
这是一个使用花式索引并将实际值作为中间值的变体：
p31 = numpy.asarray(o31)
values = p31[p31<200]
za = len(values)
回答by Ricoter
To count the number of values larger than x in any numpy array you can use:
要计算任何 numpy 数组中大于 x 的值的数量，您可以使用：
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.
布尔索引返回一个数组，该数组仅包含满足条件 (矩阵 > x) 的元素。然后 len() 计算这些值。
