Python 如果满足条件，则替换 Numpy 元素
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Replacing Numpy elements if condition is met
提问by ChrisFro
I have a large numpy array that I need to manipulate so that each element is changed to either a 1 or 0 if a condition is met (will be used as a pixel mask later). There are about 8 million elements in the array and my current method takes too long for the reduction pipeline:
我有一个大型 numpy 数组，我需要对其进行操作，以便在满足条件时将每个元素更改为 1 或 0（稍后将用作像素掩码）。数组中大约有 800 万个元素，我当前的方法对于减少管道来说花费的时间太长：
for (y,x), value in numpy.ndenumerate(mask_data):
if mask_data[y,x]<3: #Good Pixel
mask_data[y,x]=1
elif mask_data[y,x]>3: #Bad Pixel
mask_data[y,x]=0
Is there a numpy function that would speed this up?
是否有一个 numpy 函数可以加快速度？
采纳答案by Steve Barnes
>>> import numpy as np
>>> a = np.random.randint(0, 5, size=(5, 4))
>>> a
array([[4, 2, 1, 1],
[3, 0, 1, 2],
[2, 0, 1, 1],
[4, 0, 2, 3],
[0, 0, 0, 2]])
>>> b = a < 3
>>> b
array([[False, True, True, True],
[False, True, True, True],
[ True, True, True, True],
[False, True, True, False],
[ True, True, True, True]], dtype=bool)
>>>
>>> c = b.astype(int)
>>> c
array([[0, 1, 1, 1],
[0, 1, 1, 1],
[1, 1, 1, 1],
[0, 1, 1, 0],
[1, 1, 1, 1]])
You can shorten this with:
您可以通过以下方式缩短它：
>>> c = (a < 3).astype(int)
回答by ev-br
>>> a = np.random.randint(0, 5, size=(5, 4))
>>> a
array([[0, 3, 3, 2],
[4, 1, 1, 2],
[3, 4, 2, 4],
[2, 4, 3, 0],
[1, 2, 3, 4]])
>>>
>>> a[a > 3] = -101
>>> a
array([[ 0, 3, 3, 2],
[-101, 1, 1, 2],
[ 3, -101, 2, -101],
[ 2, -101, 3, 0],
[ 1, 2, 3, -101]])
>>>
See, eg, Indexing with boolean arrays.
参见，例如，使用布尔数组索引。
回答by YXD
You can create your mask array in one step like this
您可以像这样一步创建您的掩码阵列
mask_data = input_mask_data < 3
This creates a boolean array which can then be used as a pixel mask. Note that we haven't changed the input array (as in your code) but have created a new array to hold the mask data - I would recommend doing it this way.
这将创建一个布尔数组，然后可以将其用作像素掩码。请注意，我们没有更改输入数组（如您的代码中所示），而是创建了一个新数组来保存掩码数据 - 我建议这样做。
>>> input_mask_data = np.random.randint(0, 5, (3, 4))
>>> input_mask_data
array([[1, 3, 4, 0],
[4, 1, 2, 2],
[1, 2, 3, 0]])
>>> mask_data = input_mask_data < 3
>>> mask_data
array([[ True, False, False, True],
[False, True, True, True],
[ True, True, False, True]], dtype=bool)
>>>
回答by mamalos
I am not sure I understood your question, but if you write:
我不确定我是否理解你的问题，但如果你写：
mask_data[:3, :3] = 1
mask_data[3:, 3:] = 0
This will make all values of mask data whose x and y indexes are less than 3 to be equal to 1 and all rest to be equal to 0
这将使 x 和 y 索引小于 3 的掩码数据的所有值都等于 1，其余的都等于 0
回答by Markus Dutschke
The quickest (and most flexible) wayis to use np.where, which chooses between two arrays according to a mask(array of true and false values):
的最快（和最灵活的）的方式是使用np.where，它根据一个掩模（真假值的阵列）两个阵列之间选：
import numpy as np
a = np.random.randint(0, 5, size=(5, 4))
b = np.where(a<3,0,1)
print('a:',a)
print()
print('b:',b)
which will produce:
这将产生：
a: [[1 4 0 1]
[1 3 2 4]
[1 0 2 1]
[3 1 0 0]
[1 4 0 1]]
b: [[0 1 0 0]
[0 1 0 1]
[0 0 0 0]
[1 0 0 0]
[0 1 0 0]]
