我怎样才能向量化numpy的阵列的2×2子阵列的平均？ [英] How can I vectorize the averaging of 2x2 sub-arrays of numpy array?
问题描述
我有一个包含了我需要的平均2x2的子集一个非常非常大的2D numpy的数组。我正在寻找一种方式来向量化此操作。例如，给定的x
I have a very a very large 2D numpy array that contains 2x2 subsets that I need to take the average of. I am looking for a way to vectorize this operation. For example, given x:
# |- col 0 -| |- col 1 -| |- col 2 -|
x = np.array( [[ 0.0, 1.0, 2.0, 3.0, 4.0, 5.0], # row 0
[ 6.0, 7.0, 8.0, 9.0, 10.0, 11.0], # row 0
[12.0, 13.0, 14.0, 15.0, 16.0, 17.0], # row 1
[18.0, 19.0, 20.0, 21.0, 22.0, 23.0]]) # row 1
我需要用的2x3阵列均为2×2子阵的平均值，即到结束：
I need to end up with a 2x3 array which are the averages of each 2x2 sub array, i.e.:
result = np.array( [[ 3.5, 5.5, 7.5],
[15.5, 17.5, 19.5]])
所以元素[0,0]被作为是x计算[0：2,0：2]，而元件[0,1]是x的平均值[2：4,0：2]。是否numpy的已矢量/关于高效亚聚集做这样的方式？
so element [0,0] is calculated as the average of x[0:2,0:2], while element [0,1] would be the average of x[2:4, 0:2]. Does numpy have vectorized/efficient ways of doing aggregates on subsets like this?
推荐答案
如果我们形成重构矩阵 Y = x.reshape（2,2,3,2） ，那么第（i，j）的2×2子矩阵是由给Y [我，：，J：] 。例如：
If we form the reshaped matrix y = x.reshape(2,2,3,2), then the (i,j) 2x2 submatrix is given by y[i,:,j,:]. E.g.:
In [340]: x
Out[340]:
array([[ 0., 1., 2., 3., 4., 5.],
[ 6., 7., 8., 9., 10., 11.],
[ 12., 13., 14., 15., 16., 17.],
[ 18., 19., 20., 21., 22., 23.]])
In [341]: y = x.reshape(2,2,3,2)
In [342]: y[0,:,0,:]
Out[342]:
array([[ 0., 1.],
[ 6., 7.]])
In [343]: y[1,:,2,:]
Out[343]:
array([[ 16., 17.],
[ 22., 23.]])
要获得2×2子矩阵的平均值，使用的意思是法，用轴=（1,3）：
To get the mean of the 2x2 submatrices, use the mean method, with axis=(1,3):
In [344]: y.mean(axis=(1,3))
Out[344]:
array([[ 3.5, 5.5, 7.5],
[ 15.5, 17.5, 19.5]])
如果您使用numpy的一个旧版本，不使用一个元组为中轴线的支持，你可以这样做：
If you are using an older version of numpy that doesn't support using a tuple for the axis, you could do:
In [345]: y.mean(axis=1).mean(axis=-1)
Out[345]:
array([[ 3.5, 5.5, 7.5],
[ 15.5, 17.5, 19.5]])
查看评论由@dashesy给出的链接，对整形猫腻更多的背景。
See the link given by @dashesy in a comment for more background on the reshaping "trick".
要与形状（M，N），推广为一个二维数组，其中m和n是偶数，用
To generalize this to a 2-d array with shape (m, n), where m and n are even, use
y = x.reshape(x.shape[0]/2, 2, x.shape[1], 2)
是则可以PTED为2×2数组的数组间$ P $。在4-D阵列充当该选择的2×2块中的一个的索引的第一和第三索引时隙。为了得到左上的2x2块，使用 Y [0：0，：] ;到块的第二行和第三列中的块中，使用 Y [1，：，2，：] ;在一般情况下，存取权限块（J，K），使用ÿ[J，：，K：]
y can then be interpreted as an array of 2x2 arrays. The first and third index slots of the 4-d array act as the indices that select one of the 2x2 blocks. To get the upper left 2x2 block, use y[0, :, 0, :]; to the block in the second row and third column of blocks, use y[1, :, 2, :]; and in general, to acces block (j, k), use y[j, :, k, :].
要计算这些块的平均值的降低阵列，使用的意思是法，用轴=（1,3）（即平均值轴1和3）：
To compute the reduced array of averages of these blocks, use the mean method, with axis=(1, 3) (i.e. average over axes 1 and 3):
avg = y.mean(axis=(1, 3))
下面是一个例子，其中 X 具有形状（8,10），因此2×2块的平均值的阵列具有形状（4,5）：
Here's an example where x has shape (8, 10), so the array of averages of the 2x2 blocks has shape (4, 5):
In [10]: np.random.seed(123)
In [11]: x = np.random.randint(0, 4, size=(8, 10))
In [12]: x
Out[12]:
array([[2, 1, 2, 2, 0, 2, 2, 1, 3, 2],
[3, 1, 2, 1, 0, 1, 2, 3, 1, 0],
[2, 0, 3, 1, 3, 2, 1, 0, 0, 0],
[0, 1, 3, 3, 2, 0, 3, 2, 0, 3],
[0, 1, 0, 3, 1, 3, 0, 0, 0, 2],
[1, 1, 2, 2, 3, 2, 1, 0, 0, 3],
[2, 1, 0, 3, 2, 2, 2, 2, 1, 2],
[0, 3, 3, 3, 1, 0, 2, 0, 2, 1]])
In [13]: y = x.reshape(x.shape[0]/2, 2, x.shape[1]/2, 2)
一对夫妇的2×2块看看：
Take a look at a couple of the 2x2 blocks:
In [14]: y[0, :, 0, :]
Out[14]:
array([[2, 1],
[3, 1]])
In [15]: y[1, :, 2, :]
Out[15]:
array([[3, 2],
[2, 0]])
计算各块的平均值：
Compute the averages of the blocks:
In [16]: avg = y.mean(axis=(1, 3))
In [17]: avg
Out[17]:
array([[ 1.75, 1.75, 0.75, 2. , 1.5 ],
[ 0.75, 2.5 , 1.75, 1.5 , 0.75],
[ 0.75, 1.75, 2.25, 0.25, 1.25],
[ 1.5 , 2.25, 1.25, 1.5 , 1.5 ]])
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