我如何在类似于Matlab的blkproc(blockproc)函数的块中高效处理numpy数组
image-processing
matlab
numpy
python
5
0

我正在寻找一种有效的方法,用于将图像有效地划分为小区域,分别处理每个区域,然后将每个过程的结果重新组合为一个已处理的图像。 Matlab为此提供了一个名为blkproc的工具(在较新版本的Matlab中由blockproc代替)。

在理想情况下,函数或类也将支持输入矩阵中各部分之间的重叠。在Matlab帮助中,blkproc定义为:

B = blkproc(A,[mn],[mborder nborder],fun,...)

  • A是您的输入矩阵,
  • [mn]是块大小
  • [mborder,nborder]是边界区域的大小(可选)
  • fun是应用于每个块的功能

我拼凑出一种方法,但是它让我感到笨拙,我敢打赌,还有更好的方法。冒着我自己的窘境,这是我的代码:


import numpy as np

def segmented_process(M, blk_size=(16,16), overlap=(0,0), fun=None):
    rows = []
    for i in range(0, M.shape[0], blk_size[0]):
        cols = []
        for j in range(0, M.shape[1], blk_size[1]):
            cols.append(fun(M[i:i+blk_size[0], j:j+blk_size[1]]))
        rows.append(np.concatenate(cols, axis=1))
    return np.concatenate(rows, axis=0)

R = np.random.rand(128,128)
passthrough = lambda(x):x
Rprime = segmented_process(R, blk_size=(16,16), 
                           overlap=(0,0), 
                           fun=passthrough)

np.all(R==Rprime)
参考资料:
Stack Overflow
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我同时接受了输入和原始方法,并比较了结果。正如@eat正确指出的那样,结果取决于输入数据的性质。令人惊讶的是,在少数情况下,串联拍子视图处理。每种方法都有一个最佳点。这是我的基准代码:

import numpy as np
from itertools import product

def segment_and_concatenate(M, fun=None, blk_size=(16,16), overlap=(0,0)):
    # truncate M to a multiple of blk_size
    M = M[:M.shape[0]-M.shape[0]%blk_size[0], 
          :M.shape[1]-M.shape[1]%blk_size[1]]
    rows = []
    for i in range(0, M.shape[0], blk_size[0]):
        cols = []
        for j in range(0, M.shape[1], blk_size[1]):
            max_ndx = (min(i+blk_size[0], M.shape[0]),
                       min(j+blk_size[1], M.shape[1]))
            cols.append(fun(M[i:max_ndx[0], j:max_ndx[1]]))
        rows.append(np.concatenate(cols, axis=1))
    return np.concatenate(rows, axis=0)


from numpy.lib.stride_tricks import as_strided
def block_view(A, block= (3, 3)):
    """Provide a 2D block view to 2D array. No error checking made.
    Therefore meaningful (as implemented) only for blocks strictly
    compatible with the shape of A."""
    # simple shape and strides computations may seem at first strange
    # unless one is able to recognize the 'tuple additions' involved ;-)
    shape= (A.shape[0]/ block[0], A.shape[1]/ block[1])+ block
    strides= (block[0]* A.strides[0], block[1]* A.strides[1])+ A.strides
    return as_strided(A, shape= shape, strides= strides)

def segmented_stride(M, fun, blk_size=(3,3), overlap=(0,0)):
    # This is some complex function of blk_size and M.shape
    stride = blk_size
    output = np.zeros(M.shape)

    B = block_view(M, block=blk_size)
    O = block_view(output, block=blk_size)

    for b,o in zip(B, O):
        o[:,:] = fun(b);

    return output

def view_process(M, fun=None, blk_size=(16,16), overlap=None):
    # truncate M to a multiple of blk_size
    from itertools import product
    output = np.zeros(M.shape)

    dz = np.asarray(blk_size)
    shape = M.shape - (np.mod(np.asarray(M.shape), 
                          blk_size))
    for indices in product(*[range(0, stop, step) 
                        for stop,step in zip(shape, blk_size)]):
        # Don't overrun the end of the array.
        #max_ndx = np.min((np.asarray(indices) + dz, M.shape), axis=0)
        #slices = [slice(s, s + f, None) for s,f in zip(indices, dz)]
        output[indices[0]:indices[0]+dz[0], 
               indices[1]:indices[1]+dz[1]][:,:] = fun(M[indices[0]:indices[0]+dz[0], 
               indices[1]:indices[1]+dz[1]])

    return output

if __name__ == "__main__":
    R = np.random.rand(128,128)
    squareit = lambda(x):x*2

    from timeit import timeit
    t ={}
    kn = np.array(list(product((8,16,64,128), 
                               (128, 512, 2048, 4096))  ) )

    methods = ("segment_and_concatenate", 
               "view_process", 
               "segmented_stride")    
    t = np.zeros((kn.shape[0], len(methods)))

    for i, (k, N) in enumerate(kn):
        for j, method in enumerate(methods):
            t[i,j] = timeit("""Rprime = %s(R, blk_size=(%d,%d), 
                          overlap = (0,0), 
                          fun = squareit)""" % (method, k, k),
                   setup="""
from segmented_processing import %s
import numpy as np
R = np.random.rand(%d,%d)
squareit = lambda(x):x**2""" % (method, N, N),
number=5
)
        print "k =", k, "N =", N #, "time:", t[i]
        print ("    Speed up (view vs. concat, stride vs. concat): %0.4f, %0.4f" % (
                       t[i][0]/t[i][1], 
                       t[i][0]/t[i][2]))

结果如下:

在numpy中以块方式处理大型矩阵的三种方法的比较请注意,对于较小的块大小,分段的跨步方法会赢3-4倍。只有在较大的块大小(128 x 128)和非常大的矩阵(2048 x 2048及更大)下,视图处理方法才能获胜,然后仅占很小的比例。基于烘烤,看起来@eat获得了选中标记!感谢你们俩提供了很好的例子!

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按切片/视图进行处理。串联非常昂贵。

for x in xrange(0, 160, 16):
    for y in xrange(0, 160, 16):
        view = A[x:x+16, y:y+16]
        view[:,:] = fun(view)
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甚至在游戏后期。在以下位置有一个名为Bob的瑞士图像处理软件包: https ://www.idiap.ch/software/bob/它具有一些用于块的python命令,例如bob.ip.base.block,它似乎可以完成Matlab命令的所有操作'blockproc'确实如此。我还没有测试。

还有一些有趣的命令bob.ip.base.DCTFeatures,其中包含“块”功能以提取或修改图像的DCT系数。

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以下是使用块的另一种(无循环)方法的示例:

import numpy as np
from numpy.lib.stride_tricks import as_strided as ast

A= np.arange(36).reshape(6, 6)
print A
#[[ 0  1  2  3  4  5]
# [ 6  7  8  9 10 11]
# ...
# [30 31 32 33 34 35]]

# 2x2 block view
B= ast(A, shape= (3, 3, 2, 2), strides= (48, 8, 24, 4))
print B[1, 1]
#[[14 15]
# [20 21]]

# for preserving original shape
B[:, :]= np.dot(B[:, :], np.array([[0, 1], [1, 0]]))
print A
#[[ 1  0  3  2  5  4]
# [ 7  6  9  8 11 10]
# ...
# [31 30 33 32 35 34]]
print B[1, 1]
#[[15 14]
# [21 20]]

# for reducing shape, processing in 3D is enough
C= B.reshape(3, 3, -1)
print C.sum(-1)
#[[ 14  22  30]
# [ 62  70  78]
# [110 118 126]]

因此,仅尝试简单地将matlab功能复制到numpy并不是所有最佳方法。有时需要“脱帽而出”的思维。

警告
通常,基于跨步技巧的实现可能会 (但不一定需要)遭受一些性能损失。因此,请做好准备以各种方式衡量您的表现。无论如何,明智的做法是首先检查所需的功能(或足够相似,以便轻松适应)是否已全部以numpyscipy

更新
请注意,有没有真正的magic与这里涉及strides ,所以我将提供一个简单的函数来得到一个block_view任何合适的2D numpy -阵列。所以我们开始:

from numpy.lib.stride_tricks import as_strided as ast

def block_view(A, block= (3, 3)):
    """Provide a 2D block view to 2D array. No error checking made.
    Therefore meaningful (as implemented) only for blocks strictly
    compatible with the shape of A."""
    # simple shape and strides computations may seem at first strange
    # unless one is able to recognize the 'tuple additions' involved ;-)
    shape= (A.shape[0]/ block[0], A.shape[1]/ block[1])+ block
    strides= (block[0]* A.strides[0], block[1]* A.strides[1])+ A.strides
    return ast(A, shape= shape, strides= strides)

if __name__ == '__main__':
    from numpy import arange
    A= arange(144).reshape(12, 12)
    print block_view(A)[0, 0]
    #[[ 0  1  2]
    # [12 13 14]
    # [24 25 26]]
    print block_view(A, (2, 6))[0, 0]
    #[[ 0  1  2  3  4  5]
    # [12 13 14 15 16 17]]
    print block_view(A, (3, 12))[0, 0]
    #[[ 0  1  2  3  4  5  6  7  8  9 10 11]
    # [12 13 14 15 16 17 18 19 20 21 22 23]
    # [24 25 26 27 28 29 30 31 32 33 34 35]]
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游戏有点晚,但是这样做会重叠块。我在这里还没有做过,但是我认为您可以轻松地调整步长来移动窗口。

from numpy.lib.stride_tricks import as_strided
def rolling_block(A, block=(3, 3)):
    shape = (A.shape[0] - block[0] + 1, A.shape[1] - block[1] + 1) + block
    strides = (A.strides[0], A.strides[1]) + A.strides
    return as_strided(A, shape=shape, strides=strides)
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