快速将图像切割成重叠的贴片并将贴片合并到图像的方法

Question

尝试将尺寸为100x100的灰度图像切割成大小为39x39且重叠的贴片,步幅大小为1.这意味着向下/向下开始一个像素的下一个贴片仅与之前的贴片不同一个额外的列/或行.

代码的粗略轮廓:首先计算每个补丁的索引,以便能够从图像构造补丁的2D阵列,并能够从补丁构建图像:

patches = imgFlat[ind]

'patches'是一个2D数组,每列包含一个矢量形式的补丁.

处理这些补丁,每个补丁单独地和之后再次合并到具有预先计算的索引的图像.

img = np.sum(patchesWithColFlat[ind],axis=2)

由于补丁重叠,最后需要将img与预先计算的权重相乘:

imgOut = weights*imgOut

我的代码非常慢,速度是一个关键问题,因为这应该在ca. 10 ^ 8个补丁.

函数get_indices_for_un_patchify和weights_unpatchify可以预先计算一次,因此速度只是patchify和unpatchify的问题.

谢谢你的任何tipps.

卡洛斯

import numpy as np
import scipy
import collections
import random as rand


def get_indices_for_un_patchify(sImg,sP,step):
    ''' creates indices for fast patchifying and unpatchifying

    INPUTS:
      sx    image size
      sp    patch size
      step  offset between two patches (default == [1,1])

      OUTPUTS:
       patchInd             collection with indices
       patchInd.img2patch   patchifying indices
                            patch = img(patchInd.img2patch);
       patchInd.patch2img   unpatchifying indices

    NOTE: * for unpatchifying necessary to add a 0 column to the patch matrix
          * matrices are constructed row by row, as normally there are less rows than columns in the 
            patchMtx
     '''
    lImg = np.prod(sImg)
    indImg = np.reshape(range(lImg), sImg)

    # no. of patches which fit into the image
    sB = (sImg - sP + step) / step

    lb              = np.prod(sB)
    lp              = np.prod(sP)
    indImg2Patch    = np.zeros([lp, lb])
    indPatch        = np.reshape(range(lp*lb), [lp, lb])

    indPatch2Img = np.ones([sImg[0],sImg[1],lp])*(lp*lb+1)

    # default value should be last column
    iRow   = 0;
    for jCol in range(sP[1]):
        for jRow in range(sP[0]):
            tmp1 = np.array(range(0, sImg[0]-sP[0]+1, step[0]))
            tmp2 = np.array(range(0, sImg[1]-sP[1]+1, step[1]))
            sel1                    = jRow  + tmp1
            sel2                    = jCol  + tmp2
            tmpIndImg2Patch = indImg[sel1,:]          
            # do not know how to combine following 2 lines in python
            tmpIndImg2Patch = tmpIndImg2Patch[:,sel2]
            indImg2Patch[iRow, :]   = tmpIndImg2Patch.flatten()

            # next line not nice, but do not know how to implement it better
            indPatch2Img[min(sel1):max(sel1)+1, min(sel2):max(sel2)+1, iRow] = np.reshape(indPatch[iRow, :, np.newaxis], sB)
            iRow                    += 1

    pInd = collections.namedtuple
    pInd.patch2img = indPatch2Img
    pInd.img2patch = indImg2Patch

    return pInd

def weights_unpatchify(sImg,pInd):
    weights = 1./unpatchify(patchify(np.ones(sImg), pInd), pInd)
    return weights

# @profile
def patchify(img,pInd):
    imgFlat = img.flat
   # imgFlat = img.flatten()
    ind = pInd.img2patch.tolist()
    patches = imgFlat[ind]

    return patches

# @profile
def unpatchify(patches,pInd):
    # add a row of zeros to the patches matrix    
    h,w = patches.shape
    patchesWithCol = np.zeros([h+1,w])
    patchesWithCol[:-1,:] = patches
    patchesWithColFlat = patchesWithCol.flat
   #  patchesWithColFlat = patchesWithCol.flatten()
    ind = pInd.patch2img.tolist()
    img = np.sum(patchesWithColFlat[ind],axis=2)
    return img

我将这些功能称为例如随机图像

if __name__ =='__main__':
    img = np.random.randint(255,size=[100,100])
    sImg = img.shape
    sP = np.array([39,39])  # size of patch
    step = np.array([1,1])  # sliding window step size
    pInd = get_indices_for_un_patchify(sImg,sP,step)
    patches = patchify(img,pInd)
    imgOut = unpatchify(patches,pInd)
    weights = weights_unpatchify(sImg,pInd)
    imgOut = weights*imgOut

    print 'Difference of img and imgOut = %.7f' %sum(img.flatten() - imgOut.flatten())

Answer 1

"补充"数组的一种有效方法,即获取原始数组的窗口数组,就是创建一个带有自定义步长的视图,即跳转到以下元素的字节数.将numpy数组视为(荣耀的)内存块可能会有所帮助,然后strides是一种将索引映射到内存地址的方法.

例如,在

a = np.arange(10).reshape(2, 5)

a.itemsize等于4(即,每个元素为4个字节或32位)并且a.strides是(20, 4)(5个元素,1个元素),因此它a[1,2]指的是在第一个元素之后是1*20 + 2*4字节(或1*5 + 2元素)的元素:

0 1 2 3 4
5 6 7 x x

实际上,这些元素一个接一个地放在内存中,0 1 2 3 4 5 6 7 x x但是我们可以将它作为一个2D数组索引.

基于这个概念,我们可以重写patchify如下

def patchify(img, patch_shape):
    img = np.ascontiguousarray(img)  # won't make a copy if not needed
    X, Y = img.shape
    x, y = patch_shape
    shape = ((X-x+1), (Y-y+1), x, y) # number of patches, patch_shape
    # The right strides can be thought by:
    # 1) Thinking of `img` as a chunk of memory in C order
    # 2) Asking how many items through that chunk of memory are needed when indices
    #    i,j,k,l are incremented by one
    strides = img.itemsize*np.array([Y, 1, Y, 1])
    return np.lib.stride_tricks.as_strided(img, shape=shape, strides=strides)

此函数返回一个视图img,因此没有分配内存,它只运行几十微秒.输出形状并不完全符合您的要求,实际上必须将其复制才能获得该形状.

在处理比基本数组大得多的数组视图时,必须要小心,因为操作可以触发需要分配大量内存的副本.在你的情况下,由于数组不是太大而且没有那么多补丁,它应该没问题.

最后,我们可以稍微调整补丁数组:

patches = patchify(img, (39,39))
contiguous_patches = np.ascontiguousarray(patches)
contiguous_patches.shape = (-1, 39**2)

这不会重现补丁函数的输出,因为您可以按照Fortran顺序开发补丁.我建议你改用它,因为

1. 它会导致更自然的索引(即,第一个补丁是补丁[0]而不是补丁[:,0]用于您的解决方案).

2. numpy在任何地方使用C排序也更容易,因为你需要更少的输入(你可以避免使用order ='F'之类的东西,默认情况下以C顺序创建数组......).

如果你坚持"提示":strides = img.itemsize * np.array([1, Y, Y, 1])使用.reshape(..., order='F')在contiguous_patches最后它转.T