将图像分解和收集到碎片100x100 python

问题描述 投票:1回答:2

我有一个图像阵列,我需要在100x100的碎片上分解它,然后用它进行操作收集回原始图像。问题:我不能收回件,我有这样的东西,

但真实的图像是800x800

我的代码:

将图像作为数组,删除第三维

path_to_image = './la3.jpg'
image_array = plt.imread(path_to_image)
image_array = image_array[:, :, 0]

写入新的数组件100x100(工作正常):

main2_array = np.zeros(10000,)
for row in tqdm_notebook(range(0,8)):

    for col in range(0,8):
        main2_array = np.vstack((main2_array, image_array[0 + (100*row):100 + (100*row) ,0 + (100*col):100 + (100*col)].flatten()))

main2_array = np.delete(main2_array, main2_array[0] , axis=0  )

收回件(它不起作用)

main_array = np.zeros(100,)

for p in tqdm_notebook(range(0,100)):
    for i in range(0,64):
        main_array = np.vstack((main_array, main2_array[0 + (10000*i) + (100*p): 100 + (10000*i) + (100*p)]))

main_array = np.delete(main_array, main_array[0] , axis=0  )     

收集完毕后,我得到了

python arrays image numpy
2个回答
0
投票

假图像

x, y = 800,800
img_array = np.arange(x*y).reshape((x,y))

解构后,main2_array.shape是(64,10000);每行是一个扁平的100x100补丁。在解构过程中,您从左到右,从上到下穿过图像,并在前一个补丁下方滑动每个补丁。

重建逆转过程:

main_array = np.zeros((x,y))
for n, patch in enumerate(main2_array):
    patch = patch.reshape(100,100)
    # eight patches per row
    row, col = divmod(n, 8)
    row_offset, col_offset = row*100, col*100
    row_slice = slice(row_offset, 100 + row_offset)
    col_slice = slice(col_offset, 100 + col_offset)
    #print(np.all(patch == image_array[row_slice,col_slice]))
    main_array[row_slice, col_slice] = patch


>>> np.all(main_array == img_array)
True
>>> 

或者你可以重新塑造你原来的方式

>>> b = main2_array.reshape(8,8,100,100)
>>> b[0,1].shape    # row zero column 1? 
(100, 100)
>>> np.all(b[0,1] == a[0:100, 100:200])
True
>>> 
>>> c = np.swapaxes(b, 1,2)
>>> c.shape
(8, 100, 8, 100)
>>> np.all(c[0,:,1,:] == a[0:100, 100:200])    # row zero column 1? 
True
>>> d = c.reshape(800,800)
>>> np.all(d==img_array)
True
>>>

0
投票

有点晚了,但我相信这个问题应该得到一个更加图解的答案。

你不需要慢循环,你可以用numpy重塑和花式索引来完成所有这一切。

让我们从一个示例图像开始

import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt  
import skimage.transform
import skimage.data

img = skimage.data.chelsea()
# crop from (300, 451, 3) to (300, 300, 3)
img = img[:,80:380,:]
# resize to (800, 800)
img = skimage.transform.resize(img, (800,800))
plt.imshow(img)

enter image description here

64 100*100瓷砖中分解图像。新的形状是(8, 100, 8, 100, 3),你可以用img[i, :, j, :, ...]解决单个图像。除了可能更容易阅读之外,无需将它们存储在新阵列中。

img = img.reshape(8, 100, 8, 100, 3)
gs = mpl.gridspec.GridSpec(8,8)
for i in range(8):
    for j in range(8):
        ax = plt.subplot(gs[i,j])
        ax.imshow(img[i,:,j,:,...])

enter image description here

现在让我们对瓷砖进行一些操作。

清除一些随机的瓷砖

cells = np.random.randint(8, size=(20,2))
img[cells[:,0],:,cells[:,1],...] = 1

翻转并从左到右翻转

img = img[:,::-1,:,::-1,...]

添加黑色边框

img[:,:6,...] = 0
img[:,-6:,...] = 0
img[:,:,:,:6,...] = 0
img[:,:,:,-6:,...] = 0

并绘制它们

for i in range(8):
    for j in range(8):
        ax = plt.subplot(gs[i,j])
        ax.imshow(img[i,:,j,:,...])

enter image description here

现在重建你可以重塑原始形状

img = img.reshape(800, 800, 3)
plt.imshow(img)

enter image description here

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