计算方差图像python

Calculating variance image python

有没有一种简单的方法可以使用 Python/NumPy/Scipy 计算图像上的运行方差过滤器?通过运行方差图像,我的意思是计算图像中每个子窗口 I 的 sum((I – mean(I))^2)/nPixels 的结果。

由于图像非常大(12000×12000 像素),我想避免在格式之间转换数组的开销,以便能够使用不同的库然后再转换回来。

我想我可以通过使用类似

的方法找到平均值来手动执行此操作

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kernel = np.ones((winSize, winSize))/winSize**2
image_mean = scipy.ndimage.convolve(image, kernel)
diff = (image – image_mean)**2
# Calculate sum over winSize*winSize sub-images
# Subsample result

但是如果有像 Matlab 中的 stdfilt 函数这样的东西会更好。

任何人都可以指出具有此功能并支持 numpy 数组的库的方向,或者提示/提供一种在 NumPy/SciPy 中执行此操作的方法吗?


更简单的解决方案也更快:使用 SciPy//’s ndimage.uniform_filter

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import numpy as np
from scipy import ndimage
rows, cols = 500, 500
win_rows, win_cols = 5, 5

img = np.random.rand(rows, cols)
win_mean = ndimage.uniform_filter(img, (win_rows, win_cols))
win_sqr_mean = ndimage.uniform_filter(img**2, (win_rows, win_cols))
win_var = win_sqr_mean – win_mean**2

“跨步技巧”是个漂亮的技巧,但速度慢了 4 且不那么可读。
generic_filter 比步幅慢 20 倍…


您可以使用 numpy.lib.stride_tricks.as_strided 来获得图像的窗口视图:

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

rows, cols = 500, 500
win_rows, win_cols = 5, 5

img = np.random.rand(rows, cols)

win_img = as_strided(img, shape=(rows-win_rows+1, cols-win_cols+1,
                                 win_rows, win_cols),
                     strides=img.strides*2)

现在 win_img[i, j](win_rows, win_cols) 数组,左上角位于 [i, j] 位置:

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>>> img[100:105, 100:105]
array([[ 0.34150754,  0.17888323,  0.67222354,  0.9020784 ,  0.48826682],
       [ 0.68451774,  0.14887515,  0.44892615,  0.33352743,  0.22090103],
       [ 0.41114758,  0.82608407,  0.77190533,  0.42830363,  0.57300759],
       [ 0.68435626,  0.94874394,  0.55238567,  0.40367885,  0.42955156],
       [ 0.59359203,  0.62237553,  0.58428725,  0.58608119,  0.29157555]])
>>> win_img[100,100]
array([[ 0.34150754,  0.17888323,  0.67222354,  0.9020784 ,  0.48826682],
       [ 0.68451774,  0.14887515,  0.44892615,  0.33352743,  0.22090103],
       [ 0.41114758,  0.82608407,  0.77190533,  0.42830363,  0.57300759],
       [ 0.68435626,  0.94874394,  0.55238567,  0.40367885,  0.42955156],
       [ 0.59359203,  0.62237553,  0.58428725,  0.58608119,  0.29157555]])

但是,您必须小心,不要将图像的窗口视图转换为它的窗口副本:在我的示例中,这将需要 25 倍的存储空间。我相信 numpy 1.7 可以让您选择多个轴,因此您可以简单地执行以下操作:

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>>> np.var(win_img, axis=(1,2))

我被 numpy 1.6.2 卡住了,所以我无法测试它。如果我没记错我的数学,另一个选项可能会因窗口不太大而失败:

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>>> win_mean = np.sum(np.sum(win_img, axis=1), axis=1)/win_rows/win_cols
>>> win_sqr_mean = np.sum(np.sum(win_img**2, axis=1), axis=1)/win_rows/win_cols
>>> win_var = win_sqr_mean – win_mean**2

现在 win_var 是一个形状数组

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>>> win_var.shape
(496, 496)

win_var[i, j] 保存 (5, 5) 窗口的方差,左上角在 [i, j]


经过一些优化,我们想出了一个通用 3D 图像的函数:

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def variance_filter( img, VAR_FILTER_SIZE ):
  from numpy.lib.stride_tricks import as_strided

  WIN_SIZE=(2*VAR_FILTER_SIZE)+1
  if ~ VAR_FILTER_SIZE%2==1:
      print ‘Warning, VAR_FILTER_SIZE must be ODD Integer number  ‘
  # hack — this could probably be an input to the function but Alessandro is lazy
  WIN_DIMS = [ WIN_SIZE, WIN_SIZE, WIN_SIZE ]

  # Check that there is a 3D image input.
  if len( img.shape ) != 3:
      print"//t variance_filter: Are you sure that you passed me a 3D image?"
      return1
  else:
      DIMS = img.shape

  # Set up a windowed view on the data… this will have a border removed compared to the img_in
  img_strided = as_strided(img, shape=(DIMS[0]-WIN_DIMS[0]+1, DIMS[1]-WIN_DIMS[1]+1, DIMS[2]-WIN_DIMS[2]+1, WIN_DIMS[0], WIN_DIMS[1], WIN_DIMS[2] ), strides=img.strides*2)

  # Calculate variance, vectorially
  win_mean = numpy.sum(numpy.sum(numpy.sum(img_strided, axis=1), axis=1), axis=1) / (WIN_DIMS[0]*WIN_DIMS[1]*WIN_DIMS[2])

  # As per http://en.wikipedia.org/wiki/Variance, we are removing the mean from every window,
  #   then squaring the result.
  # Casting to 64 bit float inside, because the numbers (at least for our images) get pretty big
  win_var = numpy.sum(numpy.sum(numpy.sum((( img_strided.T.astype(‘<f8’) – win_mean.T.astype(‘<f8’) )**2).T, axis=1), axis=1), axis=1) / (WIN_DIMS[0]*WIN_DIMS[1]*WIN_DIMS[2])

  # Prepare an output image of the right size, in order to replace the border removed with the windowed view call
  out_img = numpy.zeros( DIMS, dtype=‘<f8’ )
  # copy borders out…
  out_img[ WIN_DIMS[0]/2:DIMS[0]-WIN_DIMS[0]+1+WIN_DIMS[0]/2, WIN_DIMS[1]/2:DIMS[1]-WIN_DIMS[1]+1+WIN_DIMS[1]/2, WIN_DIMS[2]/2:DIMS[2]-WIN_DIMS[2]+1+WIN_DIMS[2]/2, ] = win_var

  # output
  return out_img.astype(‘>f4’)


您可以使用 scipy.ndimage.generic_filter。我无法使用 matlab 进行测试,但也许这可以为您提供所需的内容:

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import numpy as np
import scipy.ndimage as ndimage    
subs = 10  # this is the size of the (square) sub-windows
img = np.random.rand(500, 500)
img_std = ndimage.filters.generic_filter(img, np.std, size=subs)

您可以使用 footprint 关键字制作任意大小的子窗口。有关示例,请参阅此问题。


原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/tech/python/268043.html

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