我试图在Python中插入两个图像。
图像具有形状(188,188)
我希望在中间插入图像'这两个图像。假设Image_1位于z = 0位置,Image_2位于z = 2位置。我希望插值图像位于z = 1。
我相信这个答案(MATLAB)包含一个类似的问题和解决方案。
Creating intermediate slices in a 3D MRI volume with MATLAB
我尝试将此代码转换为Python,如下所示:
from scipy.interpolate import interpn
from scipy.interpolate import griddata
# Construct 3D volume from images
# arr.shape = (2, 182, 182)
arr = np.r_['0,3', image_1, image_2]
slices,rows,cols = arr.shape
# Construct meshgrids
[X,Y,Z] = np.meshgrid(np.arange(cols), np.arange(rows), np.arange(slices));
[X2,Y2,Z2] = np.meshgrid(np.arange(cols), np.arange(rows), np.arange(slices*2));
# Run n-dim interpolation
Vi = interpn([X,Y,Z], arr, np.array([X1,Y1,Z1]).T)
然而,这会产生错误:
ValueError: The points in dimension 0 must be strictly ascending
我怀疑我没有正确构建我的网格,但是这种方法是否正确却有点迷失。
有什么想法吗?
----------编辑-----------
找到一些似乎可以解决这个问题的MATLAB代码:
Interpolating Between Two Planes in 3d space
我试图将其转换为Python:
from scipy.ndimage.morphology import distance_transform_edt
from scipy.interpolate import interpn
def ndgrid(*args,**kwargs):
"""
Same as calling ``meshgrid`` with *indexing* = ``'ij'`` (see
``meshgrid`` for documentation).
"""
kwargs['indexing'] = 'ij'
return np.meshgrid(*args,**kwargs)
def bwperim(bw, n=4):
"""
perim = bwperim(bw, n=4)
Find the perimeter of objects in binary images.
A pixel is part of an object perimeter if its value is one and there
is at least one zero-valued pixel in its neighborhood.
By default the neighborhood of a pixel is 4 nearest pixels, but
if `n` is set to 8 the 8 nearest pixels will be considered.
Parameters
----------
bw : A black-and-white image
n : Connectivity. Must be 4 or 8 (default: 8)
Returns
-------
perim : A boolean image
From Mahotas: http://nullege.com/codes/search/mahotas.bwperim
"""
if n not in (4,8):
raise ValueError('mahotas.bwperim: n must be 4 or 8')
rows,cols = bw.shape
# Translate image by one pixel in all directions
north = np.zeros((rows,cols))
south = np.zeros((rows,cols))
west = np.zeros((rows,cols))
east = np.zeros((rows,cols))
north[:-1,:] = bw[1:,:]
south[1:,:] = bw[:-1,:]
west[:,:-1] = bw[:,1:]
east[:,1:] = bw[:,:-1]
idx = (north == bw) & \
(south == bw) & \
(west == bw) & \
(east == bw)
if n == 8:
north_east = np.zeros((rows, cols))
north_west = np.zeros((rows, cols))
south_east = np.zeros((rows, cols))
south_west = np.zeros((rows, cols))
north_east[:-1, 1:] = bw[1:, :-1]
north_west[:-1, :-1] = bw[1:, 1:]
south_east[1:, 1:] = bw[:-1, :-1]
south_west[1:, :-1] = bw[:-1, 1:]
idx &= (north_east == bw) & \
(south_east == bw) & \
(south_west == bw) & \
(north_west == bw)
return ~idx * bw
def signed_bwdist(im):
'''
Find perim and return masked image (signed/reversed)
'''
im = -bwdist(bwperim(im))*np.logical_not(im) + bwdist(bwperim(im))*im
return im
def bwdist(im):
'''
Find distance map of image
'''
dist_im = distance_transform_edt(1-im)
return dist_im
def interp_shape(top, bottom, num):
if num<0 and round(num) == num:
print("Error: number of slices to be interpolated must be integer>0")
top = signed_bwdist(top)
bottom = signed_bwdist(bottom)
r, c = top.shape
t = num+2
print("Rows - Cols - Slices")
print(r, c, t)
print("")
# rejoin top, bottom into a single array of shape (2, r, c)
# MATLAB: cat(3,bottom,top)
top_and_bottom = np.r_['0,3', top, bottom]
#top_and_bottom = np.rollaxis(top_and_bottom, 0, 3)
# create ndgrids
x,y,z = np.mgrid[0:r, 0:c, 0:t-1] # existing data
x1,y1,z1 = np.mgrid[0:r, 0:c, 0:t] # including new slice
print("Shape x y z:", x.shape, y.shape, z.shape)
print("Shape x1 y1 z1:", x1.shape, y1.shape, z1.shape)
print(top_and_bottom.shape, len(x), len(y), len(z))
# Do interpolation
out = interpn((x,y,z), top_and_bottom, (x1,y1,z1))
# MATLAB: out = out(:,:,2:end-1)>=0;
array_lim = out[-1]-1
out[out[:,:,2:out] >= 0] = 1
return out
我这称之为:
new_image = interp_shape(image_1,image_2, 1)
我很确定这是80%的方式,但我在运行时仍然遇到此错误:
ValueError: The points in dimension 0 must be strictly ascending
同样,我可能没有正确构建我的网格。我相信np.mgrid
应该产生与MATLAB ndgrid
相同的结果。
有没有更好的方法来构建ndgrid
等价物?
答案 0 :(得分:1)
我不知道问题的解决方案,但我不认为可以使用interpn
执行此操作。
我更正了您尝试过的代码,并使用了以下输入图像:
但结果是:
以下是更正后的代码:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy import interpolate
n = 8
img1 = np.zeros((n, n))
img2 = np.zeros((n, n))
img1[2:4, 2:4] = 1
img2[4:6, 4:6] = 1
plt.figure()
plt.imshow(img1, cmap=cm.Greys)
plt.figure()
plt.imshow(img2, cmap=cm.Greys)
points = (np.r_[0, 2], np.arange(n), np.arange(n))
values = np.stack((img1, img2))
xi = np.rollaxis(np.mgrid[:n, :n], 0, 3).reshape((n**2, 2))
xi = np.c_[np.ones(n**2), xi]
values_x = interpolate.interpn(points, values, xi, method='linear')
values_x = values_x.reshape((n, n))
print(values_x)
plt.figure()
plt.imshow(values_x, cmap=cm.Greys)
plt.clim((0, 1))
plt.show()
我认为您的代码与我的代码之间的主要区别在于xi
的规范。 interpn
使用起来有点混乱,我在older answer中更详细地解释了它。如果您对我如何指定xi
的机制感到好奇,请参阅我的this answer解释我已做过的事情。
这个结果并不完全令人惊讶,因为interpn
只是在两个图像之间进行线性插值:所以在一个图像中有1个而在另一个图像中有0的部分只是变成0.5。
在这里,由于一张图片是另一张图片的翻译,因此我们很清楚我们想要一张翻译过的图像&#34;介于&#34;之间。但interpn
如何插入两个一般图像呢?如果你有一个小圆圈和一个大圆圈,那么它是否应该清楚地表明应该有一个中间尺寸的圆圈&#34;在&#34;之间。他们?在狗和猫之间进行插值怎么样?还是狗和建筑物?
我认为你基本上是在尝试画线&#34;连接两个图像的边缘,然后试图找出其间的图像。这类似于在半帧处对移动视频进行采样。您可能希望查看optical flow之类的内容,它使用向量连接相邻的帧。我不知道是否以及有哪些python包/实现可用。
答案 1 :(得分:1)
我想出来了。或者至少是产生理想结果的方法。
基于:Interpolating Between Two Planes in 3d space
def signed_bwdist(im):
'''
Find perim and return masked image (signed/reversed)
'''
im = -bwdist(bwperim(im))*np.logical_not(im) + bwdist(bwperim(im))*im
return im
def bwdist(im):
'''
Find distance map of image
'''
dist_im = distance_transform_edt(1-im)
return dist_im
def interp_shape(top, bottom, precision):
'''
Interpolate between two contours
Input: top
[X,Y] - Image of top contour (mask)
bottom
[X,Y] - Image of bottom contour (mask)
precision
float - % between the images to interpolate
Ex: num=0.5 - Interpolate the middle image between top and bottom image
Output: out
[X,Y] - Interpolated image at num (%) between top and bottom
'''
if precision>2:
print("Error: Precision must be between 0 and 1 (float)")
top = signed_bwdist(top)
bottom = signed_bwdist(bottom)
# row,cols definition
r, c = top.shape
# Reverse % indexing
precision = 1+precision
# rejoin top, bottom into a single array of shape (2, r, c)
top_and_bottom = np.stack((top, bottom))
# create ndgrids
points = (np.r_[0, 2], np.arange(r), np.arange(c))
xi = np.rollaxis(np.mgrid[:r, :c], 0, 3).reshape((r**2, 2))
xi = np.c_[np.full((r**2),precision), xi]
# Interpolate for new plane
out = interpn(points, top_and_bottom, xi)
out = out.reshape((r, c))
# Threshold distmap to values above 0
out = out > 0
return out
# Run interpolation
out = interp_shape(image_1,image_2, 0.5)
答案 2 :(得分:1)
我遇到了一个类似的问题,我需要对帧之间的偏移进行插值,其中变化不仅构成平移,而且还改变了形状本身。我通过以下方法解决了这个问题:
代码如下:
def inter(images,t):
#input:
# images: list of arrays/frames ordered according to motion
# t: parameter ranging from 0 to 1 corresponding to first and last frame
#returns: interpolated image
#direction of movement, assumed to be approx. linear
a=np.array(center_of_mass(images[0]))
b=np.array(center_of_mass(images[-1]))
#find index of two nearest frames
arr=np.array([center_of_mass(images[i]) for i in range(len(images))])
v=a+t*(b-a) #convert t into vector
idx1 = (np.linalg.norm((arr - v),axis=1)).argmin()
arr[idx1]=np.array([0,0]) #this is sloppy, should be changed if relevant values are near [0,0]
idx2 = (np.linalg.norm((arr - v),axis=1)).argmin()
if idx1>idx2:
b=np.array(center_of_mass(images[idx1])) #center of mass of nearest contour
a=np.array(center_of_mass(images[idx2])) #center of mass of second nearest contour
tstar=np.linalg.norm(v-a)/np.linalg.norm(b-a) #define parameter ranging from 0 to 1 for interpolation between two nearest frames
im1_shift=shift(images[idx2],(b-a)*tstar) #shift frame 1
im2_shift=shift(images[idx1],-(b-a)*(1-tstar)) #shift frame 2
return im1_shift+im2_shift #return average
if idx1<idx2:
b=np.array(center_of_mass(images[idx2]))
a=np.array(center_of_mass(images[idx1]))
tstar=np.linalg.norm(v-a)/np.linalg.norm(b-a)
im1_shift=shift(images[idx2],-(b-a)*(1-tstar))
im2_shift=shift(images[idx1],(b-a)*(tstar))
return im1_shift+im2_shift