[Python知识库]Matlab函数psf2otf()的python实现

本文实现Matlab函数psf2otf()重写为python版本，亲测有效

def psf2otf(psf,size):
	import numpy as np
	if not(0 in psf):
		#Pad the PSF to outsize
		psf=np.double(psf)
		psfsize=np.shape(psf)
		psfsize=np.array(psfsize)
		padsize=size-psfsize
		psf=np.lib.pad(psf,((0,padsize[0]),(0,padsize[1])),'constant')
		#Circularly shift otf so that the "center" of the PSF is at the (1,1) element of the array.
		psf=np.roll(psf,-np.array(np.floor(psfsize/2),'i'),axis=(0,1))
		#Compute the OTF
		otf=np.fft.fftn(psf,axes=(0,1))
		#Estimate the rough number of operations involved in the computation of the FFT.
		nElem=np.prod(psfsize,axis=0)
		nOps=0
		for k in range(0,np.ndim(psf)):
			nffts=nElem/psfsize[k]
			nOps=nOps+psfsize[k]*np.log2(psfsize[k])*nffts
		mx1=(abs(np.imag(otf[:])).max(0)).max(0)
		mx2=(abs(otf[:]).max(0)).max(0)
		eps= 2.2204e-16
		if mx1/mx2<=nOps*eps:
			otf=np.real(otf)
	else:
		otf=np.zeros(size)
	return otf

下面为使用样例（图像平滑）：

MATLAB版：

function S = L0Smoothing(Im, lambda, kappa)
 
if ~exist('kappa','var')
    kappa = 2.0;
end
if ~exist('lambda','var')
    lambda = 2e-2;
end
S = im2double(Im);
betamax = 1e5;
fx = [1, -1];
fy = [1; -1];
[N,M,D] = size(Im);
sizeI2D = [N,M];
otfFx = psf2otf(fx,sizeI2D);
otfFy = psf2otf(fy,sizeI2D);
Normin1 = fft2(S);
Denormin2 = abs(otfFx).^2 + abs(otfFy ).^2;
if D>1
    Denormin2 = repmat(Denormin2,[1,1,D]);
end
beta = 2*lambda;
while beta < betamax
    Denormin   = 1 + beta*Denormin2;
    % h-v subproblem
    h = [diff(S,1,2), S(:,1,:) - S(:,end,:)];
    v = [diff(S,1,1); S(1,:,:) - S(end,:,:)];
    if D==1
        t = (h.^2+v.^2)<lambda/beta;
    else
        t = sum((h.^2+v.^2),3)<lambda/beta;
        t = repmat(t,[1,1,D]);
    end
    h(t)=0; v(t)=0;
    % S subproblem
    Normin2 = [h(:,end,:) - h(:, 1,:), -diff(h,1,2)];
    Normin2 = Normin2 + [v(end,:,:) - v(1, :,:); -diff(v,1,1)];
    FS = (Normin1 + beta*fft2(Normin2))./Denormin;
    S = real(ifft2(FS));
    beta = beta*kappa;
    fprintf('.');
end
fprintf('\n');
end

Python版本：

def psf2otf(psf,size):
	import numpy as np
	if not(0 in psf):
		#Pad the PSF to outsize
		psf=np.double(psf)
		psfsize=np.shape(psf)
		psfsize=np.array(psfsize)
		padsize=size-psfsize
		psf=np.lib.pad(psf,((0,padsize[0]),(0,padsize[1])),'constant')
		#Circularly shift otf so that the "center" of the PSF is at the (1,1) element of the array.
		psf=np.roll(psf,-np.array(np.floor(psfsize/2),'i'),axis=(0,1))
		#Compute the OTF
		otf=np.fft.fftn(psf,axes=(0,1))
		#Estimate the rough number of operations involved in the computation of the FFT.
		nElem=np.prod(psfsize,axis=0)
		nOps=0
		for k in range(0,np.ndim(psf)):
			nffts=nElem/psfsize[k]
			nOps=nOps+psfsize[k]*np.log2(psfsize[k])*nffts
		mx1=(abs(np.imag(otf[:])).max(0)).max(0)
		mx2=(abs(otf[:]).max(0)).max(0)
		eps= 2.2204e-16
		if mx1/mx2<=nOps*eps:
			otf=np.real(otf)
	else:
		otf=np.zeros(size)
	return otf
 
 
def L0Smoothing(Im,lamda=2e-2,kappa=2.0):
	import numpy as np
	S=Im/255
	betamax=1e5
	fx=np.array([[1,-1]])
	fy=np.array([[1],[-1]])
	N,M,D=np.shape(Im)
	sizeI2D=np.array([N,M])
	otfFx=psf2otf(fx,sizeI2D)
	otfFy=psf2otf(fy,sizeI2D)
	Normin1=np.fft.fft2(S,axes=(0,1))
	Denormin2=abs(otfFx)**2+abs(otfFy)**2
	if D>1:
		D2=np.zeros((N,M,D),dtype=np.double)
		for i in range(D):
			D2[:,:,i]=Denormin2
		Denormin2=D2
	beta=lamda*2
	while beta<betamax:
		Denormin=1+beta*Denormin2
		#h-v subproblem
		h1=np.diff(S,1,1)
		h2=np.reshape(S[:,0],(N,1,3))-np.reshape(S[:,-1],(N,1,3))
		h=np.hstack((h1,h2))
		v1=np.diff(S,1,0)
		v2=np.reshape(S[0,:],(1,M,3))-np.reshape(S[-1,:],(1,M,3))
		v=np.vstack((v1,v2))
		if D==1:
			t=(h**2+v**2)<lamda/beta
		else:
			t=np.sum((h**2+v**2),2)<lamda/beta
			t1=np.zeros((N,M,D),dtype=np.bool)
			for i in range(D):
				t1[:,:,i]=t
			t=t1
		h[t]=0
		v[t]=0
		#S subproblem
		Normin2=np.hstack((np.reshape(h[:,-1],(N,1,3))-np.reshape(h[:,0],(N,1,3)),-np.diff(h,1,1)))
		Normin2=Normin2+np.vstack((np.reshape(v[-1,:],(1,M,3))-np.reshape(v[0,:],(1,M,3)),-np.diff(v,1,0)))
		FS=(Normin1+beta*np.fft.fft2(Normin2,axes=(0,1)))/Denormin
		S=np.real(np.fft.ifft2(FS,axes=(0,1)))
		beta*=kappa
		print('.')
	print('\n')
	return S
 
		
def main():
	import PIL
	import numpy as np
	import pylab
	Im=np.array(PIL.Image.open("文件名"),'d')
	S=L0Smoothing(Im,0.01)
	pylab.imshow(S)
	
	
main()		

该代码的处理效果：

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