我正在尝试创建水印标识符，但收到错误
AxisError：轴0超出维度0数组的范围
在我看来，这是矩阵问题，但我无法理解。谁能让我理解此错误代码
下面是我的代码：
import sys, os
import cv2
import numpy as np
import warnings
from matplotlib import pyplot as plt
import math
import numpy
import scipy, scipy.fftpack
# Variables
KERNEL_SIZE = 3
def estimate_watermark(foldername):
"""
Given a folder, estimate the watermark (grad(W) = median(grad(J)))
Also, give the list of gradients, so that further processing can be done on it
"""
if not os.path.exists(foldername):
warnings.warn("Folder does not exist.", UserWarning)
return None
images = []
for r, dirs, files in os.walk(foldername):
# Get all the images
for file in files:
img = cv2.imread(os.sep.join([r, file]))
if img is not None:
images.append(img)
else:
print("%s not found."%(file))
# Compute gradients
print("Computing gradients.")
gradx = map(lambda x: cv2.Sobel(x, cv2.CV_64F, 1, 0, ksize=KERNEL_SIZE), images)
grady = map(lambda x: cv2.Sobel(x, cv2.CV_64F, 0, 1, ksize=KERNEL_SIZE), images)
print(type(gradx))
print(type(grady))
# Compute median of grads
print("Computing median gradients.")
Wm_x = np.median(np.array(gradx), axis=0)#it occurs here
Wm_y = np.median(np.array(grady), axis=0)
return (Wm_x, Wm_y, gradx, grady)
def PlotImage(image):
"""
PlotImage: Give a normalized image matrix which can be used with implot, etc.
Maps to [0, 1]
"""
im = image.astype(float)
return (im - np.min(im))/(np.max(im) - np.min(im))
def poisson_reconstruct2(gradx, grady, boundarysrc):
# Thanks to Dr. Ramesh Raskar for providing the original matlab code from which this is derived
# Dr. Raskar's version is available here: http://web.media.mit.edu/~raskar/photo/code.pdf
# Laplacian
gyy = grady[1:,:-1] - grady[:-1,:-1]
gxx = gradx[:-1,1:] - gradx[:-1,:-1]
f = numpy.zeros(boundarysrc.shape)
f[:-1,1:] += gxx
f[1:,:-1] += gyy
# Boundary image
boundary = boundarysrc.copy()
boundary[1:-1,1:-1] = 0;
# Subtract boundary contribution
f_bp = -4*boundary[1:-1,1:-1] + boundary[1:-1,2:] + boundary[1:-1,0:-2] + boundary[2:,1:-1] + boundary[0:-2,1:-1]
f = f[1:-1,1:-1] - f_bp
# Discrete Sine Transform
tt = scipy.fftpack.dst(f, norm='ortho')
fsin = scipy.fftpack.dst(tt.T, norm='ortho').T
# Eigenvalues
(x,y) = numpy.meshgrid(range(1,f.shape[1]+1), range(1,f.shape[0]+1), copy=True)
denom = (2*numpy.cos(math.pi*x/(f.shape[1]+2))-2) + (2*numpy.cos(math.pi*y/(f.shape[0]+2)) - 2)
f = fsin/denom
# Inverse Discrete Sine Transform
tt = scipy.fftpack.idst(f, norm='ortho')
img_tt = scipy.fftpack.idst(tt.T, norm='ortho').T
# New center + old boundary
result = boundary
result[1:-1,1:-1] = img_tt
return result
def poisson_reconstruct(gradx, grady, kernel_size=KERNEL_SIZE, num_iters=100, h=0.1,
boundary_image=None, boundary_zero=True):
"""
Iterative algorithm for Poisson reconstruction.
Given the gradx and grady values, find laplacian, and solve for image
Also return the squared difference of every step.
h = convergence rate
"""
fxx = cv2.Sobel(gradx, cv2.CV_64F, 1, 0, ksize=kernel_size)
fyy = cv2.Sobel(grady, cv2.CV_64F, 0, 1, ksize=kernel_size)
laplacian = fxx + fyy
m,n,p = laplacian.shape
if boundary_zero == True:
est = np.zeros(laplacian.shape)
else:
assert(boundary_image is not None)
assert(boundary_image.shape == laplacian.shape)
est = boundary_image.copy()
est[1:-1, 1:-1, :] = np.random.random((m-2, n-2, p))
loss = []
for i in xrange(num_iters):
old_est = est.copy()
est[1:-1, 1:-1, :] = 0.25*(est[0:-2, 1:-1, :] + est[1:-1, 0:-2, :] + est[2:, 1:-1, :] + est[1:-1, 2:, :] - h*h*laplacian[1:-1, 1:-1, :])
error = np.sum(np.square(est-old_est))
loss.append(error)
return (est)
def image_threshold(image, threshold=0.5):
'''
Threshold the image to make all its elements greater than threshold*MAX = 1
'''
m, M = np.min(image), np.max(image)
im = PlotImage(image)
im[im >= threshold] = 1
im[im < 1] = 0
return im
def crop_watermark(gradx, grady, threshold=0.4, boundary_size=2):
"""
Crops the watermark by taking the edge map of magnitude of grad(W)
Assumes the gradx and grady to be in 3 channels
@param: threshold - gives the threshold param
@param: boundary_size - boundary around cropped image
"""
W_mod = np.sqrt(np.square(gradx) + np.square(grady))
W_mod = PlotImage(W_mod)
W_gray = image_threshold(np.average(W_mod, axis=2), threshold=threshold)
x, y = np.where(W_gray == 1)
xm, xM = np.min(x) - boundary_size - 1, np.max(x) + boundary_size + 1
ym, yM = np.min(y) - boundary_size - 1, np.max(y) + boundary_size + 1
return gradx[xm:xM, ym:yM, :] , grady[xm:xM, ym:yM, :]
def normalized(img):
"""
Return the image between -1 to 1 so that its easier to find out things like
correlation between images, convolutionss, etc.
Currently required for Chamfer distance for template matching.
"""
return (2*PlotImage(img)-1)
def watermark_detector(img, gx, gy, thresh_low=200, thresh_high=220, printval=False):
"""
Compute a verbose edge map using Canny edge detector, take its magnitude.
Assuming cropped values of gradients are given.
Returns image, start and end coordinates
"""
Wm = (np.average(np.sqrt(np.square(gx) + np.square(gy)), axis=2))
img_edgemap = (cv2.Canny(img, thresh_low, thresh_high))
chamfer_dist = cv2.filter2D(img_edgemap.astype(float), -1, Wm)
rect = Wm.shape
index = np.unravel_index(np.argmax(chamfer_dist), img.shape[:-1])
if printval:
print(index)
x,y = (index[0]-rect[0]/2), (index[1]-rect[1]/2)
im = img.copy()
cv2.rectangle(im, (y, x), (y+rect[1], x+rect[0]), (255, 0, 0))
return (im, (x, y), (rect[0], rect[1]))
任何人都可以帮忙...似乎是矩阵问题
File "<ipython-input-6-0e709e7207fd>", line 1, in <module>
runfile('C:/Users/prince.bhatia/Desktop/automatic-watermark-detection-master/automatic-watermark-detection-master/Watermarking.py', wdir='C:/Users/prince.bhatia/Desktop/automatic-watermark-detection-master/automatic-watermark-detection-master')
File "C:\Users\prince.bhatia\AppData\Local\Continuum\anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 710, in runfile
execfile(filename, namespace)
File "C:\Users\prince.bhatia\AppData\Local\Continuum\anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 101, in execfile
exec(compile(f.read(), filename, 'exec'), namespace)
File "C:/Users/prince.bhatia/Desktop/automatic-watermark-detection-master/automatic-watermark-detection-master/Watermarking.py", line 4, in <module>
gx, gy, gxlist, gylist = estimate_watermark('images/fotolia_processed')
File "C:\Users\prince.bhatia\Desktop\automatic-watermark-detection-master\automatic-watermark-detection-master\src\estimate_watermark.py", line 41, in estimate_watermark
Wm_x = np.median(np.array(gradx), axis=0)
File "C:\Users\prince.bhatia\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\lib\function_base.py", line 4102, in median
overwrite_input=overwrite_input)
File "C:\Users\prince.bhatia\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\lib\function_base.py", line 3997, in _ureduce
axis = _nx.normalize_axis_tuple(axis, nd)
File "C:\Users\prince.bhatia\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\core\numeric.py", line 1536, in normalize_axis_tuple
axis = tuple(normalize_axis_index(ax, ndim, argname) for ax in axis)
File "C:\Users\prince.bhatia\AppData\Local\Continuum\anaconda3\lib\site-packages\numpy\core\numeric.py", line 1536, in <genexpr>
axis = tuple(normalize_axis_index(ax, ndim, argname) for ax in axis)
AxisError: axis 0 is out of bounds for array of dimension 0
答案 0 :(得分：2)
因此错误在第一个函数中。其余所有代码都不相关。
正在接听median的{{1}}。在python3中，不会评估map；您需要map(...)。
我的猜测是list(map(...))确实median会产生0d数组对象dtype并因此产生错误。
我想我以前见过这种错误。
np.array(map....)显示？