I am trying to modify a code written by a software developer (Kyle Dickerson) and have written it up like this:
So I have this code:
from __future__ import division
## Kyle Dickerson
## kyle.dickerson#gmail.com
## Jan 15, 2008
##
## Self-organizing map using scipy
## This code is licensed and released under the GNU GPL
## This code uses a square grid rather than hexagonal grid, as scipy allows for fast square grid computation.
## I designed sompy for speed, so attempting to read the code may not be very intuitive.
## If you're trying to learn how SOMs work, I would suggest starting with Paras Chopras SOMPython code:
## http://www.paraschopra.com/sourcecode/SOM/index.php
## It has a more intuitive structure for those unfamiliar with scipy, however it is much slower.
## If you do use this code for something, please let me know, I'd like to know if has been useful to anyone.
from random import *
from math import *
import sys
import scipy
import numpy
class SOM:
def __init__(self, height=4, width=4, FV_size=3, learning_rate=0.005):
self.height = height
self.width = width
self.FV_size = FV_size
self.radius = (height+width)/3
self.learning_rate = learning_rate
self.nodes = scipy.array([[ [random()*255 for
i in range(FV_size)] for x in range(width)] for y in range(height)])
self.nodes = scipy.array([[1,2,3],[4,5,6],[4,5,6],
[4,5,6],[4,5,6], [4,5,6],[4,5,6],[4,5,6],[4,5,6],
[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6]])
print "SOM",self.nodes
def train(self, iterations=1000, train_vector=[[]]):
for t in range(len(train_vector)):
train_vector[t] = scipy.array(train_vector[t])
print "training",train_vector[t],t
time_constant = iterations/log(self.radius)
delta_nodes = scipy.array([[[0 for i in range(self.FV_size)]
for x in range(self.width)] for y in range(self.height)])
for i in range(1, iterations+1):
delta_nodes.fill(0)
radius_decaying=self.radius*exp(-1.0*i/time_constant)
rad_div_val = 2 * radius_decaying * i
learning_rate_decaying=self.learning_rate*exp(-1.0*i/time_constant)
sys.stdout.write("\rTraining Iteration:
" + str(i) + "/" + str(iterations))
for j in range(len(train_vector)):
best = self.best_match(train_vector[j])
for loc in self.find_neighborhood(best, radius_decaying):
influence = exp( (-1.0 * (loc[2]**2)) / rad_div_val)
inf_lrd = influence*learning_rate_decaying
delta_nodes[loc[0],loc[1]] += inf_lrd*
(train_vector[j]- self.nodes[loc[0],loc[1]])
self.nodes += delta_nodes
sys.stdout.write("\n")
# Returns a list of points which live within 'dist' of 'pt'
# Uses the Chessboard distance
# pt is (row, column)
def find_neighborhood(self, pt, dist):
min_y = max(int(pt[0] - dist), 0)
max_y = min(int(pt[0] + dist), self.height)
min_x = max(int(pt[1] - dist), 0)
max_x = min(int(pt[1] + dist), self.width)
neighbors = []
for y in range(min_y, max_y):
for x in range(min_x, max_x):
dist = abs(y-pt[0]) + abs(x-pt[1])
neighbors.append((y,x,dist))
return neighbors
# Returns location of best match, uses Euclidean distance
# target_FV is a scipy array
def best_match(self, target_FV):
loc = scipy.argmin((((self.nodes - target_FV)**2).sum(axis=2))**0.5)
r = 0
while loc > self.width:
loc -= self.width
r += 1
c = loc
return (r, c)
# returns the Euclidean distance between two Feature Vectors
# FV_1, FV_2 are scipy arrays
def FV_distance(self, FV_1, FV_2):
return (sum((FV_1 - FV_2)**2))**0.5
if __name__ == "__main__":
print "Initialization..."
colors = [ [0, 0, 0], [0, 0, 255], [0, 255, 0],
[0, 255, 255], [255, 0, 0], [255, 0, 255],
[255, 255, 0], [255, 255, 255]]
width = 32
height = 32
color_som = SOM(width,height,3,0.05)
print "Training colors..."
color_som.train(1000, colors)
try:
from PIL import Image
print "Saving Image: sompy_test_colors.png..."
img = Image.new("RGB", (width, height))
for r in range(height):
for c in range(width):
img.putpixel((c,r),(int(color_som.nodes[r,c,0]),
int(color_som.nodes[r,c,1]), int(color_som.nodes[r,c,2])))
print "color nodes",color_som.nodes
img = img.resize((width*10, height*10),Image.NEAREST)
img.save("sompy_test_colors.png")
except:
print "Error saving the image, do you have PIL (Python Imaging Library) installed?"
but when I try to go from
self.nodes = scipy.array([[ [random()*255
for i in range(FV_size)] for x in range(width)]
for y in range(height)])
which was in the original code to something like this:
self.nodes = scipy.array([[1,2,3],[4,5,6],[4,5,6],
[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],
[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6]])
I get the error message:
File "sompy5.py", line 112, in <module>
color_som.train(1000, colors)
File "sompy5.py", line 65, in train
best = self.best_match(train_vector[j])
File "sompy5.py", line 92, in best_match
loc = scipy.argmin((((self.nodes - target_FV)**2).sum(axis=2))**0.5)
File "/usr/lib/python2.7/dist-packages/numpy/core/_methods.py",
line 25, in _sum
out=out, keepdims=keepdims)
ValueError: 'axis' entry is out of bounds
Is there something that has to be done to get the vectors to match up?
This part is a 3-D array (3 square brackets to begin the argument):
self.nodes = scipy.array([[ [random()*255 for
i in range(FV_size)] for x in range(width)] for y in range(height)])
This part is a 2-D array:
self.nodes = scipy.array([[1,2,3],[4,5,6],[4,5,6],
[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],
[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6],[4,5,6]])
So you need to turn self.nodes into the appropriate 3-D array.
EDIT: an example of the required syntax:
self.nodes = scipy.array([[ [1,2,3],[4,5,6]] , [[7,8,9],[10,11,12]]])
print(self.nodes)
>>> array([[[ 1, 2, 3],
[ 4, 5, 6]],
[[ 7, 8, 9],
[10, 11, 12]]])
EDIT 2:
Another option is to build a linear array and then reshape():
myarray = scipy.array([1,2,3,4,5,6,7,8,9,10,11,12])
myarray = myarray.reshape( (2, 2, 3) ) ## 3 numbers for 3 dimensions, but the product must be the same as the number of elements of the original array
print(myarray)
>>> array([[[ 1, 2, 3],
[ 4, 5, 6]],
[[ 7, 8, 9],
[10, 11, 12]]])
Related
So I have 2 surfaces (PolyData in PyVista) one on top of another:
They are shaped a little differently on Z access yet whenever a top one has a Z value on X, Y plane we are sure a-bottom one has the same. So how one can merge two surfaces X, Y aligned into one solid mesh?
What I try:
import numpy as np
import pyvista as pv
import vtk
def extruder(mesh, val_z):
extrude = vtk.vtkLinearExtrusionFilter()
extrude.SetInputData(mesh)
extrude.SetVector(0, 0, val_z)
extrude.Update()
extruded_mesh = pv.wrap(extrude.GetOutput())
return extruded_mesh
# generate two sheets of input data
noise = pv.perlin_noise(2, (0.2, 0.2, 0.2), (0, 0, 0))
bounds_2d = (-10, 10, -10, 10)
dim = (40, 50, 1)
bottom, top = [
pv.sample_function(noise, dim=dim, bounds=bounds_2d + (z, z)).warp_by_scalar()
for z in [-5, 5]
]
bottom = bottom.extract_surface(nonlinear_subdivision=5)
top = top.extract_surface(nonlinear_subdivision=5)
top = extruder(top, -50).triangulate()
bottom = extruder(bottom, 50).triangulate()
intersection = bottom.boolean_cut(top)
#top = top.clip_surface(bottom, invert=False, compute_distance=True)
#top = top.extrude([0, 0, -50]).triangulate()
#bottom = bottom.extrude([0, 0, 50]).triangulate()
#intersection = bottom.boolean_cut(top).triangulate()
p = pv.Plotter()
p.add_mesh(top, cmap="hot", opacity=0.15)
p.add_mesh(bottom, cmap="RdYlBu", opacity=0.15)
p.add_mesh(intersection, cmap="Dark2", opacity=1)
p.show()
What do I get:
What I expected:
only middle to be filled.
So had to do this:
import numpy as np
import pyvista as pv
# generate two sheets of input data
noise = pv.perlin_noise(2, (0.2, 0.2, 0.2), (0, 0, 0))
bounds_2d = (-10, 10, -10, 10)
dim = (40, 50, 1)
bottom, top = [
pv.sample_function(noise, dim=dim, bounds=bounds_2d + (z, z)).warp_by_scalar()
for z in [-5, 5]
]
bottom = bottom.extract_surface()
top = top.extract_surface()
topm = top.extrude([0, 0, -50]).triangulate().clean()
bottomm = bottom.extrude([0, 0, 50]).triangulate().clean()
topm = topm.clip_surface(bottom, invert=False)
bottomm = bottomm.clip_surface(top, invert=True)
intersection = topm.boolean_add(bottomm).triangulate().clean().subdivide(2).clean()
p = pv.Plotter()
#p.add_mesh(topm, cmap="hot", opacity=0.15)
#p.add_mesh(bottomm, cmap="gnuplot2", opacity=0.15)
p.add_mesh(intersection, cmap="Dark2", opacity=1)
p.show()
the resulting mesh is really bad, yet it has desired shape and gets to be computed in usable time:
I would like to know if there is an efficient method to get sub-arrays from a larger numpy array.
What I have is an application of np.where. I iterate 'manually' over x and y as offsets and apply where with a kernel to each rectangle extracted from the larger array with proper dimensions.
But is there a more direct approach in numpy's collection of methods?
import numpy as np
example = np.arange(20).reshape((5, 4))
# e.g. a cross kernel
a_kernel = np.asarray([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
np.where(a_kernel, example[1:4, 1:4], 0)
# returns
# array([[ 0, 6, 0],
# [ 9, 10, 11],
# [ 0, 14, 0]])
def arrays_from_kernel(a, a_kernel):
width, height = a_kernel.shape
y_max, x_max = a.shape
return [np.where(a_kernel, a[y:(y + height), x:(x + width)], 0)
for y in range(y_max - height + 1)
for x in range(x_max - width + 1)]
sub_arrays = arrays_from_kernel(example, a_kernel)
This returns the arrays I need for further processing.
# [array([[0, 1, 0],
# [4, 5, 6],
# [0, 9, 0]]),
# array([[ 0, 2, 0],
# [ 5, 6, 7],
# [ 0, 10, 0]]),
# ...
# array([[ 0, 9, 0],
# [12, 13, 14],
# [ 0, 17, 0]]),
# array([[ 0, 10, 0],
# [13, 14, 15],
# [ 0, 18, 0]])]
The context: similar to 2D convolution I would like to apply a custom function on each of the subarrays (e.g. product of squared numbers).
At the moment, you're manually advancing a sliding window over the data - stride tricks to the rescue! (And no, I didn't just make that up - there's actually a submodule called stride_tricks in numpy!) Instead of manually building windows into the data, and calling np.where() on them, if you had the windows in an array, you could call np.where() just once. Stride tricks allow you to create such an array without even having to copy the data.
Let me explain. Normal slices in numpy create views into the original data instead of copies. This is done by referring to the original data, but changing the strides used to access the data (ie. how much to jump between two elements or two rows, and so on). Stride tricks allow you to modify those strides more freely than just slicing and reshaping does, so you can eg. iterate over the same data more than once, which is useful here.
Let me demonstrate:
import numpy as np
example = np.arange(20).reshape((5, 4))
a_kernel = np.array([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
def sliding_window(data, win_shape, **kwargs):
assert data.ndim == len(win_shape)
shape = tuple(dn - wn + 1 for dn, wn in zip(data.shape, win_shape)) + win_shape
strides = data.strides * 2
return np.lib.stride_tricks.as_strided(data, shape=shape, strides=strides, **kwargs)
def arrays_from_kernel(a, a_kernel):
windows = sliding_window(a, a_kernel.shape)
return np.where(a_kernel, windows, 0)
sub_arrays = arrays_from_kernel(example, a_kernel)
The scipy.ndimage module offers a number of filters -- one of which might meet your needs. If none of those filters do what you want, you could use ndimage.generic_filter
to call a custom function on each subarray. ndimage.generic_filter is not as fast as the other ndimage filters, however.
For example,
import numpy as np
example = np.arange(20).reshape((5, 4))
a_kernel = np.asarray([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
# def arrays_from_kernel(a, a_kernel):
# width, height = a_kernel.shape
# y_max, x_max = a.shape
# return [np.where(a_kernel, a[y:(y + height), x:(x + width)], 0)
# for y in range(y_max - height + 1)
# for x in range(x_max - width + 1)]
# sub_arrays = arrays_from_kernel(example, a_kernel)
# for arr in sub_arrays:
# print(arr)
# print('-'*80)
import scipy.ndimage as ndimage
def func(x):
# reject subarrays that extend beyond the border of the `example` array
if not np.isnan(x).any():
y = np.zeros_like(a_kernel, dtype=example.dtype)
np.put(y, np.flatnonzero(a_kernel), x)
print(y)
# Instead or returning 0, you can perform your desired computation on the subarray here.
# Note that you may not need the 2D array y; often, you only need the values in the 1D array x
return 0
result = ndimage.generic_filter(example, func, footprint=a_kernel, mode='constant', cval=np.nan)
For the particular problem of computing the product of squares for each subarray, you
could convert the product into a sum by taking advantage of the fact that A * B = exp(log(A)+log(B)). This would allow you to express the computation as a normal convolution. Now using ndimage.convolve can improve performance a lot. The amount of the improvement depends on the size of example:
import numpy as np
import scipy.ndimage as ndimage
import perfplot
a_kernel = np.asarray([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
def orig(example, a_kernel=a_kernel):
def arrays_from_kernel(a, a_kernel):
width, height = a_kernel.shape
y_max, x_max = a.shape
return [
np.where(a_kernel, a[y : (y + height), x : (x + width)], 1)
for y in range(y_max - height + 1)
for x in range(x_max - width + 1)
]
return [np.prod(x) ** 2 for x in arrays_from_kernel(example, a_kernel)]
def alt(example, a_kernel=a_kernel):
logged = np.log(example)
result = ndimage.convolve(logged, a_kernel, mode="constant", cval=0)[1:-1, 1:-1]
return (np.exp(result) ** 2).ravel()
def make_example(N):
return np.random.random(size=(N, N))
def check(A, B):
return np.allclose(A, B)
perfplot.show(
setup=make_example,
kernels=[orig, alt],
n_range=[2 ** k for k in range(2, 11)],
logx=True,
logy=True,
xlabel="len(example)",
equality_check=check,
)
I am using a small window to detect Mario which is represented by a red block. However, this red block is composed of 16 by 12 pixels. I want to take the pixel coordinates I found, and convert this to a normal x/y coordinate system based on the window shown in the image: Actual frame which should be 13 by 16 grid (NOT pixels).
So for example, if Mario box is in the upper left corner of screen, the coordinates should be 0,0.
I'm also not sure how to actually make the grid.
The code I'm using is as follows:
import numpy as np
from PIL import Image
class MarioPixels:
def __init__(self):
self.mario = np.array([
[[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0],
[248, 56, 0]
]]
)
self.height = len(self.mario) # specify number of pixels for columns in the frame
self.width = len(self.mario[0]) # specificy number of pixels representing a line in the frame
print(self.mario.shape)
# find difference in R, G and B values between what's in window and what's on the frame
def pixelDiff(self, p1, p2):
return abs(p1[0] - p2[0]), abs(p1[1] - p2[1]), abs(p1[2] - p2[2])
def isMario(self, window, pattern):
total = [0, 0, 0]
count = 0
for line in range(len(pattern)):
lineItem = pattern[line]
sample = window[line]
for pixelIdx in range(len(lineItem)):
count += 1
pixel1 = lineItem[pixelIdx]
pixel2 = sample[pixelIdx]
d1, d2, d3 = self.pixelDiff(pixel1, pixel2)
# print(pixelIdx)
total[0] = total[0] + d1 # sum of difference between all R values found between window and frame
total[1] = total[1] + d2 # sum of difference between all G values found between window and frame
total[2] = total[2] + d3 # sum of difference between all B values found between window and frame
# Mario has a red hat
# if line == 0 and pixelIdx == 4 and pixel2[0] != 248:
# return 1.0
rscore = total[0] / (
count * 255) # divided by count of all possible places the R difference could be calculated
gscore = total[1] / (
count * 255) # divided by count of all possible places the G difference could be calculated
bscore = total[2] / (
count * 255) # divided by count of all possible places the B difference could be calculated
return (
rscore + gscore + bscore) / 3.0 # averaged to find a value between 0 and 1. Num close to 0 means object(mario, pipe, etc.) is there,
# whereas, number close to 1 means object was not found.
def searchForMario(self, step, state, pattern):
height = self.height
width = self.width
x1 = 0
y1 = 0
x2 = width
y2 = height
imageIdx = 0
bestScore = 1.1
bestImage = None
bestx1, bestx2, besty1, besty2 = 0, 0, 0, 0
for y1 in range(0, 240 - height, 8): # steps in range row, jump by 8 rows
y2 = y1 + height
for x1 in range(0, 256 - width, 3): # jump by 3 columns
x2 = x1 + width
window = state[y1:y2, x1:x2, :]
score = self.isMario(window, pattern)
# print(imageIdx, score)
if score < bestScore:
bestScore = score
bestImageIdx = imageIdx
bestImage = Image.fromarray(window)
bestx1, bestx2, besty1, besty2 = x1, x2, y1, y2
imageIdx += 1
bestImage.save('testrgb' + str(step) + '_' + str(bestImageIdx) + '_' + str(bestScore) + '.png')
return bestx1, bestx2, besty1, besty2
It looks like you've got a pixel aspect ratio at play here, so the width and height of each "block" in pixels will be different.
Going by your code, your pixel space is 256x240 pixels, but you say that it actually represents a 13x16 grid. This means that every block in the x-domain is (256/13) or about 20 pixels, and in the y-domain (240/16) 15 pixels. This means that "Mario", at 16x12 pixels occupies less than one complete block. Looking at your image, this seems to be a possibility - bushes and clouds also occupy less than one block.
I suggest you first make sure the 13x16 grid is correct (simply because it doesn't seem to match your pixel size exactly, and because the stride sizes in your ranges imply that blocks might actually be 3x8 pixels). Then, you can try to add the grid on to the pixel image simply by setting the value of every pixel that has an x-co-ordinate exactly divisible by 20 equal to (0,0,0) for a black RGB pixel (and also a y-coordinate exactly divisible by 15 - use modulus operator %). To get the "block" co-ordinates, simply divide the x-co by 20 and the y-co by 15 and round down to the nearest whole number (or use // to do the rounding as part of the division).
I've assumed that your pixel co-ordinates also run from top left (0,0) to bottom right (256, 240).
I am having a problem with package lmfit.minimize minimization procedure. Actually, I could not create a correct objective function for my problem.
Problem definition
My function: yn = a_11*x1**2 + a_12*x2**2 + ... + a_m*xn**2,where xn- unknowns, a_m -
coefficients. n = 1..N, m = 1..M
In my case, N=5 for x1,..,x5 and M=3 for y1, y2, y3.
I need to find the optimum: x1, x2,...,x5 so that it can satisfy the y
My question:
Error: ValueError: operands could not be broadcast together with shapes (3,) (3,5).
Did I create the objective function of my problem properly in Python?
My code:
import numpy as np
from lmfit import Parameters, minimize
def func(x,a):
return np.dot(a, x**2)
def residual(pars, a, y):
vals = pars.valuesdict()
x = vals['x']
model = func(x,a)
return y - model
def main():
# simple one: a(M,N) = a(3,5)
a = np.array([ [ 0, 0, 1, 1, 1 ],
[ 1, 0, 1, 0, 1 ],
[ 0, 1, 0, 1, 0 ] ])
# true values of x
x_true = np.array([10, 13, 5, 8, 40])
# data without noise
y = func(x_true,a)
#************************************
# Apriori x0
x0 = np.array([2, 3, 1, 4, 20])
fit_params = Parameters()
fit_params.add('x', value=x0)
out = minimize(residual, fit_params, args=(a, y))
print out
if __name__ == '__main__':
main()
Directly using scipy.optimize.minimize() the code below solves this problem. Note that with more points yn you will tend to get the same result as x_true, otherwise more than one solution exists. You can minimize the effect of the ill-constrained optimization by adding boundaries (see the bounds parameter used below).
import numpy as np
from scipy.optimize import minimize
def residual(x, a, y):
s = ((y - a.dot(x**2))**2).sum()
return s
def main():
M = 3
N = 5
a = np.random.random((M, N))
x_true = np.array([10, 13, 5, 8, 40])
y = a.dot(x_true**2)
x0 = np.array([2, 3, 1, 4, 20])
bounds = [[0, None] for x in x0]
out = minimize(residual, x0=x0, args=(a, y), method='L-BFGS-B', bounds=bounds)
print(out.x)
If M>=N you could also use scipy.optimize.leastsq for this task:
import numpy as np
from scipy.optimize import leastsq
def residual(x, a, y):
return y - a.dot(x**2)
def main():
M = 5
N = 5
a = np.random.random((M, N))
x_true = np.array([10, 13, 5, 8, 40])
y = a.dot(x_true**2)
x0 = np.array([2, 3, 1, 4, 20])
out = leastsq(residual, x0=x0, args=(a, y))
print(out[0])
I'm trying to apply the Sobel filter on an image to detect edges using scipy. I'm using Python 3.2 (64 bit) and scipy 0.9.0 on Windows 7 Ultimate (64 bit). Currently my code is as follows:
import scipy
from scipy import ndimage
im = scipy.misc.imread('bike.jpg')
processed = ndimage.sobel(im, 0)
scipy.misc.imsave('sobel.jpg', processed)
I don't know what I'm doing wrong, but the processed image does not look anything like what it should. The image, 'bike.jpg' is a greyscale (mode 'L' not 'RGB') image so each pixel has only one value associated with it.
Unfortunately I can't post the images here yet (don't have enough reputation) but I've provided links below:
Original Image (bike.jpg):
http://s2.postimage.org/64q8w613j/bike.jpg
Scipy Filtered (sobel.jpg):
http://s2.postimage.org/64qajpdlb/sobel.jpg
Expected Output:
http://s1.postimage.org/5vexz7kdr/normal_sobel.jpg
I'm obviously going wrong somewhere! Can someone please tell me where. Thanks.
1) Use a higher precision. 2) You are only calculating the approximation of the derivative along the zero axis. The 2D Sobel operator is explained on Wikipedia. Try this code:
import numpy
import scipy
from scipy import ndimage
im = scipy.misc.imread('bike.jpg')
im = im.astype('int32')
dx = ndimage.sobel(im, 0) # horizontal derivative
dy = ndimage.sobel(im, 1) # vertical derivative
mag = numpy.hypot(dx, dy) # magnitude
mag *= 255.0 / numpy.max(mag) # normalize (Q&D)
scipy.misc.imsave('sobel.jpg', mag)
I couldn't comment on cgohlke's answer so I repeated his answer with a corrction. Parameter 0 is used for vertical derivative and 1 for horizontal derivative (first axis of an image array is y/vertical direction - rows, and second axis is x/horizontal direction - columns). Just wanted to warn other users, because I lost 1 hour searching for mistake in the wrong places.
import numpy
import scipy
from scipy import ndimage
im = scipy.misc.imread('bike.jpg')
im = im.astype('int32')
dx = ndimage.sobel(im, 1) # horizontal derivative
dy = ndimage.sobel(im, 0) # vertical derivative
mag = numpy.hypot(dx, dy) # magnitude
mag *= 255.0 / numpy.max(mag) # normalize (Q&D)
scipy.misc.imsave('sobel.jpg', mag)
or you can use :
def sobel_filter(im, k_size):
im = im.astype(np.float)
width, height, c = im.shape
if c > 1:
img = 0.2126 * im[:,:,0] + 0.7152 * im[:,:,1] + 0.0722 * im[:,:,2]
else:
img = im
assert(k_size == 3 or k_size == 5);
if k_size == 3:
kh = np.array([[-1, 0, 1], [-2, 0, 2], [-1, 0, 1]], dtype = np.float)
kv = np.array([[1, 2, 1], [0, 0, 0], [-1, -2, -1]], dtype = np.float)
else:
kh = np.array([[-1, -2, 0, 2, 1],
[-4, -8, 0, 8, 4],
[-6, -12, 0, 12, 6],
[-4, -8, 0, 8, 4],
[-1, -2, 0, 2, 1]], dtype = np.float)
kv = np.array([[1, 4, 6, 4, 1],
[2, 8, 12, 8, 2],
[0, 0, 0, 0, 0],
[-2, -8, -12, -8, -2],
[-1, -4, -6, -4, -1]], dtype = np.float)
gx = signal.convolve2d(img, kh, mode='same', boundary = 'symm', fillvalue=0)
gy = signal.convolve2d(img, kv, mode='same', boundary = 'symm', fillvalue=0)
g = np.sqrt(gx * gx + gy * gy)
g *= 255.0 / np.max(g)
#plt.figure()
#plt.imshow(g, cmap=plt.cm.gray)
return g
for more see here