I need to make a mask of hexagonal packed disks. The code below does the job, but I don't feel like its efficient. I'm learning python as well so I'd love to get some expert advice on how to do this more computationally efficient.
r = 0.01
X, Y = np.mgrid[0:1:1000j, 0:1:1000j]
mask = np.full(X.shape, False)
px, py = np.mgrid[r : 1 : 2 * r * np.sqrt(3), r : 1 + r + np.finfo(float).eps: 2 * r]
px = np.vstack((px, px + r * np.sqrt(3)))
py = np.vstack((py, py - r))
fig, ax = plt.subplots(figsize= (12, 12), dpi=50)
img = ax.imshow(mask * 1, cmap = 'gray', vmin = 0, vmax = 1, extent = [0, 1, 0, 1])
for i, _ in np.ndenumerate(px): #is this loop dumb and inefficient?
C = (X - px[i]) ** 2 + (Y - py[i]) ** 2
mask = mask | (C < r ** 2)
img.set_data(mask * 1)
ax.set_aspect(1)
In particular, is there a way to vectorize the for loop?
Thanks
It may be efficient to create a single tile of the pattern, and then repeat it horizontally and vertically as needed:
Create a tile:
import numpy as np
import matplotlib.pyplot as plt
r = 0.01
sq3 = np.sqrt(3)
samples = 1000
X, Y = np.mgrid[1:(1 + 2 * sq3):int(sq3 * samples) * 1j, 0:2:samples * 1j]
XY = np.c_[X.flatten(), Y.flatten()]
# coordinates of centers of disks; suffices to take disks of radius 1 here
p = np.array([[1, 1], [(1 + sq3), 0], [(1 + sq3), 2], [(1 + 2 * sq3), 1]])
# compute the distance from each point of XY to each disk center
dist = (XY**2).sum(axis=1).reshape(-1, 1) + (p**2).sum(axis=1) - 2 * (XY # p.T)
# mask points outside the disks
tile = (np.min(dist, axis=1) < 1).reshape(X.shape)
fig, ax = plt.subplots(figsize=(5, 5))
ax.set_aspect(1)
plt.imshow(tile, extent=[0, 2 * r, r, (1 + 2 * sq3) * r]);
It gives:
Repeat the tile:
# the number of times to repeat the tile in the horizontal and vertical directions
h, w = 20, 30
# donwsample the tile as needed and replicate
sample_rate = 10
mask = np.tile(tile[::sample_rate, ::sample_rate], (h, w))
fig, ax = plt.subplots(figsize=(8, 8))
ax.set_aspect(1)
ax.imshow(mask, extent=[0, 2 * r * w, 0, 2 * sq3 * r * h]);
This gives:
Related
I'm trying to implement notch-reject filtering in python for an assignment. I have tried using the notch reject filter formula from Rafael Gonzales book and all I got was a edge detected image. Then I tried ideal notch rejecting and here are the results:
Input image--Output of my program -- Expected output
Here is my code:
import cv2
import numpy as np
import matplotlib.pyplot as plt
def notch_reject_filter(shape, d0=9, u_k=0, v_k=0):
P, Q = shape
# Initialize filter with zeros
H = np.zeros((P, Q))
# Traverse through filter
for u in range(0, P):
for v in range(0, Q):
# Get euclidean distance from point D(u,v) to the center
D_uv = np.sqrt((u - P / 2 + u_k) ** 2 + (v - Q / 2 + v_k) ** 2)
D_muv = np.sqrt((u - P / 2 - u_k) ** 2 + (v - Q / 2 - v_k) ** 2)
if D_uv <= d0 or D_muv <= d0:
H[u, v] = 0.0
else:
H[u, v] = 1.0
return H
img = cv2.imread('input.png', 0)
img_shape = img.shape
original = np.fft.fft2(img)
center = np.fft.fftshift(original)
NotchRejectCenter = center * notch_reject_filter(img_shape, 32, 50, 50)
NotchReject = np.fft.ifftshift(NotchRejectCenter)
inverse_NotchReject = np.fft.ifft2(NotchReject) # Compute the inverse DFT of the result
plot_image = np.concatenate((img, np.abs(inverse_NotchReject)),axis=1)
plt.imshow(plot_image, "gray"), plt.title("Notch Reject Filter")
plt.show()
all I got was a edge detected image because your implementation was High pass filter which is a black circle in the middle, and that works as Edge detector.
Then I tried ideal notch rejecting This is correct if you applied that correctly.
The main concept is to filter the undesired Noise in the frequency domain, the noise can be seen as white spots, and your role is to suppress that white spots by multiplying them by black circles in frequency domain(known as filtering).
to improve this result add more notch filters (H5, H6, ...) to suppress the noise.
import cv2
import numpy as np
import matplotlib.pyplot as plt
#------------------------------------------------------
def notch_reject_filter(shape, d0=9, u_k=0, v_k=0):
P, Q = shape
# Initialize filter with zeros
H = np.zeros((P, Q))
# Traverse through filter
for u in range(0, P):
for v in range(0, Q):
# Get euclidean distance from point D(u,v) to the center
D_uv = np.sqrt((u - P / 2 + u_k) ** 2 + (v - Q / 2 + v_k) ** 2)
D_muv = np.sqrt((u - P / 2 - u_k) ** 2 + (v - Q / 2 - v_k) ** 2)
if D_uv <= d0 or D_muv <= d0:
H[u, v] = 0.0
else:
H[u, v] = 1.0
return H
#-----------------------------------------------------
img = cv2.imread('input.png', 0)
f = np.fft.fft2(img)
fshift = np.fft.fftshift(f)
phase_spectrumR = np.angle(fshift)
magnitude_spectrum = 20*np.log(np.abs(fshift))
img_shape = img.shape
H1 = notch_reject_filter(img_shape, 4, 38, 30)
H2 = notch_reject_filter(img_shape, 4, -42, 27)
H3 = notch_reject_filter(img_shape, 2, 80, 30)
H4 = notch_reject_filter(img_shape, 2, -82, 28)
NotchFilter = H1*H2*H3*H4
NotchRejectCenter = fshift * NotchFilter
NotchReject = np.fft.ifftshift(NotchRejectCenter)
inverse_NotchReject = np.fft.ifft2(NotchReject) # Compute the inverse DFT of the result
Result = np.abs(inverse_NotchReject)
plt.subplot(222)
plt.imshow(img, cmap='gray')
plt.title('Original')
plt.subplot(221)
plt.imshow(magnitude_spectrum, cmap='gray')
plt.title('magnitude spectrum')
plt.subplot(223)
plt.imshow(magnitude_spectrum*NotchFilter, "gray")
plt.title("Notch Reject Filter")
plt.subplot(224)
plt.imshow(Result, "gray")
plt.title("Result")
plt.show()
Drive by comment, using the for-loop for the notch filter generation is very slow. That operation can be optimized
def notch_reject_filter_vec(shape: tuple[int, int], d0: int, u_k: int, v_k: int):
(M, N) = shape
H_0_u = np.repeat(np.arange(M), N).reshape((M, N))
H_0_v = np.repeat(np.arange(N), M).reshape((N, M)).transpose()
D_uv = np.sqrt((H_0_u - M / 2 + u_k) ** 2 + (H_0_v - N / 2 + v_k) ** 2)
D_muv = np.sqrt((H_0_u - M / 2 - u_k) ** 2 + (H_0_v - N / 2 - v_k) ** 2)
selector_1 = D_uv <= d0
selector_2 = D_muv <= d0
selector = np.logical_or(selector_1, selector_2)
H = np.ones((M, N))
H[selector] = 0
return H
I am going to solve 1D FitzHugh Nagoma with homogeneous Neumann boundary conditions.
How to plot U and V separately. Here a1=2, a0=-0.03 , ep= 0.01 Du= 1, Dv=4 I was confused by plotting the figure
U_t=Du U_xx +U -U^3 - V
V_t=Dv V_xx + ep(U-a1V - a0)
import numpy as np
import matplotlib.pyplot as plt
#matplotlib inline
Du = 0.001
Dv = 0.005
tau = 1
k = -.00
ep = 0.01
a1 = 2
a0 = -0.03
L = 2
N= 10
x = np.linspace(0, L, N+1)
dx = x[1]-x[0]
T = 45 # total time
dt = .01 # time step
size = N
n = int(T / dt) # number of iterations
U = np.random.rand(size)
V = np.random.rand(size)
def laplacian(Z):
Ztop = Z[0:-2]
Zbottom = Z[2:]
Zcenter = Z[1:-1]
return (Ztop + Zbottom -
2 * Zcenter) / dx**2
def show_patterns(U, ax=None):
ax.imshow(U, cmap=plt.cm.copper,
interpolation='bilinear',
extent=[-1, 1])
ax.set_axis_off()
fig, axes = plt.subplots(3, 3, figsize=(16, 16))
step_plot = n // 9
# We simulate the PDE with the finite difference
# method.
for i in range(n):
# We compute the Laplacian of u and v.
deltaU = laplacian(U)
deltaV = laplacian(V)
# We take the values of u and v inside the grid.
Uc = U[1:-1]
Vc = V[1:-1]
# We update the variables.
U[1:-1], V[1:-1] = \
Uc + dt * (Du * deltaU + Uc - Uc**3 - Vc),\
Vc + dt * (Dv * deltaV + ep*(Uc - a1*Vc - a0)) / tau
# Neumann conditions: derivatives at the edges
# are null.
for Z in (U, V):
Z[0] = Z[1]
Z[-1] = Z[-2]
# Z[:, 0] = Z[:, 1]
# Z[:, -1] = Z[:, -2]
# We plot the state of the system at
# 9 different times.
fig, ax = plt.subplots(1, 1, figsize=(8, 8))
show_patterns(U,ax=None)
I got an error 'NoneType' object has no attribute 'imshow'
and could not solve it
This line
show_patterns(U,ax=None)
passed in a None to the ax parameter.
I don't know what ax should be but it needs to be properly initialised.
I would like to find a way to translate and add the bottom graph (from y = -20 to 0) onto the above graph (from y = 0-20) so that the final domain is between y = 0 to 20:
However, I am finding problems doing it, as the graph I used to draw the bottom graph (R12) has already a negative input, thus it will not show up on the positive y-axis. Here is my code:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
N = 1001
lower = 0
upper = 20
u = np.linspace(lower, upper, N)
t1a, t3a = np.meshgrid(u,-u)
t1, t3 = np.meshgrid(u,u)
omega = 10
delta = 5
tau = 2
mu = 1
t2 = 0.1
def g(t):
return delta * (omega ** 2) * (tau ** 2) * ((np.e ** (-t/tau))+(t/tau)-1)
R12 = 1 * (mu ** 4) * (np.e **(-1j * omega * (t3a-(t1a)))) * (np.e ** (-g(t1a)+g(t2)-g(t3a)-g(t1a+t2)-g(t2+t3a)+g(t1a+t2+t3a)))
R45 = 1 * (mu ** 4) * (np.e ** (-1j * omega * (t3+t1))) * (np.e ** (-g(t1)-g(t2)-g(t3)+g(t1+t2)+g(t2+t3)-g(t1+t2+t3)))
R12_fft = np.fft.fftshift((np.fft.fft2((R12)))) / np.sqrt(len(R12))
R45_fft = np.fft.fftshift((np.fft.fft2((R45)))) / np.sqrt(len(R45))
R_pure = (R12_fft + R45_fft)
plt.contourf(t1a,t3a,R12_fft, cmap = 'seismic')
plt.contourf(t1,t3,R45_fft, cmap = 'seismic')
plt.xlabel('${\omega}_{3}$', fontsize = 24)
plt.ylabel('${\omega}_{1}$', fontsize = 24)
plt.xlim(-20, 20)
plt.ylim(-20, 20)
plt.gca().set_aspect('equal', adjustable='box')
plt.colorbar()
plt.show()
As an example, if I try to do the simple adding:
plt.contourf(t1,t3,R_pure, cmap = 'seismic')
it basically gives me back the same shape of the graph. What I would like instead is superimposing the bottom graph onto the top and adding the output together. Is there any way I can achieve this? Thank you!
I feel like it's a bit too dumb, so it's probably wrong.
R_pure = (R12_fft + abs(R45_fft))
I'm trying to produce 2D perlin noise using numpy, but instead of something smooth I get this :
my broken perlin noise, with ugly squares everywhere
For sure, I'm mixing up my dimensions somewhere, probably when I combine the four gradients ... But I can't find it and my brain is melting right now. Anyone can help me pinpoint the problem ?
Anyway, here is the code:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
def perlin(x,y,seed=0):
# permutation table
np.random.seed(seed)
p = np.arange(256,dtype=int)
np.random.shuffle(p)
p = np.stack([p,p]).flatten()
# coordinates of the first corner
xi = x.astype(int)
yi = y.astype(int)
# internal coordinates
xf = x - xi
yf = y - yi
# fade factors
u = fade(xf)
v = fade(yf)
# noise components
n00 = gradient(p[p[xi]+yi],xf,yf)
n01 = gradient(p[p[xi]+yi+1],xf,yf-1)
n11 = gradient(p[p[xi+1]+yi+1],xf-1,yf-1)
n10 = gradient(p[p[xi+1]+yi],xf-1,yf)
# combine noises
x1 = lerp(n00,n10,u)
x2 = lerp(n10,n11,u)
return lerp(x2,x1,v)
def lerp(a,b,x):
"linear interpolation"
return a + x * (b-a)
def fade(t):
"6t^5 - 15t^4 + 10t^3"
return 6 * t**5 - 15 * t**4 + 10 * t**3
def gradient(h,x,y):
"grad converts h to the right gradient vector and return the dot product with (x,y)"
vectors = np.array([[0,1],[0,-1],[1,0],[-1,0]])
g = vectors[h%4]
return g[:,:,0] * x + g[:,:,1] * y
lin = np.linspace(0,5,100,endpoint=False)
y,x = np.meshgrid(lin,lin)
plt.imshow(perlin(x,y,seed=0))
Thanks to Paul Panzer and a good night of sleep it works now ...
import numpy as np
import matplotlib.pyplot as plt
def perlin(x, y, seed=0):
# permutation table
np.random.seed(seed)
p = np.arange(256, dtype=int)
np.random.shuffle(p)
p = np.stack([p, p]).flatten()
# coordinates of the top-left
xi, yi = x.astype(int), y.astype(int)
# internal coordinates
xf, yf = x - xi, y - yi
# fade factors
u, v = fade(xf), fade(yf)
# noise components
n00 = gradient(p[p[xi] + yi], xf, yf)
n01 = gradient(p[p[xi] + yi + 1], xf, yf - 1)
n11 = gradient(p[p[xi + 1] + yi + 1], xf - 1, yf - 1)
n10 = gradient(p[p[xi + 1] + yi], xf - 1, yf)
# combine noises
x1 = lerp(n00, n10, u)
x2 = lerp(n01, n11, u) # FIX1: I was using n10 instead of n01
return lerp(x1, x2, v) # FIX2: I also had to reverse x1 and x2 here
def lerp(a, b, x):
"linear interpolation"
return a + x * (b - a)
def fade(t):
"6t^5 - 15t^4 + 10t^3"
return 6 * t**5 - 15 * t**4 + 10 * t**3
def gradient(h, x, y):
"grad converts h to the right gradient vector and return the dot product with (x,y)"
vectors = np.array([[0, 1], [0, -1], [1, 0], [-1, 0]])
g = vectors[h % 4]
return g[:, :, 0] * x + g[:, :, 1] * y
lin = np.linspace(0, 5, 100, endpoint=False)
x, y = np.meshgrid(lin, lin) # FIX3: I thought I had to invert x and y here but it was a mistake
plt.imshow(perlin(x, y, seed=2), origin='upper')
Update
i try to follow this tutorial
but i dont know how can export to new tiff images slope/aspect with GDAL ?
the full code :
from __future__ import division
from osgeo import gdal
from matplotlib.colors import ListedColormap
from matplotlib import colors
import sys
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
import math
filename = 'dem.tif'
def getResolution(rasterfn):
raster = gdal.Open(rasterfn)
geotransform = raster.GetGeoTransform()
res = {"east-west": abs(geotransform[1]),
"north-south": abs(geotransform[5])}
return res
def raster2array(rasterfn):
raster = gdal.Open(rasterfn)
band = raster.GetRasterBand(1)
return band.ReadAsArray()
def getNoDataValue(rasterfn):
raster = gdal.Open(rasterfn)
band = raster.GetRasterBand(1)
return band.GetNoDataValue()
data_array = raster2array(filename)
nodataval = getNoDataValue(filename)
resolution = getResolution(filename)
print(resolution)
print(nodataval)
print(type(data_array))
print(data_array.shape)
num_rows = data_array.shape[0]
num_cols = data_array.shape[1]
slope_array = np.ones_like(data_array) * nodataval
aspect_array = np.ones_like(data_array) * nodataval
for i in range(1, num_rows - 1):
for j in range(1, num_cols - 1):
a = data_array[i - 1][j - 1]
b = data_array[i - 1][j]
c = data_array[i - 1][j + 1]
d = data_array[i][j - 1]
e = data_array[i][j]
f = data_array[i][j + 1]
g = data_array[i + 1][j - 1]
h = data_array[i + 1][j]
q = data_array[i + 1][j + 1]
vals = [a, b, c, d, e, f, g, h, q]
if nodataval in vals:
all_present = False
else:
all_present = True
if all_present == True:
dz_dx = (c + (2 * f) + q - a - (2 * d) - g) / (8 * resolution['east-west'])
dz_dy = (g + (2 * h) + q - a - (2 * b) - c) / (8 * resolution['north-south'])
dz_dx_sq = math.pow(dz_dx, 2)
dz_dy_sq = math.pow(dz_dy, 2)
rise_run = math.sqrt(dz_dx_sq + dz_dy_sq)
slope_array[i][j] = math.atan(rise_run) * 57.29578
aspect = math.atan2(dz_dy, (-1 * dz_dx)) * 57.29578
if aspect < 0:
aspect_array[i][j] = 90 - aspect
elif aspect > 90:
aspect_array[i][j] = 360 - aspect + 90
else:
aspect_array[i][j] = 90 - aspect
hist, bins = np.histogram(slope_array, bins=100, range=(0, np.amax(slope_array)))
width = 0.7 * (bins[1] - bins[0])
center = (bins[:-1] + bins[1:]) / 2
plt.bar(center, hist, align='center', width=width)
plt.xlabel('Slope (degrees)')
plt.ylabel('Frequency')
plt.show()
color_map = ListedColormap(['white', 'darkgreen', 'green', 'limegreen', 'lime',
'greenyellow', 'yellow', 'gold',
'orange', 'orangered', 'red'])
# range begins at negative value so that missing values are white
color_bounds = list(range(-3, math.ceil(np.amax(slope_array)), 1))
color_norm = colors.BoundaryNorm(color_bounds, color_map.N)
#Create the plot and colorbar
img = plt.imshow(slope_array, cmap = color_map, norm = color_norm)
cbar = plt.colorbar(img, cmap = color_map, norm = color_norm,
boundaries = color_bounds, ticks = color_bounds)
#Show the visualization
plt.axis('off')
plt.title("Slope (degrees)")
plt.show()
plt.close()
first trial to export i follow this tutorial,but the slope in not correct
have lower degrees from the original(using gis program)
import gdal, ogr, os, osr
import numpy as np
def array2raster(newRasterfn,rasterOrigin,pixelWidth,pixelHeight,array):
cols = array.shape[1]
rows = array.shape[0]
originX = rasterOrigin[0]
originY = rasterOrigin[1]
driver = gdal.GetDriverByName('GTiff')
outRaster = driver.Create(newRasterfn, cols, rows, 1, gdal.GDT_Byte)
outRaster.SetGeoTransform((originX, pixelWidth, 0, originY, 0, pixelHeight))
outband = outRaster.GetRasterBand(1)
outband.WriteArray(array)
outRasterSRS = osr.SpatialReference()
outRasterSRS.ImportFromEPSG(4326)
outRaster.SetProjection(outRasterSRS.ExportToWkt())
outband.FlushCache()
def main(newRasterfn,rasterOrigin,pixelWidth,pixelHeight,array):
reversed_arr = slope_array # reverse array so the tif looks like the array
array2raster(newRasterfn,rasterOrigin,pixelWidth,pixelHeight,reversed_arr) # convert array to raster
if __name__ == "__main__":
rasterOrigin = (-123.25745,45.43013)
pixelWidth = 10
pixelHeight = 10
newRasterfn = 'test.tif'
array = slope_array
main(newRasterfn,rasterOrigin,pixelWidth,pixelHeight,array)
second trial i follow this quest but i dont take some export
def array_to_raster(slope_array):
"""Array > Raster
Save a raster from a C order array.
:param array: ndarray
"""
dst_filename = 'xxx.tiff'
x_pixels = num_rows # number of pixels in x
y_pixels = num_cols # number of pixels in y
driver = gdal.GetDriverByName('GTiff')
dataset = driver.Create(
dst_filename,
x_pixels,
y_pixels,
1,
gdal.GDT_Float32, )
dataset.GetRasterBand(1).WriteArray(array)
dataset.FlushCache() # Write to disk.
return dataset, dataset.GetRasterBand(1)
any idea
The error states exactly what the problem is. You are trying to multiply an int type by a NoneType (None). The most likely case is that nodataval is None, which would occur because the NoDataValue for filename's first raster band is not defined. Your print(nodataval) command should demonstrate that. Remember, printing None doesn't appear as the string 'None'. It appears as a blank or no character.
Your edits show that nodataval is None.