Comparing noises on images, are there clever distances? #5

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2024-03-04T15:26:42+01:00

Benjamin_Loison commented

2024-03-04 15:26:42 +01:00

Other than L1 or L2? Even among both what distance is the most interesting one.

from PIL import Image
from collections import Counter
import matplotlib.pyplot as plt

def openImage(filePath):
    return Image.open(filePath).convert('L')

im1 = openImage('Wiener_Filtering-001.png')
im1Pixels = im1.load()

im2 = openImage('Wiener_Filtering-002.png')
im2Pixels = im2.load()

differences = []

for m in range(im1.size[0]):
    for n in range(im1.size[1]):
        difference = im1Pixels[m, n] - im2Pixels[m, n]
        differences += [difference]

differencesCounter = Counter(differences)
X = range(min(differencesCounter), max(differencesCounter) + 1)
Y = [differencesCounter.get(x, 0) for x in X]

# Shows a Gaussian centered in 0.
# How to deduce common parameter used to generate this Gaussian? Can verify with the source code in #10 or if the value looks like an integer.
plt.title('Number of occurrences per difference value')
plt.xlabel('Difference value')
plt.ylabel('Number of occurrences')
plt.plot(X, Y)
plt.show()

# Identical result as `rmsdiff`.
print((sum([(difference ** 2) * differencesCounter[difference] for difference in differencesCounter]) / (im1.size[0] * im1.size[1])) ** 0.5)

Related to #10.

Other than L1 or L2? Even among both what distance is the most interesting one. ```py from PIL import Image from collections import Counter import matplotlib.pyplot as plt def openImage(filePath): return Image.open(filePath).convert('L') im1 = openImage('Wiener_Filtering-001.png') im1Pixels = im1.load() im2 = openImage('Wiener_Filtering-002.png') im2Pixels = im2.load() differences = [] for m in range(im1.size[0]): for n in range(im1.size[1]): difference = im1Pixels[m, n] - im2Pixels[m, n] differences += [difference] differencesCounter = Counter(differences) X = range(min(differencesCounter), max(differencesCounter) + 1) Y = [differencesCounter.get(x, 0) for x in X] # Shows a Gaussian centered in 0. # How to deduce common parameter used to generate this Gaussian? Can verify with the source code in #10 or if the value looks like an integer. plt.title('Number of occurrences per difference value') plt.xlabel('Difference value') plt.ylabel('Number of occurrences') plt.plot(X, Y) plt.show() # Identical result as `rmsdiff`. print((sum([(difference ** 2) * differencesCounter[difference] for difference in differencesCounter]) / (im1.size[0] * im1.size[1])) ** 0.5) ``` ![Figure_1](/attachments/3b96fb18-8e1c-4388-a89d-453cb669c069) Related to #10.

Figure_1.svg

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Benjamin_Loison added the Context-Adaptive Interpolator label 2024-03-21 10:08:01 +01:00

Benjamin_Loison referenced this issue from Benjamin_Loison/Pillow

2024-03-21 11:24:50 +01:00

`PixelAccess` - `PixelAccess` to make an image difference #2

Benjamin_Loison referenced this issue

2024-03-21 11:26:59 +01:00

Wiener filter #10

Benjamin_Loison added the medium label 2024-03-26 02:17:31 +01:00

Benjamin_Loison commented

2024-04-02 10:58:54 +02:00