Extracting watermark from video The 2019 Stack Overflow Developer Survey Results Are In ...
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Extracting watermark from video
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Recording timeback seconds of video based on compared pixels from an imageExtracting Pixels From Image Byte[]Extracting information from filesExtracting lines from a bytearrayExtracting original values from cumulative sum valuesVideo Feed via TCP SocketAnalyzing the color composition of a videoGenerate video thumbnailExtracting specific words from PANDAS dataframeExtracting face features from selfies faster
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I've already wrote some code in python and in nutshell it works like that.
Extracting a few frames from video
Comparing the similarity or equality of pixel color (from two first image)
Saving to the new image
Comparing new image (conjunction of two first) and the next image etc
Is there any better and more efficient way to do that ? Or do you see any improvement which i can do in my code ?
I am asking because the results aren't good :(
Sorry for my language English I am not native
import sys
import os
import numpy as np
from PIL import Image, ImageDraw
def main(obr1,obr2):
img1= Image.open("%s" %(obr1))
img2= Image.open("%s" %(obr2))
im1 = img1.convert("RGBA")
im2 = img2.convert("RGBA")
pix1 = im1.load()
pix2 = im2.load()
im = Image.new("RGBA", (im1.width, im1.height), (0, 0, 0, 0))
draw = ImageDraw.Draw(im)
x = 0
y = 0
while y != im1.height-1 or x != im1.width-1:
if pix1[x,y] == pix2[x,y]:
draw.point((x,y),fill=pix1[x,y])
else:
p1 = np.array([(pix1[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
p2 = np.array([(pix2[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
squared_dist = np.sum(p1**2 + p2**2, axis=0)
dist = np.sqrt(squared_dist)
if dist < 200 and pix1[x,y] !=(0,0,0,0) and pix2[x,y] != (0,0,0,0):
color = (round(pix1[x,y][0]+pix2[x,y][0]/2), round(pix1[x,y][1]+pix2[x,y][1]/2), round(pix1[x,y][2]+pix2[x,y][2]/2), round(pix1[x,y][3]+pix2[x,y][3]/2))
#color=pix1[x,y]
draw.point((x,y),fill=color)
else:
draw.point((x,y),fill=(0,0,0,0))
if x == im1.width-1:
x=0
y=y+1
else:
x=x+1
im.save('test%s.png' %(z), 'PNG')
print("Zapisano obraz test%s.png" %(z))
imglist = sys.argv[1:]
z=0
while imglist != []:
exists = os.path.isfile("./test%s.png" % (z-1))
if exists:
obr1="test%s.png" % (z-1)
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
else:
obr1=imglist.pop()
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
python beginner image
asked 10 mins ago
FejorFejor
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I've already wrote some code in python and in nutshell it works like that.
Extracting a few frames from video
Comparing the similarity or equality of pixel color (from two first image)
Saving to the new image
Comparing new image (conjunction of two first) and the next image etc
Is there any better and more efficient way to do that ? Or do you see any improvement which i can do in my code ?
I am asking because the results aren't good :(
Sorry for my language English I am not native
import sys
import os
import numpy as np
from PIL import Image, ImageDraw
def main(obr1,obr2):
img1= Image.open("%s" %(obr1))
img2= Image.open("%s" %(obr2))
im1 = img1.convert("RGBA")
im2 = img2.convert("RGBA")
pix1 = im1.load()
pix2 = im2.load()
im = Image.new("RGBA", (im1.width, im1.height), (0, 0, 0, 0))
draw = ImageDraw.Draw(im)
x = 0
y = 0
while y != im1.height-1 or x != im1.width-1:
if pix1[x,y] == pix2[x,y]:
draw.point((x,y),fill=pix1[x,y])
else:
p1 = np.array([(pix1[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
p2 = np.array([(pix2[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
squared_dist = np.sum(p1**2 + p2**2, axis=0)
dist = np.sqrt(squared_dist)
if dist < 200 and pix1[x,y] !=(0,0,0,0) and pix2[x,y] != (0,0,0,0):
color = (round(pix1[x,y][0]+pix2[x,y][0]/2), round(pix1[x,y][1]+pix2[x,y][1]/2), round(pix1[x,y][2]+pix2[x,y][2]/2), round(pix1[x,y][3]+pix2[x,y][3]/2))
#color=pix1[x,y]
draw.point((x,y),fill=color)
else:
draw.point((x,y),fill=(0,0,0,0))
if x == im1.width-1:
x=0
y=y+1
else:
x=x+1
im.save('test%s.png' %(z), 'PNG')
print("Zapisano obraz test%s.png" %(z))
imglist = sys.argv[1:]
z=0
while imglist != []:
exists = os.path.isfile("./test%s.png" % (z-1))
if exists:
obr1="test%s.png" % (z-1)
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
else:
obr1=imglist.pop()
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
python beginner image
asked 10 mins ago
FejorFejor
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I've already wrote some code in python and in nutshell it works like that.
Extracting a few frames from video
Comparing the similarity or equality of pixel color (from two first image)
Saving to the new image
Comparing new image (conjunction of two first) and the next image etc
Is there any better and more efficient way to do that ? Or do you see any improvement which i can do in my code ?
I am asking because the results aren't good :(
Sorry for my language English I am not native
import sys
import os
import numpy as np
from PIL import Image, ImageDraw
def main(obr1,obr2):
img1= Image.open("%s" %(obr1))
img2= Image.open("%s" %(obr2))
im1 = img1.convert("RGBA")
im2 = img2.convert("RGBA")
pix1 = im1.load()
pix2 = im2.load()
im = Image.new("RGBA", (im1.width, im1.height), (0, 0, 0, 0))
draw = ImageDraw.Draw(im)
x = 0
y = 0
while y != im1.height-1 or x != im1.width-1:
if pix1[x,y] == pix2[x,y]:
draw.point((x,y),fill=pix1[x,y])
else:
p1 = np.array([(pix1[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
p2 = np.array([(pix2[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
squared_dist = np.sum(p1**2 + p2**2, axis=0)
dist = np.sqrt(squared_dist)
if dist < 200 and pix1[x,y] !=(0,0,0,0) and pix2[x,y] != (0,0,0,0):
color = (round(pix1[x,y][0]+pix2[x,y][0]/2), round(pix1[x,y][1]+pix2[x,y][1]/2), round(pix1[x,y][2]+pix2[x,y][2]/2), round(pix1[x,y][3]+pix2[x,y][3]/2))
#color=pix1[x,y]
draw.point((x,y),fill=color)
else:
draw.point((x,y),fill=(0,0,0,0))
if x == im1.width-1:
x=0
y=y+1
else:
x=x+1
im.save('test%s.png' %(z), 'PNG')
print("Zapisano obraz test%s.png" %(z))
imglist = sys.argv[1:]
z=0
while imglist != []:
exists = os.path.isfile("./test%s.png" % (z-1))
if exists:
obr1="test%s.png" % (z-1)
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
else:
obr1=imglist.pop()
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
python beginner image
asked 10 mins ago
FejorFejor
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've already wrote some code in python and in nutshell it works like that.
Extracting a few frames from video
Comparing the similarity or equality of pixel color (from two first image)
Saving to the new image
Comparing new image (conjunction of two first) and the next image etc
Is there any better and more efficient way to do that ? Or do you see any improvement which i can do in my code ?
I am asking because the results aren't good :(
Sorry for my language English I am not native
import sys
import os
import numpy as np
from PIL import Image, ImageDraw
def main(obr1,obr2):
img1= Image.open("%s" %(obr1))
img2= Image.open("%s" %(obr2))
im1 = img1.convert("RGBA")
im2 = img2.convert("RGBA")
pix1 = im1.load()
pix2 = im2.load()
im = Image.new("RGBA", (im1.width, im1.height), (0, 0, 0, 0))
draw = ImageDraw.Draw(im)
x = 0
y = 0
while y != im1.height-1 or x != im1.width-1:
if pix1[x,y] == pix2[x,y]:
draw.point((x,y),fill=pix1[x,y])
else:
p1 = np.array([(pix1[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
p2 = np.array([(pix2[x,y][0]),(pix1[x,y][1]),(pix1[x,y][2])])
squared_dist = np.sum(p1**2 + p2**2, axis=0)
dist = np.sqrt(squared_dist)
if dist < 200 and pix1[x,y] !=(0,0,0,0) and pix2[x,y] != (0,0,0,0):
color = (round(pix1[x,y][0]+pix2[x,y][0]/2), round(pix1[x,y][1]+pix2[x,y][1]/2), round(pix1[x,y][2]+pix2[x,y][2]/2), round(pix1[x,y][3]+pix2[x,y][3]/2))
#color=pix1[x,y]
draw.point((x,y),fill=color)
else:
draw.point((x,y),fill=(0,0,0,0))
if x == im1.width-1:
x=0
y=y+1
else:
x=x+1
im.save('test%s.png' %(z), 'PNG')
print("Zapisano obraz test%s.png" %(z))
imglist = sys.argv[1:]
z=0
while imglist != []:
exists = os.path.isfile("./test%s.png" % (z-1))
if exists:
obr1="test%s.png" % (z-1)
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
else:
obr1=imglist.pop()
obr2=imglist.pop()
print("Porównywanie obraza %s i %s" % (obr1,obr2))
main(obr1,obr2)
print("Analiza skończona")
z=z+1
python beginner image
python beginner image
asked 10 mins ago
FejorFejor
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 mins ago
FejorFejor
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 mins ago
FejorFejor
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 mins ago
FejorFejor
1
asked 10 mins ago
FejorFejor
1
1
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fejor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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