How to fit image right in homemade rotation algorithm in python












0














I have made this algorithm myself based on how to rotate an image.



import cv2
import math
import numpy as np

class rotation:
angle = 60.0
x = 100
y = 100
img = cv2.imread('FNAF.png',0)
width,height = img.shape
def showImage(name, self):
cv2.imshow(name, self.img)
def printWidthHeight(self):
print(self.width)
print(self.height)
def rotateImage(self):
ForwardMap = np.zeros((self.width,self.height),dtype="uint8")
BackwardMap = np.zeros((self.width,self.height),dtype="uint8")

for i in range(self.width):
for j in range(self.height):
# forward mapping
for_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) - (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
for_y = (i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
for_x = int(for_x)
for_y = int(for_y)
# backward mapping should change the forward mapping to the original image
back_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) + (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
back_y = -(i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
back_x = int(back_x)
back_y = int(back_y)
if for_x in range(self.width) and for_y in range(self.height):
ForwardMap[i, j] = self.img[for_x, for_y]
else:
pass
if back_x in range(self.width) and back_y in range(self.height):
BackwardMap[i, j] = self.img[back_x, back_y]
else:
pass
cv2.imshow('Forward Mapping', ForwardMap)
cv2.imshow('Backward Mapping', BackwardMap)
def demo():
rotation.showImage('normal', rotation)
rotation.printWidthHeight(rotation)
rotation.rotateImage(rotation)
cv2.waitKey(0)
cv2.destroyAllWindows

if __name__ == '__main__':
demo()


My problem is now that I want to make the rotated pictures (both for forward and backwards mapping) fit JUST right so there is no unnecessary use of space. Can anybody help me with this? Much appreciated.



If you have any suggestions to optimize my code, feel free to comment on that as well.










share|improve this question






















  • Could you show a picture - what you want to do? About optimization - you are using opencv that rotates images very fast.
    – MBo
    Nov 21 at 18:09
















0














I have made this algorithm myself based on how to rotate an image.



import cv2
import math
import numpy as np

class rotation:
angle = 60.0
x = 100
y = 100
img = cv2.imread('FNAF.png',0)
width,height = img.shape
def showImage(name, self):
cv2.imshow(name, self.img)
def printWidthHeight(self):
print(self.width)
print(self.height)
def rotateImage(self):
ForwardMap = np.zeros((self.width,self.height),dtype="uint8")
BackwardMap = np.zeros((self.width,self.height),dtype="uint8")

for i in range(self.width):
for j in range(self.height):
# forward mapping
for_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) - (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
for_y = (i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
for_x = int(for_x)
for_y = int(for_y)
# backward mapping should change the forward mapping to the original image
back_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) + (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
back_y = -(i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
back_x = int(back_x)
back_y = int(back_y)
if for_x in range(self.width) and for_y in range(self.height):
ForwardMap[i, j] = self.img[for_x, for_y]
else:
pass
if back_x in range(self.width) and back_y in range(self.height):
BackwardMap[i, j] = self.img[back_x, back_y]
else:
pass
cv2.imshow('Forward Mapping', ForwardMap)
cv2.imshow('Backward Mapping', BackwardMap)
def demo():
rotation.showImage('normal', rotation)
rotation.printWidthHeight(rotation)
rotation.rotateImage(rotation)
cv2.waitKey(0)
cv2.destroyAllWindows

if __name__ == '__main__':
demo()


My problem is now that I want to make the rotated pictures (both for forward and backwards mapping) fit JUST right so there is no unnecessary use of space. Can anybody help me with this? Much appreciated.



If you have any suggestions to optimize my code, feel free to comment on that as well.










share|improve this question






















  • Could you show a picture - what you want to do? About optimization - you are using opencv that rotates images very fast.
    – MBo
    Nov 21 at 18:09














0












0








0







I have made this algorithm myself based on how to rotate an image.



import cv2
import math
import numpy as np

class rotation:
angle = 60.0
x = 100
y = 100
img = cv2.imread('FNAF.png',0)
width,height = img.shape
def showImage(name, self):
cv2.imshow(name, self.img)
def printWidthHeight(self):
print(self.width)
print(self.height)
def rotateImage(self):
ForwardMap = np.zeros((self.width,self.height),dtype="uint8")
BackwardMap = np.zeros((self.width,self.height),dtype="uint8")

for i in range(self.width):
for j in range(self.height):
# forward mapping
for_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) - (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
for_y = (i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
for_x = int(for_x)
for_y = int(for_y)
# backward mapping should change the forward mapping to the original image
back_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) + (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
back_y = -(i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
back_x = int(back_x)
back_y = int(back_y)
if for_x in range(self.width) and for_y in range(self.height):
ForwardMap[i, j] = self.img[for_x, for_y]
else:
pass
if back_x in range(self.width) and back_y in range(self.height):
BackwardMap[i, j] = self.img[back_x, back_y]
else:
pass
cv2.imshow('Forward Mapping', ForwardMap)
cv2.imshow('Backward Mapping', BackwardMap)
def demo():
rotation.showImage('normal', rotation)
rotation.printWidthHeight(rotation)
rotation.rotateImage(rotation)
cv2.waitKey(0)
cv2.destroyAllWindows

if __name__ == '__main__':
demo()


My problem is now that I want to make the rotated pictures (both for forward and backwards mapping) fit JUST right so there is no unnecessary use of space. Can anybody help me with this? Much appreciated.



If you have any suggestions to optimize my code, feel free to comment on that as well.










share|improve this question













I have made this algorithm myself based on how to rotate an image.



import cv2
import math
import numpy as np

class rotation:
angle = 60.0
x = 100
y = 100
img = cv2.imread('FNAF.png',0)
width,height = img.shape
def showImage(name, self):
cv2.imshow(name, self.img)
def printWidthHeight(self):
print(self.width)
print(self.height)
def rotateImage(self):
ForwardMap = np.zeros((self.width,self.height),dtype="uint8")
BackwardMap = np.zeros((self.width,self.height),dtype="uint8")

for i in range(self.width):
for j in range(self.height):
# forward mapping
for_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) - (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
for_y = (i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
for_x = int(for_x)
for_y = int(for_y)
# backward mapping should change the forward mapping to the original image
back_x = (i - self.x) * math.cos(self.angle*(math.pi/180)) + (j - self.y) * math.sin(self.angle*(math.pi/180)) + self.x
back_y = -(i - self.x) * math.sin(self.angle*(math.pi/180)) + (j - self.y) * math.cos(self.angle*(math.pi/180)) + self.x
back_x = int(back_x)
back_y = int(back_y)
if for_x in range(self.width) and for_y in range(self.height):
ForwardMap[i, j] = self.img[for_x, for_y]
else:
pass
if back_x in range(self.width) and back_y in range(self.height):
BackwardMap[i, j] = self.img[back_x, back_y]
else:
pass
cv2.imshow('Forward Mapping', ForwardMap)
cv2.imshow('Backward Mapping', BackwardMap)
def demo():
rotation.showImage('normal', rotation)
rotation.printWidthHeight(rotation)
rotation.rotateImage(rotation)
cv2.waitKey(0)
cv2.destroyAllWindows

if __name__ == '__main__':
demo()


My problem is now that I want to make the rotated pictures (both for forward and backwards mapping) fit JUST right so there is no unnecessary use of space. Can anybody help me with this? Much appreciated.



If you have any suggestions to optimize my code, feel free to comment on that as well.







python image opencv math image-processing






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share|improve this question











share|improve this question




share|improve this question










asked Nov 21 at 15:50









ACommonDane01

132




132












  • Could you show a picture - what you want to do? About optimization - you are using opencv that rotates images very fast.
    – MBo
    Nov 21 at 18:09


















  • Could you show a picture - what you want to do? About optimization - you are using opencv that rotates images very fast.
    – MBo
    Nov 21 at 18:09
















Could you show a picture - what you want to do? About optimization - you are using opencv that rotates images very fast.
– MBo
Nov 21 at 18:09




Could you show a picture - what you want to do? About optimization - you are using opencv that rotates images very fast.
– MBo
Nov 21 at 18:09












1 Answer
1






active

oldest

votes


















1














Your original image has 4 corners, coordinates: (0,0), (w-1, 0), (0,h-1) and (w-1, h-1). Since your transformation is affine, why not transform these coordinates into destination coordinates?




  • (0,0) → (x1, y1)

  • (w-1, 0) → (x2, y2)

  • (0, h-1) → (x3, y3)

  • (w-1, h-1) → (x4, y4)


Your destination image size is then:



width  = max(x1, x2, x3, x4) - min(x1, x2, x3, x4)
height = max(y1, y2, y3, y4) - min(y1, y2, y3, y4)





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    1 Answer
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    oldest

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    active

    oldest

    votes









    1














    Your original image has 4 corners, coordinates: (0,0), (w-1, 0), (0,h-1) and (w-1, h-1). Since your transformation is affine, why not transform these coordinates into destination coordinates?




    • (0,0) → (x1, y1)

    • (w-1, 0) → (x2, y2)

    • (0, h-1) → (x3, y3)

    • (w-1, h-1) → (x4, y4)


    Your destination image size is then:



    width  = max(x1, x2, x3, x4) - min(x1, x2, x3, x4)
    height = max(y1, y2, y3, y4) - min(y1, y2, y3, y4)





    share|improve this answer


























      1














      Your original image has 4 corners, coordinates: (0,0), (w-1, 0), (0,h-1) and (w-1, h-1). Since your transformation is affine, why not transform these coordinates into destination coordinates?




      • (0,0) → (x1, y1)

      • (w-1, 0) → (x2, y2)

      • (0, h-1) → (x3, y3)

      • (w-1, h-1) → (x4, y4)


      Your destination image size is then:



      width  = max(x1, x2, x3, x4) - min(x1, x2, x3, x4)
      height = max(y1, y2, y3, y4) - min(y1, y2, y3, y4)





      share|improve this answer
























        1












        1








        1






        Your original image has 4 corners, coordinates: (0,0), (w-1, 0), (0,h-1) and (w-1, h-1). Since your transformation is affine, why not transform these coordinates into destination coordinates?




        • (0,0) → (x1, y1)

        • (w-1, 0) → (x2, y2)

        • (0, h-1) → (x3, y3)

        • (w-1, h-1) → (x4, y4)


        Your destination image size is then:



        width  = max(x1, x2, x3, x4) - min(x1, x2, x3, x4)
        height = max(y1, y2, y3, y4) - min(y1, y2, y3, y4)





        share|improve this answer












        Your original image has 4 corners, coordinates: (0,0), (w-1, 0), (0,h-1) and (w-1, h-1). Since your transformation is affine, why not transform these coordinates into destination coordinates?




        • (0,0) → (x1, y1)

        • (w-1, 0) → (x2, y2)

        • (0, h-1) → (x3, y3)

        • (w-1, h-1) → (x4, y4)


        Your destination image size is then:



        width  = max(x1, x2, x3, x4) - min(x1, x2, x3, x4)
        height = max(y1, y2, y3, y4) - min(y1, y2, y3, y4)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 22 at 21:20









        AJNeufeld

        6,73911337




        6,73911337






























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