Inverse mapping with bilinear interpolation on an image
1
$begingroup$
The following is my solution for an inverse mapping with bilinear interpolation on an image. The original image is img and newmatrix is the transformed image. invRot is the inverse transformation matrix.
How can I optimize the nested for loops (or remove them altogether) to give me a better time complexity for large values of row and col?
newmatrix = np.zeros((row, col, 3), np.uint8)
for r in range(row):
for c in range(col):
if offset > 0:
offset = -1 * offset
pt = np.array([r+offset,c,1]) #Adjust the offset.
newpt = np.matmul(invRot, pt) #Reverse map by reverse rotation and pick up color.
#Check the bounds of the inverse pts we got and if they lie in the original image,
#then copy the color from that original pt to the new matrix/image.
if (newpt[0] >= 0 and newpt[0] < (yLen - 1) and newpt[1] >= 0 and newpt[1] < (xLen - 1)):
x = np.asarray(newpt[1])
y = np.asarray(newpt[0])
x0 = np.floor(x).astype(int)
x1 = x0 + 1
y0 = np.floor(y).astype(int)
y1 = y0 + 1
Ia = img[y0, x0]
Ib = img[y1, x0]
Ic = img[y0, x1]
Id = img[y1, x1]
color1 = (x1-x) * (y1-y) * Ia
color2 = (x1-x) * (y-y0) * Ib
color3 = (x-x0) * (y1-y) * Ic
color4 = (x-x0) * (y-y0) * Id
weightedAvgColor = color1 + color2 + color3 + color4
newmatrix[r][c] = weightedAvgColor
python performance image matrix numpy
edited 13 mins ago
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asked 10 hours ago
The White WolfThe White Wolf
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add a comment |
1
$begingroup$
The following is my solution for an inverse mapping with bilinear interpolation on an image. The original image is img and newmatrix is the transformed image. invRot is the inverse transformation matrix.
How can I optimize the nested for loops (or remove them altogether) to give me a better time complexity for large values of row and col?
newmatrix = np.zeros((row, col, 3), np.uint8)
for r in range(row):
for c in range(col):
if offset > 0:
offset = -1 * offset
pt = np.array([r+offset,c,1]) #Adjust the offset.
newpt = np.matmul(invRot, pt) #Reverse map by reverse rotation and pick up color.
#Check the bounds of the inverse pts we got and if they lie in the original image,
#then copy the color from that original pt to the new matrix/image.
if (newpt[0] >= 0 and newpt[0] < (yLen - 1) and newpt[1] >= 0 and newpt[1] < (xLen - 1)):
x = np.asarray(newpt[1])
y = np.asarray(newpt[0])
x0 = np.floor(x).astype(int)
x1 = x0 + 1
y0 = np.floor(y).astype(int)
y1 = y0 + 1
Ia = img[y0, x0]
Ib = img[y1, x0]
Ic = img[y0, x1]
Id = img[y1, x1]
color1 = (x1-x) * (y1-y) * Ia
color2 = (x1-x) * (y-y0) * Ib
color3 = (x-x0) * (y1-y) * Ic
color4 = (x-x0) * (y-y0) * Id
weightedAvgColor = color1 + color2 + color3 + color4
newmatrix[r][c] = weightedAvgColor
python performance image matrix numpy
edited 13 mins ago
200_success
129k15153415
asked 10 hours ago
The White WolfThe White Wolf
61
New contributor
The White Wolf 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 |
1
1
1
$begingroup$
The following is my solution for an inverse mapping with bilinear interpolation on an image. The original image is img and newmatrix is the transformed image. invRot is the inverse transformation matrix.
How can I optimize the nested for loops (or remove them altogether) to give me a better time complexity for large values of row and col?
newmatrix = np.zeros((row, col, 3), np.uint8)
for r in range(row):
for c in range(col):
if offset > 0:
offset = -1 * offset
pt = np.array([r+offset,c,1]) #Adjust the offset.
newpt = np.matmul(invRot, pt) #Reverse map by reverse rotation and pick up color.
#Check the bounds of the inverse pts we got and if they lie in the original image,
#then copy the color from that original pt to the new matrix/image.
if (newpt[0] >= 0 and newpt[0] < (yLen - 1) and newpt[1] >= 0 and newpt[1] < (xLen - 1)):
x = np.asarray(newpt[1])
y = np.asarray(newpt[0])
x0 = np.floor(x).astype(int)
x1 = x0 + 1
y0 = np.floor(y).astype(int)
y1 = y0 + 1
Ia = img[y0, x0]
Ib = img[y1, x0]
Ic = img[y0, x1]
Id = img[y1, x1]
color1 = (x1-x) * (y1-y) * Ia
color2 = (x1-x) * (y-y0) * Ib
color3 = (x-x0) * (y1-y) * Ic
color4 = (x-x0) * (y-y0) * Id
weightedAvgColor = color1 + color2 + color3 + color4
newmatrix[r][c] = weightedAvgColor
python performance image matrix numpy
edited 13 mins ago
200_success
129k15153415
asked 10 hours ago
The White WolfThe White Wolf
61
New contributor
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The following is my solution for an inverse mapping with bilinear interpolation on an image. The original image is img and newmatrix is the transformed image. invRot is the inverse transformation matrix.
How can I optimize the nested for loops (or remove them altogether) to give me a better time complexity for large values of row and col?
newmatrix = np.zeros((row, col, 3), np.uint8)
for r in range(row):
for c in range(col):
if offset > 0:
offset = -1 * offset
pt = np.array([r+offset,c,1]) #Adjust the offset.
newpt = np.matmul(invRot, pt) #Reverse map by reverse rotation and pick up color.
#Check the bounds of the inverse pts we got and if they lie in the original image,
#then copy the color from that original pt to the new matrix/image.
if (newpt[0] >= 0 and newpt[0] < (yLen - 1) and newpt[1] >= 0 and newpt[1] < (xLen - 1)):
x = np.asarray(newpt[1])
y = np.asarray(newpt[0])
x0 = np.floor(x).astype(int)
x1 = x0 + 1
y0 = np.floor(y).astype(int)
y1 = y0 + 1
Ia = img[y0, x0]
Ib = img[y1, x0]
Ic = img[y0, x1]
Id = img[y1, x1]
color1 = (x1-x) * (y1-y) * Ia
color2 = (x1-x) * (y-y0) * Ib
color3 = (x-x0) * (y1-y) * Ic
color4 = (x-x0) * (y-y0) * Id
weightedAvgColor = color1 + color2 + color3 + color4
newmatrix[r][c] = weightedAvgColor
python performance image matrix numpy
python performance image matrix numpy
edited 13 mins ago
200_success
129k15153415
asked 10 hours ago
The White WolfThe White Wolf
61
New contributor
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 mins ago
200_success
129k15153415
asked 10 hours ago
The White WolfThe White Wolf
61
New contributor
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 mins ago
200_success
129k15153415
edited 13 mins ago
200_success
129k15153415
edited 13 mins ago
200_success
129k15153415
129k15153415
asked 10 hours ago
The White WolfThe White Wolf
61
New contributor
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 10 hours ago
The White WolfThe White Wolf
61
asked 10 hours ago
The White WolfThe White Wolf
61
61
New contributor
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The White Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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