2
\$\begingroup\$

My code takes 3 images taken with a polarising filter rotated 45° between them and transform them into a single image encoding the polarization parameters as HSV. what can be improved?

import cv2
import numpy as np
import matplotlib.pyplot as plt
from PIL import Image 
from IPython.display import Image
%matplotlib inline
import glob
import math    
imagefiles=glob.glob(r"C:\Users\HP\My Python stuff\openCV\Polarization\IMG*")
imagefiles.sort()
images=[]
for filename in imagefiles:
    img=cv2.imread(filename)
    img=cv2.cvtColor(img,cv2.COLOR_BGR2RGB)
    images.append(img)
num_images=len(images)
pixel0x,pixel0y,pixel0z=cv2.split(images[0])
pixel45x,pixel45y,pixel45z=cv2.split(images[1])
pixel90x,pixel90y,pixel90z=cv2.split(images[2])
i0 =  pixel0x * 0.3 +  pixel0y * 0.59 +  pixel0z * 0.11
i45 = pixel45x * 0.3 + pixel45y * 0.59 + pixel45z * 0.11
i90 = pixel90x * 0.3 + pixel90y * 0.59 + pixel90z * 0.11
stokesI = i0 + i90
stokesQ = i0 - i90
stokesU = (np.ones(stokesI.shape)*(2.0 * i45))- stokesI
polint = np.sqrt(stokesQ*stokesQ+stokesU*stokesU)
poldolp = polint/(stokesI+((np.ones(stokesI.shape)*0.001)))
polaop = 0.5 * np.arctan(stokesU, stokesQ)
h=(polaop+(np.ones(polaop.shape)*(3.1416/2.0)))/3.1416
s=poldolp
v=polint
hsvpolar=cv2.merge((h,s,v))
hsvpolar=np.multiply(hsvpolar, 255)
rgbimg = cv2.cvtColor(np.clip(hsvpolar, 0,255).astype('uint8'),cv2.COLOR_HSV2RGB)
plt.imshow(rgbimg)
cv2.imwrite("hsvpolarized.jpg",rgbimg)

these are the input images: input images

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Please embed sample input images, and show all of your imports \$\endgroup\$
    – Reinderien
    Oct 7, 2022 at 12:38
  • 1
    \$\begingroup\$ That's... Not really what I meant. The images need to be separate and not in a combined screenshot, so that we can feed them to your code. \$\endgroup\$
    – Reinderien
    Oct 8, 2022 at 3:34

1 Answer 1

1
\$\begingroup\$

Not great use of Numpy - for instance, rather than writing your coefficients out in triplicate, this needs to be a single tensor dot product. ones() is never called for in this program if you correctly use broadcasting.

Put your code in functions.

Don't hard-code system-specific file paths.

Prefer pathlib over os, for glob etc.

Don't write a (truncated) literal for pi; instead write np.pi.

Most importantly, your results are basically nonsense because the images don't only change polarisation - they change alignment. You need to apply a matching, homography and perspective de-warp to your images for this to make any sense, and OpenCV offers these to you out of the box. The demonstration code below follows https://docs.opencv.org/4.6.0/d1/de0/tutorial_py_feature_homography.html .

Suggested

from pathlib import Path
from typing import Iterable

import cv2
import matplotlib.pyplot as plt
import numpy as np


def warp_align(images: list[np.ndarray]) -> None:
    sift = cv2.SIFT_create()
    keys: list[tuple] = []
    descriptors: list[np.ndarray] = []
    for image in images:
        key, descriptor = sift.detectAndCompute(image, mask=None)
        keys.append(key)
        descriptors.append(descriptor)

    FLANN_INDEX_KDTREE = 1
    flann = cv2.FlannBasedMatcher(
        indexParams={'algorithm': FLANN_INDEX_KDTREE, 'trees': 5},
        searchParams={'checks': 50},
    )

    LOWES_RATIO = 0.7
    train_desc, *query_descs = descriptors
    matches = [
        [
            m
            for m, n in flann.knnMatch(query_desc, train_desc, k=2)
            if m.distance < LOWES_RATIO*n.distance
        ]
        for query_desc in query_descs
    ]

    def keys_to_points(matched_keys: Iterable[tuple[float, float]]) -> np.ndarray:
        return np.array(tuple(matched_keys), dtype=np.float32)

    train_key, *query_keys = keys
    for query_key, target_matches, image in zip(query_keys, matches, images[1:]):
        query_points = keys_to_points(query_key[m.queryIdx].pt for m in target_matches)
        train_points = keys_to_points(train_key[m.trainIdx].pt for m in target_matches)
        M, mask = cv2.findHomography(query_points, train_points, method=cv2.RANSAC, ransacReprojThreshold=5)
        print(M, '\n')
        cv2.warpPerspective(src=image, dst=image, M=M, dsize=image.shape[1::-1])


def calculate_polarimetry(images: list[np.ndarray]) -> np.ndarray:
    all_images = np.stack(images)
    coefficients = np.array((0.30, 0.59, 0.11))
    i00, i45, i90 = np.tensordot(all_images,  coefficients, axes=[3, 0])

    stokes_i = i00 + i90
    stokes_q = i00 - i90
    stokes_u = 2*i45 - stokes_i

    polint = np.sqrt(stokes_q*stokes_q + stokes_u*stokes_u)
    poldolp = polint/(stokes_i + 0.001)
    polaop = 0.5 * np.arctan(stokes_u, stokes_q)

    h = polaop/np.pi + 0.5
    s = poldolp
    v = polint

    hsv_polar = 255*np.clip(cv2.merge((h, s, v)), 0, 1)
    return cv2.cvtColor(hsv_polar.astype('uint8'), cv2.COLOR_HSV2RGB)


def main() -> None:
    paths = sorted(Path.cwd().glob("pol*.png"))
    images = [
        cv2.cvtColor(cv2.imread(str(path)), cv2.COLOR_BGR2RGB)
        for path in paths
    ]
    warp_align(images)
    rgb_img = calculate_polarimetry(images)

    fig, ((tl, tr), (bl, br)) = plt.subplots(nrows=2, ncols=2)
    tl.imshow(images[0])
    tr.imshow(images[1])
    bl.imshow(images[2])
    br.imshow(rgb_img)
    plt.show()
    # cv2.imwrite("hsvpolarized.jpg", rgbimg)


if __name__ == '__main__':
    main()

Prior to homographic correction

noncorrected

After homographic correction

corrected

\$\endgroup\$
3
  • \$\begingroup\$ The result is not nonsense. They do change polarization. The images are taken strapping a polarising filter to the camera, aligning it to horizontal polarization, taking a picture. Then rotating it 45°, taking another picture and then rotating again and taking the third picture, as described in hackaday.io/project/6958-dolpi-raspi-polarization-camera \$\endgroup\$ Oct 8, 2022 at 20:00
  • 1
    \$\begingroup\$ I'm not sure that you understood what I mean. The polarisation is important, and spatial displacement is not - so any serious analysis of polarisation requires that the spatial displacement be removed; otherwise the results will not make sense. \$\endgroup\$
    – Reinderien
    Oct 8, 2022 at 20:06
  • 1
    \$\begingroup\$ Ohhh. Yes! You're absolutely right. I thought you meant that there was no polarization information in the images. That's why I explained how they were acquired. Sorry \$\endgroup\$ Oct 8, 2022 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.