Numpy rot90 for 2d points

Question

I have an image and several 2D points [(x, y), ...]. I would like to rotate both the image and the points by 90 degrees.

For the image I can simply use np.rot90. For the points I figured I have to rotate them 90 degrees around the center of the image to make them align correctly (for this I used https://stackoverflow.com/a/58781388/2445065)

Original image (note: red point is not part of the image):

After rotating the image and the point:

Whereas I would expect the red dot to be at the bottom left corner of the yellow square... Does anyone know what I'm doing wrong? Thanks in advance

Yes, that's true, but that would only work when I know the rectangle height. In this case, I used the rectangle only for demonstration — Steven, Dec 22 '21 at 15:08

score 1 · Answer 1 · answered Jan 03 '22 at 13:53

I finally found it:

To align the points and additional translation is require. np.rot90 does not equal a translation around the center but rather a rotation around the center PLUS a translation such that the top left corner of the rotated image is located at (0, 0).

You can get the translation by subtracting |origin_rotated.x - origin_original.x| from the x-coordinate and |origin_rotated.y - origin_original.y| from the y-coordinate.

score 1 · Answer 2 · answered Jun 09 '22 at 11:51

This python function will do rot90 for points:

def rot90points(x, y, k, hw):
    """
        a = np.zeros((5, 12), dtype="uint8");
        x, y = np.array([2, 5, 7]), np.array([4, 4, 2])
        a[y, x] = 200;
        for k in range(10):  # test
            a1 = np.rot90(a, k);
            x1, y1 = rot90points(x, y, k, a.shape);
            assert np.all(a1[y1, x1] == 200)
    """
    k = k % 4
    if k == 0: return x, y
    elif k == 1: return y, hw[1] - 1 - x
    elif k == 2: return hw[1] - 1 - x, hw[0] - 1 - y
    elif k == 3: return hw[0] - 1 - y, x
    else: raise ValueError(f"k error {k}")

Numpy rot90 for 2d points

2 Answers2