How to read a hdr image quickly in the RGBE format in Python?

Question

I would like to know how to read a HDR image (.hdr) by obtaining pixel values in the RGBE format quickly and efficiently in Python.

These are somethings I tried:

    import imageio
    img = imageio.imread(hdr_path, format="HDR-FI")

alternatively:

    import cv2
    img = cv2.imread(hdr_path, flags=cv2.IMREAD_ANYDEPTH)

This read the image, but gave the values in a RGB format.

How do you obtain the 4rth channel, the "E" channel for every pixel, without altered RGB values? I would prefer a solution involving only imageio, as i am restricted to use only that module.

Answer 1

If you prefer the RGBE representation over the float representation you can convert between the two

def float_to_rgbe(image, *, channel_axis=-1):

    # ensure channel-last
    image = np.moveaxis(image, channel_axis, -1)

    max_float = np.max(image, axis=-1)
    
    scale, exponent = np.frexp(max_float)
    scale *= 256.0/max_float

    image_rgbe = np.empty((*image.shape[:-1], 4)
    image_rgbe[..., :3] = image * scale
    image_rgbe[..., -1] = exponent + 128

    image_rgbe[scale < 1e-32, :] = 0
    
    # restore original axis order
    image_rgbe = np.moveaxis(image_rgbe, -1, channel_axis)

    return image_rgbe

(Note: this is based on the RGBE reference implementation (found here) and can be further optimized if it actually is the bottleneck.)

In your comment, you mention "If i parse the numpy array manually and split the channels into an E channel, it takes too much time...", but it is hard to tell why that is the case without seeing the code. The above is O(height*width), which seems reasonable for a pixel-level image processing method.