OSError: Unable to determine file type or type not supported for bip image

I have fluorescent microscopy data saved as BIP file and I would like to load it to python. I was following the documentation with:

from spectral import *
img = open_image('path_to_file/Frgrnd.bip')

but I’m getting an error:

~/anaconda3/lib/python3.8/site-packages/spectral/spectral.py in open_image(file)
    117         pass
    118 
--> 119     raise IOError('Unable to determine file type or type not supported.')
    120 
    121 

OSError: Unable to determine file type or type not supported.

Where is the problem and how can I fix it? I’m using: spectral 0.22.1 last updated: Fri Jan 29 2021

CPython 3.8.5 IPython 7.18.1 watermark 2.0.2

Thank you.

[tboggs]

“.bip” isn’t a recognized image file extension so try to open it explicitly like this:

img = spy.envi.open('/path_to_file/Frgrnd.hdr', image='/path_to_file/Frgrnd.bip')

[tboggs]

If the file extension is “.bip”, you should stick with BIP, though in the case of a single band, it doesn’t even matter which interleave you choose. An offset of 2463 seems highly unlikely (I would expect it to be a multiple of 4). For now, I would stick with either 0 or 29132 as the offset.

The image you displayed doesn’t look inherently wrong but it appears to be only be showing two colors so maybe the image is just saturated. You should take a look at the histogram of the data to see if there is a large range in the data and possibly some extreme outliers. If you use the spectral.imshow function, you can try clipping the histogram using the “stretch” kwarg to see if that makes a difference. You want to set the stretch limits to where most of the “good” data resides. For example, to apply a linear stretch between the 0.1 and 0.9 percentiles of the data, you could use

spy.imshow(img[:, :, 0], stretch=(0.1, 0.9))

Depending on how many of the pixels lie in the tails of the distribution, you may need to adjust the stretch limits accordingly.