[Doc]: replace usages of .imread with PIL.Image.open

[story645]

Documentation Link

https://matplotlib.org/devdocs/api/image_api.html#matplotlib.image.imread

Problem

We flag the imread function with a note that it's not recommended, so we should follow our own guidance and replace usages of imread with PIL when possible.

Suggested improvement

Marking this as a good first issue since it's replacing the usage of imread in the code and examples with PIL.Image.open when possible.

[timhoffm]

T.b.d.: This is often used in a context manager:

    with cbook.get_sample_data('grace_hopper.jpg') as image_file:
        photo = plt.imread(image_file)

It should be simpler to only pass the path to Image.open(), which currently reads:

photo = PIL.image.open(cbook.get_sample_data('grace_hopper.jpg', asfileobj=False))

or

photo = PIL.image.open(Path(matplotlib.get_data_path()) / 'grace_hopper.jpg'))

Both of them are still not elegant.

I thus propose to extend cbook.get_sample_data() to support image files (.jpg, .png) natively by using Image.open internally, so that we can directly write:

photo = cbook.get_sample_data('grace_hopper.jpg')

[QuLogic]

get_sample_data returns more data than just images, and @tacaswell doesn't like return type instability (and how it would decide wasn't specified; a parameter or extension sniffing?), so you may want a new get_sample_image instead.

[story645]

Pulling good first issue until we shake out what we want to do but I like 'sample_image'

[timhoffm]

Thinking about it, it may be an educational aspect to not hide PIL.Image.open() call. It is not obvious how to transfer get_sample_image() to your own image file. So

photo = PIL.image.open(Path(matplotlib.get_data_path()) / 'grace_hopper.jpg'))

seems like a reasonable solution. It would be slightly nicer if get_data_path() would return a pathlib.Path object, but we're not there and migrating is not straight forward.