Re: [Matplotlib-users] Best way to display image from URL in Python3

[Thomas C. <tca...@gm...>]

Despite my grumping earlier, a PR that makes URLs just work is probably a
good idea and would be merged.

Tom

On Fri, Mar 20, 2015 at 3:59 PM Jerzy Karczmarczuk <
jer...@un...> wrote:

>
> Le 20/03/2015 16:57, Ryan Nelson a écrit :
> > For me, if I change the script from the PR to what is shown below,
> > everything works fine in both Python 2.7 and 3.4 (Anaconda
> > environments, everything updated):
> > ##################
> > url = 'http://www.libpng.org/pub/png/img_png/pngnow.png'
> > try:
> >     import urllib2
> >     data = urllib2.urlopen(url)
> > except Exception:
> >     import urllib.request
> >     from io import BytesIO
> >     data = BytesIO(urllib.request.urlopen(url).read())
> >
> > from matplotlib import pyplot
> >
> > image = pyplot.imread(data)  # crash on py3.x
> > pyplot.imshow(image)
> > pyplot.show()
> > #################
> > But as you can see, the Python 3 version requires the addition of
> > BytesIO and read(). I take it that this is not supposed to be the case.
>
> It works for X.png, not for X.jpg. The call of imread() fails then.
> Tested also under 3.4/Anaconda.
>
> Jerzy Karczmarczuk
>
>
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