Hi Paul,
Thanks. I tried that and I get the same results as you --- running
"./version.py" returns 2.7.1 but "python version.py" returns 2.7.2. I think
it's because the python installed at /usr/bin/python is the one that came with
the OS; when I installed python 2.7.2 using macports, it installed it to
/usr/bin/python2.7, then set the environment to point to that. So, if I change
the first line from #! /usr/bin/python to #! /usr/bin/env python, I get the
same result when I run version.py either way.
Anyway, are you saying that having the two pythons installed is affecting
pylab.show()? If so, how can I fix it?
Thanks,
Collin
On Nov 18, 2011, at 10:54 AM, Paul de Beurs wrote:
Collin,
I had the same kind of trouble. OSX Lion comes with Python 2.7.1.
I have this little script named 'version.py':
#!/usr/bin/python
import sys
print 'Python', sys.version
In a Terminal you can start version.py in two different ways.
when I typed: python version.py
I got:
Python 2.7.2 (v2.7.2:8527427914a2, Jun 11 2011, 14:13:39)
[GCC 4.0.1 (Apple Inc. build 5493)]
when I typed: version.py
I got:
Python 2.7.1 (r271:86832, Jun 16 2011, 16:59:05)
[GCC 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2335.15.00)]
When I change the first line in version.py in
#!/Library/Frameworks/Python.framework/Versions/2.7/bin/python
the results of the two test are the same.
Maybe this helps you. Please give me your results.
Paul
2011/11/16 Collin Capano <cdcap...@syr.edu<mailto:cdcap...@syr.edu>>
Hi,
I've installed matplotlib on a new computer running OSX Lion 10.7.2 (Xcode
version 4.2). When I open ipython and try to run:
In [1]: import pylab
In [2]: pylab.figure(); pylab.plot([0,1],[2,2]); pylab.show()
nothing happens. I can, however, save the plot using pylab.savefig. I am using
matplotlib version 1.1.0, with Python 2.7.2 and ipython version 0.11. I
installed all of these using MacPorts (specifically, the python27,
py27-matplotlib, and py27-ipython ports).
Any help would be greatly appreciated, as interactive plotting is important for
my work.
Thanks,
Collin
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