From: <kai...@t-...> - 2003-10-28 09:30:25
|
Can any Windows users help? The problem reported below seems to be pretty common among Windows users. I believe it is caused by Gnuplot.py not being able to find the pgnuplot.exe executable because it is not in the PATH or not installed. Are there any Windows users of Gnuplot.py who can come up with a way to diagnose the problem within the Gnuplot.py package so that users are not confronted with such a strange error message? It might help people recognize and fix the problem on their own. For example, is there a way to tell if a call to popen() fails under Windows? Ideally, it should work for the old python versions, too, where win32pipe.popen() as well as the new versions where os.popen() is used. But even a partial solution would be an improvement. Ideally, there would be a test in the gp_win32.GnuplotProcess constructor which generated a more descriptive exception in the case of errors. Otherwise, I guess I have to add a FAQ.txt question about this issue. Thanks, Michael Roger Frye wrote: > I can't get started with version 1.7 on Windows XP Professional > I am using the following packages: > Python-2.3.2-1.exe > win32all-157.exe > Numeric-23.1.win32-py2.3.exe > gp373w32.zip > gnuplot-py-1.7.zip > > Here is a transcript of my attempt to run the demo from the Command Line: > --------------------------------------------------- > >C:\Python23\python demo.py > gnuplot> set terminal windows > gnuplot> set title "A simple example" > gnuplot> set data style linespoints > gnuplot> plot 'c:\docume~1\rogerf~1\locals~1\temp\tmphzycgx' notitle > Please press return to continue... > > Traceback (most recent call last): > File "demo.py", line 113, in ? > demo() > File "demo.py", line 39, in demo > g.reset() > File "C:\Python23\Lib\site-packages\Gnuplot\_Gnuplot.py", line 355, > in reset > self('reset') > File "C:\Python23\Lib\site-packages\Gnuplot\_Gnuplot.py", line 199, > in __call_ > _ > self.gnuplot(s) > File "C:\Python23\Lib\site-packages\Gnuplot\gp_win32.py", line 125, > in __call_ > _ > self.write(s + '\n') > IOError: [Errno 22] Invalid argument > -------------------------------------------------- -- Michael Haggerty mh...@al... |