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I just installed IPython on Win 7 with Python 2.7.3 and on launching ipython I get following error. I looked at some other post and concluded that this issue was fixed/closed.
Do I still need to change pickleshare.py?
Thanks
Sachin
Traceback (most recent call last):
File "C:\Python27\Scripts\ipython-script.py", line 9, in
load_entry_point('ipython==0.13.1', 'console_scripts', 'ipython')()
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\ipapp.py", line 388, in launch_new_instance
app.initialize()
File "", line 2, in initialize
File "C:\Python27\lib\site-packages\IPython\config\application.py", line 84, in catch_config_error
return method(app, _args, *_kwargs)
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\ipapp.py", line 324, in initialize
self.init_shell()
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\ipapp.py", line 340, in init_shell
ipython_dir=self.ipython_dir)
File "C:\Python27\lib\site-packages\IPython\config\configurable.py", line 318, in instance
inst = cls(_args, *_kwargs)
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\interactiveshell.py", line 360, in init
user_module=user_module, custom_exceptions=custom_exceptions
File "C:\Python27\lib\site-packages\IPython\core\interactiveshell.py", line 434, in init
self.db = PickleShareDB(os.path.join(self.profile_dir.location, 'db'))
File "C:\Python27\lib\site-packages\IPython\utils\pickleshare.py", line 52, in init
if not self.root.isdir():
TypeError: _isdir() takes exactly 1 argument (0 given)
Hi,
I just installed IPython on Win 7 with Python 2.7.3 and on launching ipython I get following error. I looked at some other post and concluded that this issue was fixed/closed.
Do I still need to change pickleshare.py?
Thanks
Sachin
Traceback (most recent call last):
File "C:\Python27\Scripts\ipython-script.py", line 9, in
load_entry_point('ipython==0.13.1', 'console_scripts', 'ipython')()
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\ipapp.py", line 388, in launch_new_instance
app.initialize()
File "", line 2, in initialize
File "C:\Python27\lib\site-packages\IPython\config\application.py", line 84, in catch_config_error
return method(app, _args, *_kwargs)
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\ipapp.py", line 324, in initialize
self.init_shell()
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\ipapp.py", line 340, in init_shell
ipython_dir=self.ipython_dir)
File "C:\Python27\lib\site-packages\IPython\config\configurable.py", line 318, in instance
inst = cls(_args, *_kwargs)
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\interactiveshell.py", line 360, in init
user_module=user_module, custom_exceptions=custom_exceptions
File "C:\Python27\lib\site-packages\IPython\core\interactiveshell.py", line 434, in init
self.db = PickleShareDB(os.path.join(self.profile_dir.location, 'db'))
File "C:\Python27\lib\site-packages\IPython\utils\pickleshare.py", line 52, in init
if not self.root.isdir():
TypeError: _isdir() takes exactly 1 argument (0 given)
If you suspect this is an IPython bug, please report it at:
https://github.com/ipython/ipython/issues
or send an email to the mailing list at ipython-dev@scipy.org
You can print a more detailed traceback right now with "%tb", or use "%debug"
to interactively debug it.
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