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Error on launching IPython on Win 7 and Python 2.7.3 #2624

shensac opened this Issue · 2 comments

3 participants



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


Traceback (most recent call last):
File "C:\Python27\Scripts\", line 9, in
load_entry_point('ipython==0.13.1', 'console_scripts', 'ipython')()
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\", line 388, in launch_new_instance
File "", line 2, in initialize
File "C:\Python27\lib\site-packages\IPython\config\", line 84, in catch_config_error
return method(app, args, *kwargs)
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\", line 324, in initialize
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\", line 340, in init_shell
File "C:\Python27\lib\site-packages\IPython\config\", line 318, in instance
inst = cls(args, *kwargs)
File "C:\Python27\lib\site-packages\IPython\frontend\terminal\", line 360, in init
user_module=user_module, custom_exceptions=custom_exceptions
File "C:\Python27\lib\site-packages\IPython\core\", line 434, in init
self.db = PickleShareDB(os.path.join(self.profile_dir.location, 'db'))
File "C:\Python27\lib\site-packages\IPython\utils\", 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:
or send an email to the mailing list at

You can print a more detailed traceback right now with "%tb", or use "%debug"
to interactively debug it.


I guess it was bug #737 that you saw. It should be resolved by now - can you make sure you don't have an older version of installed?


Yes you are right. I seem to have missed that point. I was on path 2.2, upgraded to 2.4.1 and it looks good now. Thanks for your help.

@shensac shensac closed this
@minrk minrk added this to the no action milestone
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