python pdb lambda function global name error

I was testing a fix using pdb.set_trace() to make sure it worked the way I expected before implementing it and kept getting a weird error.

(Pdb) test = [1,2,3,4,4,3,2,1]
(Pdb) max(range(len(test)),key=lambda i: test[i])
*** NameError: global name 'test' is not defined

So I moved to my localmachine to make sure I could reproduce as simply as possible before asking for help. In ipython I get the exact same behavior.

In [1]: test = [1,2,3,4,4,3,2,1]

In [2]: max(range(len(test)),key=lambda i: test[i])
Out[2]: 3

In [3]: import pdb; pdb.set_trace()
--Call--
> /Users/tristanmatthews/anaconda/lib/python2.7/site-packages/IPython/core/displayhook.py(237)__call__()
-> def __call__(self, result=None):
(Pdb) test = [1,2,3,4,4,3,2,1]
(Pdb) max(range(len(test)),key=lambda i: test[i])
*** NameError: global name 'test' is not defined

But at the normal command line it works just fine:

tristan:~$ python
Python 2.7.6 |Anaconda 1.8.0 (x86_64)| (default, Nov 11 2013, 10:49:09) 
[GCC 4.0.1 (Apple Inc. build 5493)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> test = [1,2,3,4,4,3,2,1]
>>> max(range(len(test)),key=lambda i: test[i])
3
>>> import pdb; pdb.set_trace()
--Return--
> <stdin>(1)<module>()->None
(Pdb) test = [1,2,3,4,4,3,2,1]
(Pdb) max(range(len(test)),key=lambda i: test[i])
3

If anyone has any idea what is going on here I would REALLY love to know.

For the record the fix works fine in my code, just not in the debugger.

For reference my versions of python are: Original error:

'2.7.3 (default, Apr 10 2013, 06:20:15) \n[GCC 4.6.3]'

Local Machine both ipython and the command line are the same:

In [5]: sys.version
Out[5]: '2.7.6 |Anaconda 1.8.0 (x86_64)| (default, Nov 11 2013, 10:49:09) \n[GCC 4.0.1 (Apple Inc. build 5493)]'
>>> sys.version
'2.7.6 |Anaconda 1.8.0 (x86_64)| (default, Nov 11 2013, 10:49:09) \n[GCC 4.0.1 (Apple Inc. build 5493)]'

回答1:

I can confirm this issue with Python 2.7. There is a bug report for Python 3 which suggests a workaround: interact at the pdb prompt drops you into an interactive session which is populated with globals() and locals() and your lambda should work as expected.

回答2:

pdb isn't a full python shell, and intercepts a lot of things. But adding a print in front of it should work:

print max(range(len(test)),key=lambda i: test[i])