External variables in lambda functions in Python

Question

Inspired by the construction in this answer, I am trying to do the following:

values = range(3)
vector = np.random.randint(3, size=(5,))
f = lambda x: x in values
result = [f(a) for a in values]

but I get global name 'values' is not defined.

I get the same error if I try the solution that I linked to above, i.e.:

A = [[0,1,2], [1,2,3], [2,3,4]]
v = [1,2]
B = [map(lambda val: val in v) for a in A]

Did Python change since that solution was posted? (I am working with 2.7.4). If so, how can I access an external variable within a lambda function? Should I declare it as global? pass it as another input?

Update 1:

I am only noticing this problem within an embedded shell in IPython (1.0).

Update 2:

There is an IPython ticket on GitHub on the topic, but it's unclear if the problem has been solved.

Update 3(Not from the OP):

The error is re-produceable in django's shell (thanks @Ashwini)

$ ./manage.py shell
Python 2.7.4 (default, Apr 19 2013, 18:32:33) 
Type "copyright", "credits" or "license" for more information.

IPython 0.13.2 -- An enhanced Interactive Python.

In [1]: import numpy as np
In [2]: values = range(3)
In [3]: vector = np.random.randint(3, size=(5,))
In [4]: f = lambda x: x in values
In [5]: result = [f(a) for a in values]
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
/usr/local/lib/python2.7/dist-packages/django/core/management/commands/shell.pyc in <module>()
----> 1 result = [f(a) for a in values]
/usr/local/lib/python2.7/dist-packages/django/core/management/commands/shell.pyc in <lambda>(x)
----> 1 f = lambda x: x in values

NameError: global name 'values' is not defined
In [6]: values
Out[6]: [0, 1, 2]

The code in your first block works for me on python2.7 as well as on python3.3. So I'm uncertain about what your actual problem is — inspectorG4dget, Aug 10 '13 at 02:52
@user815423426 Your code works fine in py2.x,3,x and ipy2.x, 3.x. But, django's shell(ipy 2.7) raises the same error. Weird! — Ashwini Chaudhary, Aug 10 '13 at 02:59
See [Global variables undefined in interactive use of embedded ipython shell](https://github.com/ipython/ipython/issues/62). — falsetru, Aug 10 '13 at 03:49

score 1 · Answer 1 · answered Aug 13 '13 at 05:17

I know that the code in your example and in the bug report does work successfully in the interactive shell of the pyramid framework (pshell, with ipython 0.12!), though I remember facing this issue before. The key is that with ipython >= 0.11 it uses different code. As far as I know, the 0.10 code would still have this bug.

This is a simplified excerpt from the pyramid pshell.py

def make_ipython_v0_11_shell():
    try:
        from IPython.frontend.terminal.embed import (
            InteractiveShellEmbed)
        IPShellFactory = InteractiveShellEmbed
    except ImportError:
        return None

    def shell(env, help):
        IPShell = IPShellFactory(banner2=help + '\n', user_ns=env)
        IPShell()

   return shell

def make_ipython_v0_10_shell():
    try:
        from IPython.Shell import IPShellEmbed
        IPShellFactory = IPShellEmbed
    except ImportError:
        return None

    def shell(env, help):
        IPShell = IPShellFactory(argv=[], user_ns=env)
        IPShell.set_banner(IPShell.IP.BANNER + '\n' + help + '\n')
        IPShell()

    return shell

Thanks! This is great to know. The behavior I described above occurs sort of sporadically in IPython v1.0 though. Have you noticed this with 1.0? — Amelio Vazquez-Reina, Aug 13 '13 at 14:10

External variables in lambda functions in Python

Update 1:

Update 2:

Update 3(Not from the OP):

1 Answers1