Simply call list on the generator.

lst = list(gen)
lst

Be aware that this affects the generator which will not return any further items.

You also cannot directly call list in IPython, as it conflicts with a command for listing lines of code.

Tested on this file:

def gen():
    yield 1
    yield 2
    yield 3
    yield 4
    yield 5
import ipdb
ipdb.set_trace()

g1 = gen()

text = "aha" + "bebe"

mylst = range(10, 20)

which when run:

$ python code.py 
> /home/javl/sandbox/so/debug/code.py(10)<module>()
      9 
---> 10 g1 = gen()
     11 

ipdb> n
> /home/javl/sandbox/so/debug/code.py(12)<module>()
     11 
---> 12 text = "aha" + "bebe"
     13 

ipdb> lst = list(g1)
ipdb> lst
[1, 2, 3, 4, 5]
ipdb> q
Exiting Debugger.

General method for escaping function/variable/debugger name conflicts

There are debugger commands p and pp that will print and prettyprint any expression following them.

So you could use it as follows:

$ python code.py 
> /home/javl/sandbox/so/debug/code.py(10)<module>()
      9 
---> 10 g1 = gen()
     11 

ipdb> n
> /home/javl/sandbox/so/debug/code.py(12)<module>()
     11 
---> 12 text = "aha" + "bebe"
     13 

ipdb> p list(g1)
[1, 2, 3, 4, 5]
ipdb> c

There is also an exec command, called by prefixing your expression with !, which forces debugger to take your expression as Python one.

ipdb> !list(g1)
[]

For more details see help p, help pp and help exec when in debugger.

ipdb> help exec
(!) statement
Execute the (one-line) statement in the context of
the current stack frame.
The exclamation point can be omitted unless the first word
of the statement resembles a debugger command.
To assign to a global variable you must always prefix the
command with a 'global' command, e.g.:
(Pdb) global list_options; list_options = ['-l']