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']
