python: dynamically create and execute existing variable names

Question

I need to create a ORDERED list that contains a series of sub-lists. But I'm not sure how to iterate through known variables.

Here's an example. Let's say I have the following sub-lists:

list_0 = [0, 1, 2]
list_1 = [3, 4, 5, 6]
list_2 = [7, 8, 9]

I'm trying to append the lists to a larger list list_all:

list_all = [ [0, 1, 2], [3, 4, 5, 6], [7, 8, 9] ]

So I'm trying to iterate through the variable names but I'm not sure this is possible?

list_all = [list_{num}.format(num=i) for i in range(3)]
# SyntaxError

This seems to work if list_all contains strings of the known variables but I don't need the variables obviously, I need the data within those variables.

list_all = ['list_{num}'.format(num=i) for i in range(3)]
print list_all
# ['list_0', 'list_1', 'list_2']

So I guess the question is how can I dynamically create and execute python variable names? I need to do this because the number of lists changes with each use case.

What you should do is use a list (or other data structure) instead of naming your variables "foo_#". Then you can very simply and naturally iterate through your list to do what you want. — dsh
– dsh, Commented Jan 25, 2017 at 17:02
don't create variables list_0, list_1 but list or dictionary list[0], list[1] — furas
– furas, Commented Jan 25, 2017 at 17:05

tobias_k · Accepted Answer · 2017-01-25 17:11:37Z

1

If you really, really need this, you could get the variables and their values from the globals() or locals() dictionaries.

>>> list_0 = [0, 1, 2]
>>> list_1 = [3, 4, 5, 6]
>>> list_2 = [7, 8, 9]
>>> [globals()["list_{}".format(i)] for i in range(3)]
[[0, 1, 2], [3, 4, 5, 6], [7, 8, 9]]

However, almost certainly it would be a better idea to store the individual lists as part of list_all to begin with, and access them as list_all[0] etc. instead of list_0:

>>> list_all = []
>>> list_all.append([0, 1, 2])
>>> list_all.append([3, 4, 5, 6])
>>> list_all.append([7, 8, 9])
>>> list_all[0]
[0, 1, 2]

answered Jan 25, 2017 at 17:11

tobias_k

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1 Comment

Johnny Metz Over a year ago

Great answer thanks! I also figured out eval() works in this case too

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