Convert a list of list's names(str) into a list of lists in python

Question

I am new to python and I have search a lot about this issue. I know there is a way of converting tuples to a list, but somehow it doesn't work for me. Here is my issue.

Say I have:

l_1 = ['a','b','c']
l_2 = ['e','f','g']
l_3 = ['g','h','i']

Then say I have another list of:

l_all = ['l_1','l_2','l_3']

How can I convert it(l_all) to a list of

[['a','b','c'],['e','f','g'],['g','h','i']]

I tried ast package using ast.literal_eval, but I received this error:

ValueError: malformed node or string: <_ast.Name object at 0x00000172CA3C5278>

I also tried to use json package, still no luck.

I tried just output ast.literal_eval('l_1'), not working either.

I'd really appreciate if anyone can help on this.

Thanks a lot!

I don't see any tuples, what is the relation to your problem? — Ulrich Eckhardt, Mar 17 '18 at 17:09
BTW you have a typo `1_1` => `l_1`, but you got it right afterwards, so I took the liberty of fixing it. — Jean-François Fabre, Mar 17 '18 at 17:13
@cricket_007 actually it's the opposite. OP doesn't want to create variable, OP wants to _evaluate_ variables. — Jean-François Fabre, Mar 17 '18 at 17:20
@StephenRauch apparently they have like more then 50 lists( like l_1 to l_50) and each list has different values. Technically I can just do [l_1, l_2,....l_50] but it looks ugly. So I used a loop to get ['l_1', 'l_2' ....'l_50'], but now its all strings not list object. yeah thats how I got myself into this lol — kingslucifer, Mar 17 '18 at 17:44
@kingslucifer, As you now know naming your variables `l_1`, `l_2` ... is not a good idea in Python. Use a list or a dict instead. — Stephen Rauch, Mar 17 '18 at 17:45
@StephenRauch ohhh thanks for the reminder. I didn't name them as this haha. Just using it as example :) — kingslucifer, Mar 17 '18 at 17:48
>>> l_1 = ['a','b','c'] >>> l_2 = ['e','f','g'] >>> l_3 = ['g','h','i'] >>> l_all = ['l_1','l_2','l_3'] >>> [eval(i) for i in l_all] [['a', 'b', 'c'], ['e', 'f', 'g'], ['g', 'h', 'i']] — , Mar 18 '18 at 00:05
@Jean-FrançoisFabre I am not sure I would agree with *irrelevant*, but it may not be perfect, please feel free to reopen it — Stephen Rauch, Mar 18 '18 at 17:05
never mind. The question title sucks (and the question is not very good as well) but some of the answers address the `globals()` function, and this is solved anyway. I'm not going to pretend that this question is super-original. Leave it be :) — Jean-François Fabre, Mar 18 '18 at 17:14

Jean-François Fabre · Answer 1 · 2018-03-17T17:18:41.177

That sounds like a problem that should be fixed upstream

ast.literal_eval evaluates literals. eval is just 1) cheating and 2) so dangerous I wouldn't recommend it at all.

Anyway, you could scan global then local variables using global & local dicts in a list comprehension:

l_1 = ['a','b','c']
l_2 = ['e','f','g']
l_3 = ['g','h','i']
l_all = ['l_1','l_2','l_3']

l_all = [globals().get(x,locals().get(x)) for x in l_all]

result:

[['a', 'b', 'c'], ['e', 'f', 'g'], ['g', 'h', 'i']]

globals().get(x,locals().get(x)) is a quick & dirty code to first look in global vars & fallback to local vars if not found. It could be overcomplicated for your needs.

omg life savers!!!!! thanks a lot!!!!! – kingslucifer Mar 17 '18 at 17:32 — kingslucifer, Mar 17 '18 at 17:32

Ajax1234 · Answer 2 · 2018-03-17T17:22:46.750

2

You can use a dictionary to store the list names and each associated list:

d = {'l_2': ['e', 'f', 'g'], 'l_3': ['g', 'h', 'i'], 'l_1': ['a', 'b', 'c']}
l_all = ['l_1','l_2','l_3']
final_results = [d[i] for i in l_all]

Output:

[['a', 'b', 'c'], ['e', 'f', 'g'], ['g', 'h', 'i']]

However, to actually access the lists via variable name, you would have to use globals:

l_1 = ['a','b','c']
l_2 = ['e','f','g']
l_3 = ['g','h','i']
l_all = ['l_1','l_2','l_3']
new_l = [globals()[i] for i in l_all]

Output:

[['a', 'b', 'c'], ['e', 'f', 'g'], ['g', 'h', 'i']]

edited Mar 17 '18 at 17:22

answered Mar 17 '18 at 17:09

Ajax1234

69,937
8
61
102

not my downvote. But your first version doesn't answer the question. OP has data as he posted. And also you copied the typo from OP (that I just fixed). so none of your solutions work. – Jean-François Fabre Mar 17 '18 at 17:16
can you elaborate why using `globals()` is not recommended? – Jean-François Fabre Mar 17 '18 at 17:19
@Jean-FrançoisFabre My understand of `globals()` is that accessing `globals()['something']` with new data could produce unintended output if `'something` is actually intended to give different result, but when passed to `globals()` returns something entirely different. – Ajax1234 Mar 17 '18 at 17:25
you mean like `1_1 `? :) – Jean-François Fabre Mar 17 '18 at 17:26
Thanks lot!!!!! I tried this one and it works as well. But I decided to use the one from @Jean-FrançoisFabre too keep my code short. Many thanks again! – kingslucifer Mar 17 '18 at 17:34
note that Ajax solution is actually shorter :) – Jean-François Fabre Mar 17 '18 at 17:37

score 1 · Answer 3 · edited Jun 20 '20 at 09:12

1

Then say I have another list of:

l_all = ['l_1','l_2','l_3']

Let's say you don't create this using strings, and use the list variables directly

You can then get the wanted output

l_all = [l_1,l_2, l_3]

edited Jun 20 '20 at 09:12

Community

1
1

answered Mar 17 '18 at 17:26

OneCricketeer

179,855
19
132
245

oh, cuz i have a lot of those lists and those are defined by users. I want to keep them more organized so I used a loop to create l_all. – kingslucifer Mar 17 '18 at 17:36
Okay, so something like `l_all = [user.get_list() for user in users]` would not be possible? – OneCricketeer Mar 17 '18 at 17:37
My point here is that "numbered variables" are a sign that you need a better design. For example, you can use a SQLite database for storing user input rather than just an in-memory list. Then you can query all rows of the database to get a list of all lists – OneCricketeer Mar 17 '18 at 17:39
yeah i totally agree on this. I know its dumb...this project is for my company and theres data control sigh – kingslucifer Mar 17 '18 at 17:49

score 1 · Answer 4 · answered Mar 17 '18 at 17:40

1

You can simply do:

l_1 = ['a','b','c']
l_2 = ['e','f','g']
l_3 = ['g','h','i']
l_all = ['l_1','l_2','l_3']

print(list(map(lambda x:globals()[x],l_all)))

output:

[['a', 'b', 'c'], ['e', 'f', 'g'], ['g', 'h', 'i']]

answered Mar 17 '18 at 17:40

Aaditya Ura

12,007
7
50
88

score 0 · Answer 5 · answered Mar 17 '18 at 17:11

You can use locals()

l_1 = ['a','b','c']
l_2 = ['e','f','g']
l_3 = ['g','h','i']

l_all = ['l_1', 'l_2', 'l_3']

all_local_variables = locals()
l_all_values = [all_local_variables[i] for i in l_all]

And you should be aware that you can get KeyError if no such variable present in current scope, for that you can all_local_variables.get(i), that will return None if not present or set default as all_local_variables.get(i, 'default')

rth · Answer 6 · 2018-03-17T17:28:16.210

0

The simplest way to do this is to evaluate your strings in l_all. Something like this

>>> l_1 = ['a','b','c']
>>> l_2 = ['e','f','g']
>>> l_3 = ['g','h','i']
>>> l_all = ['l_1','l_2','l_3']
>>> [ eval(x) for x in l_all ]

The output is

[['a', 'b', 'c'], ['e', 'f', 'g'], ['g', 'h', 'i']]

edited Mar 17 '18 at 17:28

answered Mar 17 '18 at 17:18

rth

2,946
1
22
27

1

Why the downvote? It is the exact answer to the question! Please leave a comment if you downvote. – rth Mar 17 '18 at 17:26

Convert a list of list's names(str) into a list of lists in python

6 Answers6