Access variables through their name as string in Python

Question

I would like to run a function with a list parameter :

param = ['a', 'b', 'c']
myFunction(param)

But a, b and c must be dynamic so I would like to remove the quotes to get this :

param = [a, b, c]
myFunction(param)

How can I do that ?

PS : I tried the solutions of this question but it didn't work Removing quotation marks from list items

Are you saying that `a`, `b` and `c` are variables that are defined in your code, and `param` should be a list containing those 3 variables? — Aran-Fey, Oct 04 '18 at 12:37
Please rephrase your question to avoid misleading future readers. What you need is to evaluate variables using their names, not remove quotes — Ivaylo Strandjev, Oct 04 '18 at 12:43
your answer is [here](https://stackoverflow.com/a/5445983/4092588) — Antonino, Oct 04 '18 at 12:45
Sounds like you want a dictionary like `{'a': 3, 'b': 2}`, not separate variables `a` and `b`. — chepner, Oct 04 '18 at 12:51

norok2 · Accepted Answer · 2018-10-04T12:48:46.007

3

If I understood correctly, you want to actually call myFunction with parameters a, b and c (assuming these are actually being defined earlier).

One way of doing this is:

a, b, c = 1, 2, 3
param = ['a', 'b', 'c']

def myFunction(x, y, z):
    return x + y + z

myFunction(*[globals()[s] for s in param])
# returns: 6

EDIT:

As people have suggested, a safe way of doing this would be through globals() (or locals() using a slightly different construct). Otherwise, you could use eval() which is more powerful, but also potentially less safe.

edited Oct 04 '18 at 12:48

answered Oct 04 '18 at 12:42

norok2

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*Please* don't use `eval` when `globals` would be sufficient. Beginners will copy your `eval` without thinking and open themselves up to code injection. – Aran-Fey Oct 04 '18 at 12:44
In my case, there is no code injection issue, good for me ;) – Charles R Oct 04 '18 at 12:46
@CharlesR this still is very bad practice, have a look at the duplicate, use `locals()`. It is exactly what it is meant for – Olivier Melançon Oct 04 '18 at 12:48
@Aran-Fey you are right, I just typed faster than I should have. @OlivierMelançon `locals()` would not work inside the list comprehension. – norok2 Oct 04 '18 at 12:51

vash_the_stampede · Answer 2 · 2018-10-04T12:49:40.137

1

'str''s cannot be converted to variable names, however if you do have th variables and your goal is to pass each variable to a function this is possible

def addition(x):
    return x + 1

a = 1 b = 2 c = 3

param = [a, b, c] 
print([addition(i) for i in param])
# [2, 3, 4]

The entire list could be passed as well like so

def addition(x):
    return sum(x)

a = 1
b = 2
c = 3

param = [a, b, c]
print(addition(param))
# 6

edited Oct 04 '18 at 12:49

answered Oct 04 '18 at 12:43

vash_the_stampede

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It is wrong that a string cannot be converted to variable. You can use `locals()['a']`. Although, I agree OP just should not rely on strings – Olivier Melançon Oct 04 '18 at 12:46
@OlivierMelançon noted, every lookup I did involved a tedious process to convert a `str` to variable name will look into this, ty – vash_the_stampede Oct 04 '18 at 12:48

Access variables through their name as string in Python

2 Answers2

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