Python : pass variable name as argument

Question

I have a function f(x) in which many local variables are created. x is a string with the same name as one of these local variables and I would like to change this local variable by changing x. What is the clean way to do this? Currently I am using a lot of if/elif statements.

Some dummy code to represent my problem:

def f(x):
         a = [1,2,3]
         b = [2,3,4]
         c = [3,4,5]
         if x == "a":
                     a[0] = 10
         elif x == "b":
                     b[0] = 10
         elif x == "c":
                     c[0] = 10
         return a,b,c

I would like for the right variable to change value but using all these if/elif statements feels a bit redundant.

Can you be more specific - e.g. post some code. Are you saying you want a variable inside f to take a value you pass in - but it depends on a string? Is this heading towards key word args? — doctorlove
– doctorlove, Commented Sep 26, 2017 at 14:41
why do you need 2 variables with the same name with different purposes??? or i did not understand your question - please copy your code to the question — ddor254
– ddor254, Commented Sep 26, 2017 at 14:41

hyperneutrino · Accepted Answer · 2017-09-26 14:48:38Z

5

Use a `dict`

Simply, use a dict:

def f(x):
    values = {
        "a": [1,2,3],
        "b": [2,3,4],
        "c": [3,4,5]
    }
    values[x][0] = 10
    return values["a"], values["b"], values["c"]

If you really really want, use your original code and do locals()[x][0] = 10 but that's not really recommended because you could cause unwanted issues if the argument is the name of some other variable you don't want changed.

answered Sep 26, 2017 at 14:48

hyperneutrino

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Comments

ddor254 · Accepted Answer · 2017-09-26 14:49:57Z

4

use dictionary like this:

def f(x):
     d = {"a" :[1,2,3],"b" : [2,3,4],"c" : [3,4,5]}
     d[x][0] = 10

     return d

answered Sep 26, 2017 at 14:49

ddor254

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Comments

zipa · Accepted Answer · 2017-09-26 14:43:43Z

1

If your input is a string and you want to refer to a variable that matches that string you can use globals() like this:

globals()['x']

This way you can get the value and/or edit its contents.

answered Sep 26, 2017 at 14:43

zipa

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4 Comments

Kevin Over a year ago

Ok, so, 1) using globals() is a bad idea 99.999% of the time, and 2) you don't know whether x is a name at the global scope anyway. what if x is first created inside a main function, which then calls f? globals() won't have an 'x' entry in that case.

zipa Over a year ago

How else is he going to access the variable, except maybe using locals()? Sure he can rewrite the code, but this answer is for this specific situation.

Chris_Rands Over a year ago

This is a classic X-Y problem, you answer their question too literally, use a dict instead

Kevin Over a year ago

I guess it's a moot point now, since OP's edited post clarifies that it really doesn't have anything to do with differently scoped variables.

B 7 · Accepted Answer · 2017-09-26 14:52:25Z

-2

You can use exec() and put your code in argument as a string

def f(x):
    a = [1,2,3]
    b = [2,3,4]
    c = [3,4,5]
    exec(x + "[0] = 10")

    return a,b,c

print f("a")
# ([10, 2, 3], [2, 3, 4], [3, 4, 5])

answered Sep 26, 2017 at 14:52

B 7

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1 Comment

Guillaume Over a year ago

Sure you can, but you shouldn't. exec is both a security hazard and a performance hog. See for example stackoverflow.com/questions/2220699/…

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