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Here is my script:

# I have 100 variables
x0 = 3.14
x1 = 2.72
x2 = 1.41
x3 = 2.33
.... (omit this part)
x100 = 7.77

# xi corresponds to the value that the index i of a list needs to subtract, 
# now I want to loop through the list
for i in range(100):
    lst[i] -= 'x{}'.format(i)

This clearly won't work, since the variable is not a string. So how should I string formatting a variable?

  • You should seriously rethink your design. Why having 100 variables, when you can have a single variable referring to a list and access each value by index? Even if it's generated code, then better generate a list! – Óscar López May 09 '20 at 12:12
  • I understand what you are saying. But this is an over-simplified version. In my real case, I need to use this string-formatting (if possible) trick to avoid an additional for loop. If there are better solutions, I would not ask this question. –  May 09 '20 at 12:32
  • Yes, you can be sure there are better solutions. And what's wrong with having an additional `for` if it helps you avoid having 100 variables? This looks like an [XY problem](https://meta.stackexchange.com/questions/66377/what-is-the-xy-problem), you're asking something to get around a problem, but the real issue is different - and we don't know all of the context to be able to effectively help you. – Óscar López May 09 '20 at 12:38
  • I understand this may look like an XY problem, but I am acutually a computational chemist and I am using some Python packages particularly for my field. It doesn't make any sense if I put my actual script here. –  May 09 '20 at 13:06

3 Answers3

1

You should instead use a list here.

x = [...] (where x has a len of 100)

Then for your loop:

for i in range(100):
    lst[i] -= x[i]

(Renamed list to lst to avoid name collision with the built-in type)

Mario Ishac
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    This is not the answer I am looking for. This script is oversimplified, and for my real case, the ```x1``` to ```x100``` are also all variables, and is generated on-the-fly. So basically my purpose is to update the ```list[i]``` AS SOON AS the ```xi``` is generated. –  May 09 '20 at 12:07
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    if the variables are dynamically generated and assigned, and you want to update `list[i]` as soon as the corresponding x value is generated, you should look into using iterators or generators. Point is, having this many variables when you could have just one containing/generating all the values is going to be very hard to maintain. – Mario Ishac May 09 '20 at 12:10
  • The OP is asking for a technique to solve their specific problem. @Arhiliuc Cristina gave a good answer. Whether using ‘eval()‘ is a good idea is a different story. – Janos May 09 '20 at 12:43
1

You can access these variables using locals:

lst[i] -= locals()['x{}'.format(i)]
match
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    This ```locals()``` function works fine. I will accept your answer since your answer is not finding another way around and don't use ```eval()``` which seems to be a bad practice. –  May 09 '20 at 13:16
1

In order to get the value of the variable, you can use Python's eval function

eval('x{}'.format(i))

And please, don't ever call your list variable list.

Edit: While this solution works in this case, it is recommended to avoid eval as much as possible because it allows code injections in a way you wouldn't expect.

Arhiliuc Cristina
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