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I want to pass some data to another python script and do something there. But the data i send conflicts if i run script multiple times at the same time with different arguments. How do i separate them?

Example code:

main.py

import otherscript

list_a = [1,2,3] # from arguments
otherscript.append_to_another_list(list_a)

otherscript.py

another_list = []
def append_to_another_list(list):
    another_list.append(list)
    print(another_list)

if i run main.py twice at the same time with arguments 1,2,3 and 4,5,6 it prints both of them in the same list like [1,2,3,4,5,6]. I hope i made this clear

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9
  • What do you mean at the same time? How do you run it at the same time? Commented Apr 27, 2019 at 12:34
  • i run main.py -l 123 and then main.py -l 456 on another terminal window, and second one prints both of them Commented Apr 27, 2019 at 12:36
  • 1
    what is -l flag ? You should have included that in your main.py example Commented Apr 27, 2019 at 12:37
  • @DeveshKumarSingh it's very hard to include actual code here so i wanted to simplify question. Main issue is otherscript 's another_list variable is shared between instances and it should be independent for each main.py file is running which is a infinite loop in my example Commented Apr 27, 2019 at 12:45
  • @ggnoredo another_list.append(list) this code appends new list into 'another_list`. Could you let me know more detail what you want to solve? Commented Apr 27, 2019 at 12:51

2 Answers 2

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3

Of you invoke this twice from the OS command line - say, bash - you would expect them to be totally independent, not showing the behaviour the OP describes.

On the other hand, within a single Python interpreter, a module is only initialised the once, so the list in your otherscript module (which is a module rather than a script) will stick around, and keep being appended to.

In any case, perhaps your best option for finer control would be a class.

class ListKeeper:
    def __init__(self):
        self.another_list = []

    def append_to_another_list(self, list):
        self.another_list.append(list)
        print(another_list)

Your main.py would look like:

import otherscript

list_a = [1,2,3] # from arguments
keeper1 = otherscript.ListKeeper()
keeper1.append_to_another_list(list_a)

You can create as many instances as you need, all independent of one another, and all keeping their own state.

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Comments

1

I just simplified your main.py as follows

import otherscript

import sys
list_a = [int(item) for item in sys.argv[1:]]
otherscript.append_to_another_list(list_a)

And then when I run them together using python3.7 main.py 1 2 3 && python3.7 main.py 4 5 6 I get the output

[[1, 2, 3]]
[[4, 5, 6]]

In addition, if you open the same terminal and run the append_to_another_list command twice, the output will change, since you are referring to the same list!

In [2]: import otherscript                                                      

In [3]: otherscript.append_to_another_list([1,2,3])                             
[[1, 2, 3]]

In [4]: otherscript.append_to_another_list([4,5,6])                             
[[1, 2, 3], [4, 5, 6]]
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2 Comments

I agree, running the OP's code as shown from the command line behaves like this. I wonder if they are running their code some other way, not from the raw command line, such that both are within a single Python process.
Yes, you can never run two commands at the same terminal, as the OP states in the comments as well!

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