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Home/ Questions/Q 8683473
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T22:03:16+00:00 2026-06-12T22:03:16+00:00

I am writing a program that executes a set of tasks sequentially. Each task

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I am writing a program that executes a set of tasks sequentially. Each task is a function that outputs a new file, but any given task should not execute if the filename already exists. I find myself writing this kind of code over and over again:

task1_fname = task1()
# come up with filename for next task
task2_fname = "task2.txt"
if not os.path.isfile(task2_fname):
  # task2 might depend on task1's output, so it gets task1's filename
  task2(task1_fname)
task3_fname = "task3.txt"
if not os.path.isfile(task3_fname):
  task3(...)

The basic idea is that if a file is present (and ideally nonempty) then you should not execute the task that generates this file.

What’s the best Pythonic way to express this without having to write os.path.isfile calls everytime? Could decorators express this more concisely? Or something along the lines of:

with task2(task1_fname):
  # continue to next task

any ideas?

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  1. Editorial Team
    Editorial Team
    2026-06-12T22:03:17+00:00Added an answer on June 12, 2026 at 10:03 pm

    Are you looking for something like this?

    def preserve_output(f):
        def wrap(input, output):
            if not os.path.isfile(output):
                f(input, output)
        return wrap
    
    @preserve_output
    def task1(input, output):
        ...
    
    @preserve_output
    def task2(input, output):
        ...
    
    task1('input', 'output_1')
    task2('output_1', 'output_2')
    task3('output_2', 'output_3') etc
    
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