How execute a pipeline in bash using a python wrapper? - bash

Question

Scenario:
I have a pipeline in a bash script and a list of some process along with their arguments. I want to run a python script after the execution of each process (executable) in the pipeline if the process is in my list.
(I use Python 2.7)
My proposed solution:
Using a python wrapper script. I have replaced all executable in the pipeline with my custom python script which:
1) checks the process if is in the list then FLAG=True
2) execute the process by the original executable using subprocess.Popen(process.command, shell=True).communicate()
3) if FLAG==True then do something.
Problem:
Using the current solution when I run the process using
subprocess.Popen().communicate(), the processes will execute separately
and they cannot get the output of inner process (child process) to the outer process (parent).
For example:
#!/bin/bash
Mean=`P1 $Image1 -M`
P2 "$Image2" $Mean -F
We have not output value of Mean in the second line execution.
Second line will execute like:
subprocess.Popen("P2 $Image2 \nP1 $Image1 -M -F" , shell=True).communicate()
Therefore, it returns an error!
Is there a better way in python to execute process like this?
Please let me know if there is any other suggestion for this scenario (I'm a very beginner in bash).

Answer 1

There's no need to use bash at all.
Assuming modern Python 3.x:
#!/usr/bin/env python
import subprocess
image1 = sys.argv[1]
image2 = sys.argv[2]
p1 = subprocess.run(['P1', image1, '-M'], check=True, capture_output=True)
p2 = subprocess.run(['P2', image2, p1.stdout.strip(), '-F'], check=True, capture_output=True)
print(p2_result.stdout)
See here that we refer to p1.stdout.strip() where we need the mean value in p2's arguments.