Python – Invoke python script through make command

makepython

I have a tool with 5 python scripts say A.py, B.py, C.py, D.py and E.py. I have to run this program in Linux. A.py is the one program I run which imports other files and uses them. For running A.py, I use the command

$ python A.py

Until now the input and output paths are hardcoded in A.py. Now I am supposed to write a makefile with all the file structure and dependencies. This makefile should also include the command for executing python scrip A.py. The required input and output paths has to be understood when make makefile command is invoked and the tool should start running. How to do this?

EDITED :

The makefile should be something like this holding all these details.

INPUT_PATH = list of all input paths
FILES = list of .c files located in the above specified paths
OUTPUT_PATH = output path where generated file has to be stored
command to execute python scrip:
   A.py inputpath+filename outputpath

so whenever there a new input .c files, they will be added to the list in makefile. Any sample makefile of this kind that you could suggest?

Best Answer

In python you get the commandline parameters from sys.argv so instead of

#!/usr/bin/env python
base_dir = /hard/coded/path

you do:

#!/usr/bin/env python
import sys
base_dir = sys.argv[1]

(putting in checks might be appropriate). If you explicitly start the program with python, you can leave out the first line, but it won't hurt to be there in case you make A.py executable, you will need it.

In your makefile you now can specify:

default:
        python A.py /hard/coded/path

(make sure you have a Tabbefore python, not 8 spaces.

And then you can run make (no need to do make makefile unless you have a Makefile or with Gnu make a GNUmakefile in the same directory)

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To write this Pseudocode with Regex

If I understand correctly, you're looking for something like this (in bash):

#!/usr/bin/env bash

## avoid errors if a directory has no *tex files
shopt -s nullglob

directories=("Cardiology" "Rheumatology" "Surgery");

## Change this to set whichever options you want.
printf "%s\n%s\n" "\documentclass{YOURCLASS}" "\begin{document}"

for directory in ${directories[@]}
do
    ## Reset the counter, avoid empty sections.
    c=0;
    for file in "$directory"/*tex
    do
        let c++
        [ "$c" -eq 1 ] && printf "\n%s\n" "\section{$directory}"
        ## Extract the wanted lines
        perl -lne '$a=1 && print "" if /\\begin{question}/; 
                  print if $a==1;
                  $a=0 if /\\end{question}/;' "$file" 
        echo ""
    done
done
echo "\end{document}"

If you run that script from the directory that contains Cardiology etc, it should provide output like this:

Shell – pass arguments to python program in shell

From the SyntaxError you have pasted, it sounds like you are running the script under the wrong version of python, specifically Python 2 instead of Python 3 (because the SyntaxError may be complaining about the annotations, which only exist in Python 3).

If you read the error you actually got, you can see that this really doesn't have anything to do with passing arguments (the title of your question) at all.

The correct solution is to not invoke the script by prefixing it with the interpreter name, but to let it invoke its own interpreter:

./main.py 1.0 0.1 10

instead of

python main.py 1.0 0.1 10

If it's properly written, the script will have a shebang line at the start that specifies what interpreter it needs to run under, and that will be python3 or something similar.

By letting the script specify its own interpreter, the calling script need not know or care which interpreter is required. The script can even be rewritten to use a different language altogether (or it can be replaced with a compiled binary that isn't a script at all) and the callers don't have to be modified to call it differently.

Best Answer

Related Solutions

To write this Pseudocode with Regex

Shell – pass arguments to python program in shell

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