Execute a file with arguments in Python shell

Question

I would like to run a command in Python Shell to execute a file with an argument.

For example: execfile("abc.py") but how to add 2 arguments?

Answer 1

try this:

import sys
sys.argv = ['arg1', 'arg2']
execfile('abc.py')

Note that when abc.py finishes, control will be returned to the calling program. Note too that abc.py can call quit() if indeed finished.

Answer 2

Actually, wouldn't we want to do this?

import sys
sys.argv = ['abc.py','arg1', 'arg2']
execfile('abc.py')

Answer 3

execfile runs a Python file, but by loading it, not as a script. You can only pass in variable bindings, not arguments.

If you want to run a program from within Python, use subprocess.call . Eg

import subprocess
subprocess.call(['./abc.py', arg1, arg2])

Answer 4

import sys
import subprocess

subprocess.call([sys.executable, 'abc.py', 'argument1', 'argument2'])

Answer 5

For more interesting scenarios, you could also look at the runpy module. Since python 2.7, it has the run_path function. Eg:

import runpy
import sys

# argv[0] will be replaced by runpy
# You could also skip this if you get sys.argv populated
# via other means
sys.argv = ['', 'arg1' 'arg2']
runpy.run_path('./abc.py', run_name='__main__')

Answer 6

You're confusing loading a module into the current interpreter process and calling a Python script externally.

The former can be done by import ing the file you're interested in. execfile is similar to importing but it simply evaluates the file rather than creates a module out of it. Similar to "sourcing" in a shell script.

The latter can be done using the subprocess module. You spawn off another instance of the interpreter and pass whatever parameters you want to that. This is similar to shelling out in a shell script using backticks.

Answer 7

You can't pass command line arguments with execfile() . Look at subprocess instead.

Answer 8

If you set PYTHONINSPECT in the python file you want to execute

[repl.py]

import os
import sys
from time import time 
os.environ['PYTHONINSPECT'] = 'True'
t=time()
argv=sys.argv[1:len(sys.argv)]

there is no need to use execfile , and you can directly run the file with arguments as usual in the shell:

python repl.py one two 3
>>> t
1513989378.880822
>>> argv
['one', 'two', '3']

Answer 9

If you want to run the scripts in parallel and give them different arguments you can do like below.

import os
os.system("python script.py arg1 arg2 & python script.py arg11 arg22")

Answer 10

This works for me :

import subprocess
subprocess.call(['python.exe', './abc.py', arg1, arg2])

Answer 11

Besides subprocess.call , you can also use subprocess.Popen . Like the following

subprocess.Popen(['./script', arg1, arg2])

Answer 12

这有效：

subprocess.call("python abc.py arg1 arg2", shell=True)

Answer 13

runfile('abc.py', ['arg1', 'arg2'])