如何在pypy中的.py文件中打开python中的.py文件？

Question

My program currently consists of 2 .py files.

I run the main part of the code in pypy (which is much faster) and then I open a second file in python that plots my data using matplotlib.pyplot .

I have managed to open the using:

subprocess.Popen(['C:\\python26\\python.exe ','main_plot.py',])

which opens my second file...

import matplotlib.pyplot as pyplot
def plot_function(NUMBER):
    '''some code that uses the argument NUMBER'''
    pyplot.figure()
    ---plot some data---
    pyplot.show()

However, I would like to be able to pass arguments to the plot_function that opens in python. Is that possible?

Answer 1

Yes, the Popen constructor takes a list of length n. See the note here. So just add the arguments to main_plot.py to your list:

subprocess.Popen(['C:\\python26\\python.exe ','main_plot.py','-n',1234])

EDIT (to respond to your edit):

You need to modify main_plot.py to accept a command line argument to call your function. This will do it:

import matplotlib.pyplot as pyplot
def plot_function(NUMBER):
    '''some code that uses the argument NUMBER'''
    pyplot.figure()
    ---plot some data---
    pyplot.show()

import argparse
if __name__=="__main__":
    argp=argparse.ArgumentParser("plot my function")
    argp.add_argument("-n","--number",type=int,default=0,required=True,help="some argument NUMBER, change type and default accordingly")
    args=argp.parse_args()
    plot_function(args.number)

Answer 2

最简单的方法是os.system("main_plot.py arg")