Call another file in python code

Question

I have written the following code to create a gui in python and then using lambda i am calling another file. This code is showing some indent problem . roshan/pre.sh is the path of shell script from where I am calling another python file. I want to create multiple buttons like this and using a function I will call a different shell script.

There should be 5 buttons (named GRAPH1, GRAPH2, GRAPH3, GRAPH4, GRAPH5) and on each button click I wish to load another file where i will put my graphs.

For example, when I click on the GRAPH1 button another file which i have named image1.py will load and will show graphs.

#!/usr/bin/python

import Tkinter, subprocess

class simpleapp_tk(Tkinter.Tk):
    def __init__(self,parent):
        Tkinter.Tk.__init__(self,parent)
        self.parent = parent
        self.initialize()

    def initialize(self):
        self.grid()

    clusterB = Tkinter.Button(self, text ="preprocessing",width=13,font = "Georgia 10 bold")
    clusterB.grid(row=5,sticky=Tkinter.W,padx=3)
    clusterB.config(command = lambda:clusters())

    def clusters():
    process = subprocess.Popen(["roshan/pre.sh"],shell=False,
        stdout=subprocess.PIPE,stderr=subprocess.STDOUT)
    while True:
        out = process.stdout.readline()
        if out == '' and process.poll() is not None:
            break
        print out

    return
if __name__ == "__main__":
    app = simpleapp_tk(None)
    app.title('Results and Graphs')
    app.mainloop()

Please edit the question to include this information (and then we can all delete our comments) — holdenweb
– holdenweb, Commented Sep 2, 2015 at 7:55
@BobMarshall currently i am getting this error Traceback (most recent call last): File "./graph.py", line 61, in <module> app = simpleapp_tk(None) File "./graph.py", line 10, in init self.initialize() File "./graph.py", line 23, in initialize out = process.stdout.readline() NameError: global name 'process' is not defined — Roshan Bagdiya
– Roshan Bagdiya, Commented Sep 2, 2015 at 9:09
why are you using lambda? It's not related to the problem, but it's unnecessary extra code. — Bryan Oakley
– Bryan Oakley, Commented Sep 2, 2015 at 11:19

Bob Marshall · Accepted Answer · 2015-09-02 09:20:55Z

2

Instead of using a shell to open a python file, you can always just import it if it's in the same directory and use it just like a module.

import image1
thing = image1.function_in_image1()

You can also choose to run it directly by using os.

import os
os.system(r"C:\path\to\image1.py")

In your example, I found indenting problems. Change this:

def clusters():
process = subprocess.Popen(["roshan/pre.sh"],shell=False,
    stdout=subprocess.PIPE,stderr=subprocess.STDOUT)
while True:
    out = process.stdout.readline()
    if out == '' and process.poll() is not None:
        break
    print out

return

to this:

def clusters():
    process = subprocess.Popen(["roshan/pre.sh"],shell=False,
        stdout=subprocess.PIPE,stderr=subprocess.STDOUT)
    while True:
        out = process.stdout.readline()
        if out == '' and process.poll() is not None:
            break
        print out

    return

edited Sep 2, 2015 at 9:20

answered Sep 2, 2015 at 9:08

Bob Marshall

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