How to execute a file within the python interpreter?

The Question :

308 people think this question is useful

I’m trying to execute a file with python commands from within the interpreter.

EDIT: I’m trying to use variables and settings from that file, not to invoke a separate process.

The Question Comments :
  • os module, and I’m checking the other answers.
  • Suggested os.system; And deleted answer when I read the EDIT :-/

The Answer 1

248 people think this answer is useful

Several ways.

From the shell

python someFile.py

From inside IDLE, hit F5.

If you’re typing interactively, try this: (Python 2 only!)

>>> variables= {}
>>> execfile( "someFile.py", variables )
>>> print variables # globals from the someFile module

For Python3, use:

>>> exec(open("filename.py").read())

The Answer 2

259 people think this answer is useful

For Python 2:

>>> execfile('filename.py')

For Python 3:

>>> exec(open("filename.py").read())
# or
>>> from pathlib import Path
>>> exec(Path("filename.py").read_text())

See the documentation. If you are using Python 3.0, see this question.

See answer by @S.Lott for an example of how you access globals from filename.py after executing it.

The Answer 3

102 people think this answer is useful

Python 2 + Python 3

exec(open("./path/to/script.py").read(), globals())

This will execute a script and put all it’s global variables in the interpreter’s global scope (the normal behavior in most scripting environments).

Python 3 exec Documentation

The Answer 4

52 people think this answer is useful

Surprised I haven’t seen this yet. You can execute a file and then leave the interpreter open after execution terminates using the -i option:

| foo.py |
----------
testvar = 10

def bar(bing):
  return bing*3

--------



$ python -i foo.py
>>> testvar 
10
>>> bar(6)
18

The Answer 5

30 people think this answer is useful

I’m trying to use variables and settings from that file, not to invoke a separate process.

Well, simply importing the file with import filename (minus .py, needs to be in the same directory or on your PYTHONPATH) will run the file, making its variables, functions, classes, etc. available in the filename.variable namespace.

So if you have cheddar.py with the variable spam and the function eggs – you can import them with import cheddar, access the variable with cheddar.spam and run the function by calling cheddar.eggs()

If you have code in cheddar.py that is outside a function, it will be run immediately, but building applications that runs stuff on import is going to make it hard to reuse your code. If a all possible, put everything inside functions or classes.

The Answer 6

15 people think this answer is useful

From my view, the best way is:

import yourfile

and after modifying yourfile.py

reload(yourfile)   

or

import imp; 
imp.reload(yourfile) in python3

but this will make the function and classes looks like that: yourfile.function1, yourfile.class1…..

If you cannot accept those, the finally solution is:

reload(yourfile)
from yourfile import *

The Answer 7

11 people think this answer is useful

Just do,

from my_file import *

Make sure not to add .py extension. If your .py file in subdirectory use,

from my_dir.my_file import *

The Answer 8

9 people think this answer is useful

I am not an expert but this is what I noticed:

if your code is mycode.py for instance, and you type just ‘import mycode’, Python will execute it but it will not make all your variables available to the interpreter. I found that you should type actually ‘from mycode import *’ if you want to make all variables available to the interpreter.

The Answer 9

9 people think this answer is useful

For Python 3:

>>> exec(open("helloworld.py").read())

Make sure that you’re in the correct directory before running the command.

To run a file from a different directory, you can use the below command:

with open ("C:\\Users\\UserName\\SomeFolder\\helloworld.py", "r") as file:
    exec(file.read())

The Answer 10

8 people think this answer is useful

For python3 use either with xxxx = name of yourfile.

exec(open('./xxxx.py').read())

The Answer 11

1 people think this answer is useful

Supposing you desire the following features:

  1. Source file behaves properly in your debugger (filename shows in stack, etc)
  2. __name__ == '__main__' is True so scripts behave properly as scripts.

The exec(open('foo.py').read()) fails feature 1 The import foo strategy fails feature 2

To get both, you need this:

    source = open(filename).read()
    code = compile(source, filename, 'exec')
    exec(code)

The Answer 12

0 people think this answer is useful
python -c "exec(open('main.py').read())"

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