Tell if Python is in interactive mode

Question

In a Python script, is there any way to tell if the interpreter is in interactive mode? This would be useful so that, for instance, when you run an interactive Python session and import a module, slightly different code is executed (for example, logging is turned off).

I've looked at tell whether python is in -i mode and tried the code there, however, that function only returns true if Python has been invoked with the -i flag and not when the command used to invoke interactive mode is python with no arguments.

What I mean is something like this:

if __name__=="__main__":
    #do stuff
elif __pythonIsInteractive__:
    #do other stuff
else:
    exit()

Consider using the `logging` module for logging in libraries, so users can control it with a shared configuration. — Bachsau, Jun 17 '19 at 11:46

score 70 · Accepted Answer · answered Mar 01 '10 at 14:25

70

__main__.__file__ doesn't exist in the interactive interpreter:

import __main__ as main
print hasattr(main, '__file__')

This also goes for code run via python -c, but not python -m.

answered Mar 01 '10 at 14:25

Ignacio Vazquez-Abrams

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5

This is also the case in, for example, py2exe executables. – fmark Jul 13 '10 at 13:46
6

Unfortunately this doesn't work inside an embedded shell. I.e. a shell started with IPython.embed() – titusjan Oct 30 '15 at 11:27
9

`bool(getattr(sys, 'ps1', sys.flags.interactive))`, py2.6+ – Mr. B Dec 09 '17 at 23:15
1

So far none of the solutions seems to be able to detect `PYTHONSTARTUP=script.py python3`. The expression `'readline' in sys.modules` can, as well as `python3 -i script.py` and `python3 -ic 'import script'` (tested with Python 3.8.2 on Ubuntu focal), but it's highly non-portable. – photon.engine Jul 08 '20 at 05:27
This is the only answer that worked for `ptpython` interactive session started via `python -m ptpython`. – JamesThomasMoon Sep 11 '21 at 01:55

pyjamas · Answer 2 · 2020-10-25T12:53:19.750

50

I compared all the methods I found and made a table of results. The best one seems to be this:

hasattr(sys, 'ps1')

If anyone has other scenarios that might differ, comment and I'll add it

edited Oct 25 '20 at 12:53

answered Oct 25 '20 at 12:38

pyjamas

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Hey thank you for doing the hard work, would you be able to check Windows vs Linux vs Mac? – Fred Dufresne Feb 17 '22 at 22:48
`hasattr(sys, 'ps1')` returns `False` in PyCharm 2022.1.4 (CE) console on Windows 11 – N. Maks Oct 14 '22 at 08:32
I i forgot to mention that the interpreter is Python 3.10.6 – N. Maks Oct 14 '22 at 08:41

score 29 · Answer 3 · edited Jul 13 '14 at 16:19

29

sys.ps1 and sys.ps2 are only defined in interactive mode.

edited Jul 13 '14 at 16:19

Cristian Ciupitu

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answered Mar 01 '10 at 14:27

ChristopheD

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@Keith seems to work fine for me in iPython 3.2.1 REPL using python 2.7.9 – hobs Oct 09 '15 at 21:43
2

Agreed, I just checked in 4.0 and it works there as well. May have been an issue with older versions of IPython. – Keith Hughitt Oct 10 '15 at 11:28
Seems to me that `sys.ps1` and `sys.ps2` are defined even when iPython is run *not* in interactive mode. – Jasha Mar 01 '20 at 06:40

score 17 · Answer 4 · edited Jul 13 '14 at 16:26

17

Use sys.flags:

if sys.flags.interactive:
    #interactive
else:
    #not interactive

edited Jul 13 '14 at 16:26

Cristian Ciupitu

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answered Jul 29 '11 at 21:09

Pych

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This only checks for command line argument `python -i` and is not a test of being in the Python interactive mode by typing `python` alone. – the wolf Jun 12 '12 at 15:59
10

`bool(getattr(sys, 'ps1', sys.flags.interactive))` – Mr. B Dec 09 '17 at 23:13

score 7 · Answer 5 · answered Mar 01 '10 at 14:30

7

From TFM: If no interface option is given, -i is implied, sys.argv[0] is an empty string ("") and the current directory will be added to the start of sys.path.

If the user invoked the interpreter with python and no arguments, as you mentioned, you could test this with if sys.argv[0] == ''. This also returns true if started with python -i, but according to the docs, they're functionally the same.

answered Mar 01 '10 at 14:30

wersimmon

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Uh oh. Direct violation of the Zen of Python, then :) – Tim Pietzcker Mar 01 '10 at 14:33
Heh... Though I think @echoback's version is the only obvious(ish) one. I didn't accept this simply because in C et al., it is theoretically possible that `argv[0]` is `NULL` or an empty string and I don't really feel like debugging any potential errors caused by that... – Chinmay Kanchi Mar 01 '10 at 14:59
5

This may be problematic for other interpreters, however. For example, when using IPython, sys.argv = ['/usr/bin/ipython'] – Keith Hughitt Aug 03 '11 at 13:12

jdines · Answer 6 · 2014-07-13T16:19:45.713

1

The following works both with and without the -i switch:

#!/usr/bin/python
import sys
# Set the interpreter bool
try:
    if sys.ps1: interpreter = True
except AttributeError:
    interpreter = False
    if sys.flags.interactive: interpreter = True

# Use the interpreter bool
if interpreter: print 'We are in the Interpreter'
else: print 'We are running from the command line'

edited Jul 13 '14 at 16:19

answered Jul 13 '14 at 16:13

jdines

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`if sys.ps1: interpreter = True` => `interpreter = sys.ps1` or `interpreter = bool(sys.ps1)`. – Cristian Ciupitu Jul 13 '14 at 16:25
@CristianCiupitu: You might want to actually test your code before you post it. Even if it was valid Python it would throw an AttribeError exception when run from the command line. – jdines Jul 13 '14 at 18:02
Did I recommend removing the `try ... except` statement? I only recommended replacing an `if` with a plain assignment. – Cristian Ciupitu Jul 13 '14 at 18:10
@CristianCiupitu: You made no such recommendation. Apparently you thought that the => communicated a whole lot more than it actually does. Of course, once one understands what you meant to say rather than what you said, your solution is longer and more compute intensive, but whatever floats your boat I suppose. – jdines Jul 13 '14 at 18:15
1

My bad for being elliptic, but I fail to understand how 27 characters are longer than 30 or why is it more compute intensive when `if` also needs the boolean value of `sys.ps1`. – Cristian Ciupitu Jul 13 '14 at 18:21
I mistook the last "or" as a Python statement rather than an English word, so as you say "my bad" on that. Now that I see that, your first proposal will assign ">>> " to sys.ps1 thereby causing a cast every time you use it, and also introducing an inconsistency where the type of "interpreter" is different depending upon runtime mode. On my system the pyc file for my solution is 3 bytes smaller than your second solution: You can try it yourself to verify, obviously. – jdines Jul 13 '14 at 19:40
You're right about that inconsistency, that's why I also proposed a second variant. I have python-2.7.5-13.fc20.x86_64 and the corresponding **pyc** is indeed bigger with 5 bytes for the `bool` variant. But if I also change the `except` code to `interpreter = bool(sys.flags.interactive)`, the file is smaller with 27 bytes :-D – Cristian Ciupitu Jul 13 '14 at 20:28
2

NECROMANCY!: `bool(getattr(sys, 'ps1', sys.flags.interactive))` – Mr. B Dec 09 '17 at 23:11

score -4 · Answer 7 · answered Jul 07 '11 at 18:03

Here's something that would work. Put the following code snippet in a file, and assign the path to that file to the PYTHONSTARTUP environment variable.

__pythonIsInteractive__ = None

And then you can use

if __name__=="__main__":
    #do stuff
elif '__pythonIsInteractive__' in globals():
    #do other stuff
else:
    exit()

http://docs.python.org/tutorial/interpreter.html#the-interactive-startup-file

Tell if Python is in interactive mode

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