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I need to determine whether the shell which invoked my Python script was in interactive mode or not. If it was in interactive mode, the program should pipe output to less(1) for easy reading. If not, it should simply print its output to stdout, to allow it to be piped away to a printer, file, or a different pager.

In a shell script, I would have checked if the prompt variable $PS1 was defined, or looked for the -i option among the flags stored in the $- variable.

What is the preferred method for testing interactivity from within Python?

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5 Answers 5

Reset to default
49

This is often works well enough

import os, sys
if os.isatty(sys.stdout.fileno()):
    ...
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7 Comments

sys.stdout.isatty() is shorter.
And sys.__stdin__.isatty() more reliable. (not affected by stdin redirect and allows piping out)
@Evpok: The OP asked for stdout, stdin is totally unrelated. In order to find out whether to page output or not, stdin is really the wrong stream to check, because the user can pipe something into the program (thus redirection stdin and disconnecting it from the tty) and still want to see the output in a pager (because stdout is not redirected).
@Evpok, do you mean sys.__stdout__.isatty(). It's possible for the output to be a tty when the input isn't, and the OP seems to be interested in the output
@lunaryom Yes, I know. But he did write "If it was in interactive mode, the program should pipe output to less(1) for easy reading" for me it implies that he checks stdin and redirects stdout according to the nature of it.
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2

From this link you can use the same way and test if stdin is associated to a terminate(tty), you can do this using os.isatty(), example:

>>> os.isatty(0)
True

N.B: From the same link this will fails when you invoke the command remotely via ssh, the solution given is to test if stdin is associated to a pipe.

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5 Comments

I prefer the more obvious sys.__stdin__.isatty()
@Evpok: Yes agree, i forgot about the file object method isatty(), thanks :)
This seems to be equivalent to gnibbler's answer, except that the reader must know enough unix to know that file 0 is stdin. Is there a difference between checking isatty() on stdin or stdout, do you think?
@jforberg: yes agree, and i think the best one is @Evpok comment, sys.__stdin__.isatty(),and AFAIK interactive mode mean stdin is associate to terminal not stdout, correct me if i'm wrong :)
Yes, Evpok's solution is the most elegant and readable in my eyes. Thanks!
-5

If you already have a dependency on matplotlib, or you don't mind introducing one, you can always just call matplotlib.is_interactive()

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1 Comment

Look at line 1049 of github.com/matplotlib/matplotlib/blob/master/lib/matplotlib/… . is_interactive() just reads a global variable in matplotlib. I see no guarantees that it would refer to the interactivity of the parent shell.
-5 
if sys.flags.interactive:
    #interactive
else:
    #not interactive

http://docs.python.org/library/sys.html#sys.flags

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1 Comment

This only results in true if Python was started with the -i option. Starting it with the -i option will drop Python into interpreter mode after the script is run. sys.flags.interactive can not be used to determine the nature of the current shell environment.
-9

I make a cover class for testing.

For example you have :

class SuperInteractiveClass(object):
   def get_data_from_stdin(self):
      '... a lot of code here ...'
   '... and a lot of other function'

I make a second class, just for testing

class TestSuperInteractiveClass(SuperInteractiveClass):
    prepared_data = []
    def add_prepared_data(self,data):
        self.prepared_data.append(data)
    def get_data_from_stdin(self):
          return self.prepared_data.pop(0)
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2 Comments

I don't understand. What's the point with this? How does this check interactivity of invoker?
It doesn't, at least not in any way I can understand. It seems he has implemented a stack.

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