Python syntax for "if a or b or c but not all of them"

Question

If you mean a minimal form, go with this:

if (not a or not b or not c) and (a or b or c):

Which translates the title of your question.

UPDATE: as correctly said by Volatility and Supr, you can apply De Morgan's law and obtain equivalent:

if (a or b or c) and not (a and b and c):

My advice is to use whichever form is more significant to you and to other programmers. The first means "there is something false, but also something true", the second "There is something true, but not everything". If I were to optimize or do this in hardware, I would choose the second, here just choose the most readable (also taking in consideration the conditions you will be testing and their names). I picked the first.

How about:

conditions = [a, b, c]
if any(conditions) and not all(conditions):
   ...

Other variant:

if 1 <= sum(map(bool, conditions)) <= 2:
   ...

This question already had many highly upvoted answers and an accepted answer, but all of them so far were distracted by various ways to express the boolean problem and missed a crucial point:

I have a python script that can receive either zero or three command line arguments. (Either it runs on default behavior or needs all three values specified)

This logic should not be the responsibility of your code in the first place, rather it should be handled by argparse module. Don't bother writing a complex if statement, instead prefer to setup your argument parser something like this:

#!/usr/bin/env python
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('--foo', nargs=3, default=['x', 'y', 'z'])
args = parser.parse_args()
print(args.foo)

And yes, it should be an option not a positional argument, because it is after all optional.

---

edited: To address the concern of LarsH in the comments, below is an example of how you could write it if you were certain you wanted the interface with either 3 or 0 positional args. I am of the opinion that the previous interface is better style, because optional arguments should be options, but here's an alternative approach for the sake of completeness. Note the overriding kwarg usage when creating your parser, because argparse will auto-generate a misleading usage message otherwise!

#!/usr/bin/env python
import argparse
parser = argparse.ArgumentParser(usage='%(prog)s [-h] [a b c]\n')
parser.add_argument('abc', nargs='*', help='specify 3 or 0 items', default=['x', 'y', 'z'])
args = parser.parse_args()
if len(args.abc) != 3:
  parser.error('expected 3 arguments')
print(args.abc)

Here are some usage examples:

# default case
wim@wim-zenbook:/tmp$ ./three_or_none.py 
['x', 'y', 'z']

# explicit case
wim@wim-zenbook:/tmp$ ./three_or_none.py 1 2 3
['1', '2', '3']

# example failure mode
wim@wim-zenbook:/tmp$ ./three_or_none.py 1 2 
usage: three_or_none.py [-h] [a b c]
three_or_none.py: error: expected 3 arguments

I'd go for:

conds = iter([a, b, c])
if any(conds) and not any(conds):
    # okay...

I think this should short-circuit fairly efficiently

Explanation

By making conds an iterator, the first use of any will short circuit and leave the iterator pointing to the next element if any item is true; otherwise, it will consume the entire list and be False. The next any takes the remaining items in the iterable, and makes sure than there aren't any other true values... If there are, the whole statement can't be true, thus there isn't one unique element (so short circuits again). The last any will either return False or will exhaust the iterable and be True.

note: the above checks if only a single condition is set

---

If you want to check if one or more items, but not every item is set, then you can use:

not all(conds) and any(conds)