Given 2 int values, return True if one is negative and other is positive

Question

def logical_xor(a, b): # for example, -1 and 1
    print (a < 0) # evaluates to True
    print (b < 0) # evaluates to False
    print (a < 0 != b < 0) # EVALUATES TO FALSE! why??? it's True != False
    return (a < 0 != b < 0) # returns False when it should return True

print ( logical_xor(-1, 1) ) # returns FALSE!

# now for clarification

print ( True != False) # PRINTS TRUE!

Could someone explain what is happening? I'm trying to make a one liner:

lambda a, b: (a < 0 != b < 0)

There's a lesson to be learned here: Use parentheses. Or, alternatively, use `print (a*b < 0)`. — David Hammen, Apr 22 '15 at 08:18
`print ( True != False) # PRINTS TRUE!` why are you so surprised by that? It _is_ true that true is not equal to false. — Mike G, Apr 22 '15 at 18:02
@mikeTheLiar: The OP is *not* surprised by that; it's the behavior (s)he originally expected. — ruakh, Apr 23 '15 at 02:03
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tzaman · Accepted Answer · 2015-04-22T13:48:15.180

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All comparison operators in Python have the same precedence. In addition, Python does chained comparisons. Thus,

(a < 0 != b < 0)

breaks down as:

(a < 0) and (0 != b) and (b < 0)

If any one of these is false, the total result of the expression will be False.

What you want to do is evaluate each condition separately, like so:

(a < 0) != (b < 0)

Other variants, from comments:

(a < 0) is not (b < 0) # True and False are singletons so identity-comparison works

(a < 0) ^ (b < 0) # bitwise-xor does too, as long as both sides are boolean

(a ^ b < 0) # or you could directly bitwise-xor the integers; 
            # the sign bit will only be set if your condition holds
            # this one fails when you mix ints and floats though

(a * b < 0) # perhaps most straightforward, just multiply them and check the sign

edited Apr 22 '15 at 13:48

answered Apr 22 '15 at 07:37

tzaman

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Note that `(a < 0) ^ (b < 0)` also works, and it's using the xor operator. :) Alternatively, `(a < 0) is not (b < 0)`, since `True` and `False` are singletons, although some might object to that syntax: see here for a discussion: http://stackoverflow.com/q/9494404/4014959 – PM 2Ring Apr 22 '15 at 07:53
`^` for *xor* does work if booleans are compared but it won't work for all values like `and` and `or` would... – Antti Haapala -- Слава Україні Apr 22 '15 at 12:00
OP specifically tagged Python 3.x so I linked to Python 3 manuals. – Antti Haapala -- Слава Україні Apr 22 '15 at 12:04
I personally like a * b < 0, but one needs to keep in mind that in other languages this can cause integer overflow – Paradise Apr 22 '15 at 15:49
@OverlordAlex: Why could it not cause integer overflow in Python? – supercat Apr 22 '15 at 17:37
@supercat Python integers are unlimited precision. – tzaman Apr 22 '15 at 17:44
@supercat, to expand on tzaman's answer, python automatically promotes integers to longs when they would overflow, and longs in python are unlimited length (restricted only by available memory). At worst you'll get a float where you might expect an int. In python 3 there aren't any integers/longs, but rather a hybrid general "number", that is arbitrary length, so you dont even have to worry about long vs int – Paradise Apr 22 '15 at 20:17
@OverlordAlex: That makes sense. On the other hand, I was just noticing that while the OP's code seems to have been trying to identify if one number is negative and one number is non-negative, your formula tests for the condition described in the question title. If the question title relates what's actually desired, I don't remember if Python has a function to convert values into +1/0/-1 based on whether they're pos/neg/zero, but if so, perhaps applying that to the operands before the multiplication might be a good idea. – supercat Apr 22 '15 at 21:10
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@supercat: Unfortunately python doesn't have a sign() method. One was proposed for the math package, but there wasn't agreement on edge-cases, so it wasn't integrated (according to a quick google). A workaround is using cmp(number, 0), which will return +1/0/-1. This would indeed work for the question – Paradise Apr 23 '15 at 08:10

score 7 · Answer 2 · edited Jun 20 '20 at 09:12

7

Your code doesn't work as intended because != takes higher precedence than a < 0 and b < 0. As itzmeontv suggests in his answer, you can simply decide the precedence yourself by surrounding logical components with parentheses:

(a < 0) != (b < 0)

Your code attempts to evaluate a < (0 != b) < 0

[EDIT]

As tzaman rightly points out, the operators have the same precedence, but your code is attempting to evaluate (a < 0) and (0 != b) and (b < 0). Surrounding your logical components with parentheses will resolve this:

(a < 0) != (b < 0)

Operator precedence: https://docs.python.org/3/reference/expressions.html#operator-precedence

Comparisons (i.a. chaining): https://docs.python.org/3/reference/expressions.html#not-in

edited Jun 20 '20 at 09:12

Community

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answered Apr 22 '15 at 07:32

EvenLisle

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They're actually the **same** precedence; the failure is because it's treated as a chained comparison. See [docs](https://docs.python.org/2/reference/expressions.html#not-in). – tzaman Apr 22 '15 at 07:34
@Xis88: Yes, it's equivalent to `(a < 0) and (0 != b) and (b < 0)`. – PM 2Ring Apr 22 '15 at 07:43
Op specifically tagged Python 3 so please link to Python 3 manuals. – Antti Haapala -- Слава Україні Apr 22 '15 at 12:03

itzMEonTV · Answer 3 · 2015-04-22T12:18:03.810

4

You can use this

return (a < 0) != (b < 0)

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

So it becomes

(a < 0) and (0 != b) and (b < 0)

See https://docs.python.org/3/reference/expressions.html#not-in

edited Apr 22 '15 at 12:18

answered Apr 22 '15 at 07:30

itzMEonTV

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May be it of part of precedence of `!=` and `<` – itzMEonTV Apr 22 '15 at 07:31
`<>` is not "considered obsolescent", it was removed from Python 3.0 altogether. – Antti Haapala -- Слава Україні Apr 22 '15 at 12:01
(do note that OP tagged the question with [python-3.x]) – Antti Haapala -- Слава Україні Apr 22 '15 at 12:07

score 0 · Answer 4 · edited Apr 22 '15 at 16:26

In Python, comparison operators are of the same precedence, and they are non-associative. There is a separate rule for sequences of comparison operators, the chaining rule. Python documentation states about that:

if a, b, c, ..., y, z are expressions and op1, op2, ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z, except that each expression is evaluated at most once.

Further, a op1 b and b op2 c and ... y opN z evaluates left to right.

 a < 0 and 0 != b and b < 0

a < 0 will evaluated to False, and the further evaluation will be stopped due to short-circuit evaluation. So, the whole expression will be evaluated as False.

This is wrong! The documentation you're quoting right there states that the comparisons are chained. *NOT* the same thing as left-associative. — tzaman, Apr 22 '15 at 13:56

Given 2 int values, return True if one is negative and other is positive

4 Answers4