Detect alternating signs

Is there a nice and short way to tell whether a python list (or numpy array) contains numbers with alternating signs? In other words:

is_alternating_signs([1, -1, 1, -1, 1]) == True
is_alternating_signs([-1, 1, -1, 1, -1]) == True
is_alternating_signs([1, -1, 1, -1, -1]开发者_开发技巧) == False

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OK, thanks to SO "related" feature. I found this question and adopted the answer by ianalis and the comment by lazyr

def is_alternating_signs(a):
    return numpy.all(numpy.abs(numpy.diff(numpy.sign(a))) == 2)



print is_alternating_signs([1, -1, 1, -1, 1]) 
print is_alternating_signs([-1, 1, -1, 1, -1]) 
print is_alternating_signs([1, -1, 1, -1, -1]) 

The output is

True
True
False

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You could check every even member is negative and every odd member is positive by taking a slice of every second item, starting at either the beginning or from position one. Also test the reverse to cover both possibilities.

so:

def is_alternating_signs(l):
    return ( (all(x<0 for x in l[::2]) and all(x>=0 for x in l[1::2])) or
             (all(x>=0 for x in l[::2]) and all(x<0 for x in l[1::2])))

---

Using decimal module and is_signed method:

from decimal import Decimal

a = [1, -1, 1, -1, 1]
b = [-1, 1, -1, 1, -1]
c = [1, -1, 1, -1, -1]

def is_alternating_signs(values):
    lVals = [Decimal(val).is_signed() for val in values]
    prevVal = lVals.pop(0)
    for val in lVals:
        if prevVal == val:
            return False
        prevVal = val
    return True

is_alternating_signs(a)
is_alternating_signs(b)
is_alternating_signs(c)

---

I like pairwise:

from itertools import izip, tee

def pairwise(iterable):
    a, b = tee(iterable)
    next(b)
    return izip(a, b)

def is_alternating_signs(iterable):
    return all(x < 0 < y or x > 0 > y for x, y in pairwise(iterable))

If there are no zeros in iterable this also works:

def is_alternating_signs(iterable):
    return all((x < 0) == (0 < y) for x, y in pairwise(iterable))

---

how about something like...

def is_alternating_signs(aList):
    return all( (aList[i]^aList[i-1])<0 for i in range(1,len(aList)) )

---

What about just the straight-forward solution by looping through it once and testing? Quite possibly the fastest, too, because many of the other solutions loop through the list multiple times.

def signs_are_alternating(numbers):
    """Return True if numbers in given list have alternating signs, False
    otherwise. If given list has less than 2 elements, return False.

    >>> signs_are_alternating([1, -1, 1, -1, 1])
    True
    >>> signs_are_alternating([-1, 1, -1, 1, -1])
    True
    >>> signs_are_alternating([1, -1, 1, -1, -1])
    False

    """
    if len(numbers) < 2:
        return False
    previous_positive = (numbers[0] < 0)  # Pretend it starts alternating
    for number in numbers:
        this_positive = (number >= 0)
        if previous_positive == this_positive:
            return False
        previous_positive = this_positive
    return True

Note that I wasn't quite sure what the intended behaviour is if the input list has less than 2 elements.

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Here's my one-liner, which is probably less efficient than some of the other suggestions:

def is_alternating_signs(lst):
    return all(x * y < 0 for x, y in zip(lst, lst[1:]))