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]) == Tru

Solution 1:

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

TrueTrueFalse

Solution 2:

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:

defis_alternating_signs(l):
    return ( (all(x<0for x in l[::2]) andall(x>=0for x in l[1::2])) or
             (all(x>=0for x in l[::2]) andall(x<0for x in l[1::2])))

Solution 3:

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() forvalin values]
    prevVal = lVals.pop(0)
    forvalin lVals:
        if prevVal == val:
            return False
        prevVal = valreturn True

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

Solution 4:

I like pairwise:

from itertools import izip, tee

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

defis_alternating_signs(iterable):
    returnall(x < 0 < y or x > 0 > y for x, y in pairwise(iterable))

If there are no zeros in iterable this also works:

defis_alternating_signs(iterable):
    returnall((x < 0) == (0 < y) for x, y in pairwise(iterable))

Solution 5:

how about something like...

defis_alternating_signs(aList):
    returnall( (aList[i]^aList[i-1])<0for i inrange(1,len(aList)) )