Count Sign Change

user_17264516 0
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I have this array:

X = [0,2,-1,2,1,2]

And the sign changes two times (0 is positive here).

Example:

2 to -1 (first time)

-1 to 2 (second times)

last_sign = 1
sign_changes = 0

for x in X:
    if x == 0:
        sign = -1
    else:
        sign = x / abs(x)

    if sign == -last_sign:
        sign_changes = sign_changes + 1
        last_sign = sign

print(sign_changes)

But when I print the result I get 4. Why?

math python
2 Answers

Unlike what you described, you are assigning negative sign to zero: note that you put sign = -1 at if x == 0:. This is the culprit here. (Also your current code assumes that the list starts with a non-negative number. A list starting with a negative number won't produce an expected outcome.)

You can try instead:

X = [0,2,-1,2,1,2]

print(sum((a >= 0) != (b >= 0) for a, b in zip(X, X[1:]))) # 2
user_114508200

My suggestion, iterate pairwise and check if the pairs have different signs:

from itertools import tee


def pairwise(iterable):
    # pairwise('ABCDEFG') --> AB BC CD DE EF FG
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)


X = [0, 2, -1, 2, 1, 2]

res = sum((before < 0) ^ (after < 0) for before, after in pairwise(X))
print(res)

Output

2

As an alternative, without any libraries, use:

res = sum((before < 0) ^ (after < 0) for before, after in zip(X, X[1:]))

The expression:

(before < 0) ^ (after < 0)

is an exclusive or, it will only be True (1) if the boolean values are different.

user_40015920
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