Count how often sign changes in array.

### result
number from `0` to ... . Empty array returns `0`

### example

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def catch_sign_change(lst):
	

Solutions:
def catch_sign_change(lst):
	count = 0
	for i in range(1, len(lst)):
		if lst[i] < 0 and lst[i - 1] >= 0:
			count += 1
		if lst[i] >= 0 and lst[i - 1] < 0:
			count += 1
	return count


def catch_sign_change(lst):
	return sum(((x >= 0) != (y >= 0) for (x, y) in zip(lst, lst[1:])))


def catch_sign_change(lst):
	return sum(((a or 1) * (b or 1) < 0 for (a, b) in zip(lst, lst[1:])))


def catch_sign_change(lst):
	count = 0
	countNumber = 0
	i = 0
	for n in lst:
		if countNumber >= 0 and n < 0 and (i != 0):
			count = count + 1
		elif countNumber < 0 and n >= 0 and (i != 0):
			count = count + 1
		countNumber = n
		i = i + 1
	return count


def catch_sign_change(lst):
	return sum(((m < 0) == (n > -1) for (m, n) in zip(lst, lst[1:])))


def catch_sign_change(lst):
	res = 0
	for i in range(len(lst) - 1):
		if lst[i] >= 0 and lst[i + 1] < 0 or (lst[i] < 0 and lst[i + 1] >= 0):
			res += 1
	return res


def catch_sign_change(lst):
	result = 0
	for i in range(len(lst) - 1):
		a = lst[i] if lst[i] != 0 else 1
		b = lst[i + 1] if lst[i + 1] != 0 else 1
		if a * b < 0:
			result += 1
	return result