Calculate the number of items in a vector that appear at the same index in each vector, with the same value. 

```python
   vector_affinity([1, 2, 3, 4, 5], [1, 2, 2, 4, 3]) # => 0.6
   vector_affinity([1, 2, 3], [1, 2, 3]) # => 1.0
```

Affinity value should be realized on a scale of 0.0 to 1.0, with 1.0 being absolutely identical. Two identical sets should always be evaulated as having an affinity or 1.0.

Hint: The last example test case holds a significant clue to calculating the affinity correctly.

Difficulty: EASY
Starter code: def vector_affinity(a, b):
	

Solutions:
def vector_affinity(a, b):
	longer = len(a) if len(a) > len(b) else len(b)
	return len([i for (i, j) in zip(a, b) if i == j]) / float(longer) if longer > 0 else 1.0


def vector_affinity(a, b):
	return sum([1.0 for (x, y) in zip(a, b) if x == y]) / max(len(a), len(b)) if max(len(a), len(b)) else 1


from itertools import zip_longest, starmap
from statistics import mean
from operator import eq

def vector_affinity(a, b):
	return not (a or b) or mean(starmap(eq, zip_longest(a, b, fillvalue='¯\\_(ツ)_/¯')))


def vector_affinity(a, b):
	sizeA = len(a)
	sizeB = len(b)
	if sizeA + sizeB == 0:
		return 1.0
	if sizeA >= sizeB:
		den = sizeA
		numIter = sizeB
	else:
		den = sizeB
		numIter = sizeA
	numMatch = 0
	for i in range(numIter):
		if a[i] == b[i]:
			numMatch += 1
	affinity = numMatch / den
	return affinity


def vector_affinity(a, b):
	return 1.0 if a == b else sum((1 for (m, n) in zip(a, b) if m == n)) / max(len(a), len(b))


def vector_affinity(*args):
	(l1, l2) = list(map(len, args))
	return sum((x == y for (x, y) in zip(*args))) / (1.0 * max(l1, l2)) if l1 or l2 else 1.0


def vector_affinity(a, b):
	denom = max(len(a), len(b))
	return sum((l == r for (l, r) in zip(a, b))) / float(denom) if denom else 1


def vector_affinity(a, b):
	return 1.0 if a == b else sum((1 for (i, j) in zip(a, b) if i == j)) / float(max(len(a), len(b)))


def vector_affinity(a, b):
	return sum([an == bn for (an, bn) in zip(a, b)]) / float(max(len(a), len(b))) if len(a) or len(b) else 1