How to make a function that counts how many times each element is equal to 2 elements to its right

I know i need to use a list comprehension but for the life of me I cannot figure out what would be the correct way to denote this. An example of this running right would be for "evening" the output being 2, once 开发者_如何学运维 for 'e' and once for 'n'

---

The list comprehension gives the letters that have the same letter two places to the right. We simply take the length of the resulting list:

s = "evening"
ans = len([x for x in xrange(len(s)-2) if s[x] == s[x+2]])
print ans

---

s='evening'
print len([x for x,y in zip(s, s[2:]) if x==y])

Output:

2

---

I would love to see someone more expert than me turn this into a lc but my basic sollution would be

zz='evening'
for numb, letter in enumerate(zz):
if numb+2==len(zz):
    break
if letter==zz[numb+2]:
    count+=1

Well after seeing Mike's answer and thinking about it how about this if the input is a list

foo = ['e', 'v', 'e', 'n', 'i', 'n', 'g']
new=[item for numb, item in enumerate(foo[0:-2]) if item==foo[numb+2]]
answer=len(new)

silly me, I can also work with a string and I think this is still cleaner

testString='evening'
new=[letter for numb, letter in enumerate(testString[0:-2]) if letter==testString[numb+2]]
ans=len(new)