title: How to find the symmetrical portion of a python list of integers and get the sum of that symmetrical portion?
body: <p>What the code does: Takes a Python list of integers as input and searches for a 'symmetrical' inner-portion of the list.</p>
  <p>Examples of what i want:</p>
  <pre><code>symmetrical_sum([10,11,12,11,12]) == ([11, 12, 11], 34)
  symmetrical_sum([9,99,88,8,77,7,77,8,88,10,100]) == ([88, 8, 77, 7, 77, 8, 88], 353)
  symmetrical_sum([10,8,7,5,9,8,15]) == ([8, 7, 5, 9, 8], 37) 
  </code></pre>
  <p>Symmetry occurs if the value of the ith element from the start of the list is equal to the value of the ith element from the end of the list.</p>
  <p>My Code:</p>
  <pre><code>def symmetrical_sum(a):
  
  #extract duplicate value
      dupe = [x for n, x in enumerate(a) if x in a[:n]] 
  
  #if no duplicate values found, do the following:
      if dupe == []:
          middle = float(len(a))/2
      if middle % 2 != 0:
          sym = a[int(middle - .5):int(middle + .5)]
          ans = a[int(middle - .5)]
          tuple1 = (sym,ans)
      elif middle % 2 == 0:
          sym = a[int(middle - 1):int(middle + 1)]
          ans = sum(a[int(middle - 1):int(middle + 1)])//2
          tuple1 = (sym,ans)
  
      return tuple1
  
  else:
      d_to_i = int(&quot;&quot;.join(map(str, dupe))) #convert duplicate value to integer
      p1 = a.index(d_to_i) #get index of first duplicate
      p2 = a.index(d_to_i, p1+1) #get index of second duplicate
      sym = a[p1:p2+1] #[symmetrical-portion]
      ans = sum(sym) #sum-of-symmetrical-portion
      tuple2 = (sym, ans)
  
  return tuple2
  </code></pre>
  <p>My code works but it would be great if someone can post a way shorter solution for efficiency purposes, please.</p>