How to create a dictionary from list without the repeating elements

Question

I need to create a dictionary of elements without the elements that shows more than once and i want to keep the starting index. Example:

a = ['a', 'b' 'a', 'c']

expected = {'b': 1, 'c': 3}

To solve the problem i tried this way:

def unique_array(foo):
   correct, duplicates = {}, set()
   for value in foo:
       if index, value not in enumerate(correct):
          correct[value] = index
       else:
          duplicates.add(value)
   return {k: correct[k] for k in set(correct) - duplicates

There is a better and more efficient way to do this?

Answer 1

One solution is to use a Counter and two passes. (It's still O(n).)

>>> from collections import Counter 
>>> a = ['a', 'b', 'a', 'c']                                                    
>>> counts = Counter(a)                                                         
>>> {k:idx for idx, k in enumerate(a) if counts[k] == 1}                        
{'b': 1, 'c': 3}

Answer 2

You have two imbricated loops, which makes the time complexity of your code O(n^2).

An O(n) solution could be:

a = ['a', 'b', 'a', 'c']

out = {}
for index, val in enumerate(a):
    if val not in out:
        out[val] = index
    else:
        out[val] = None

out = {item:val for item, val in out.items() if val is not None}



print(out)
# {'b': 1, 'c': 3}

We first create the output, marking the duplicates with the value None when we encounter them. At the end, we filter out the keys with None value. Both operations are O(n).

Answer 3

you can do this by using a single loop

alphabets = ['a', 'b', 'a', 'c']
alphabetDict = {}
index = 0
for alphabet in alphabets:
    count = alphabets.count(alphabet)
    if count == 1:
        alphabetDict[alphabet] = index
        index += 1
    else:
        index += 1
        continue

print(alphabetDict)