Untitled

"""
Data una lista l di interi, si vuole modificarla cancellando tutti gli elementi i cui valori compaiono esattamente una volta, mantenendo lo stesso ordine che gli elementi avevano inizialmente. (p.e. se l'ingresso e' l = [6, 1, 6, 1, 4, 1, 1, 3, 5] allora il risultato e' L = [6, 1, 6, 1, 1, 1]). Si scriva una procedura utilizzando gli operatori per le liste visti lezione.
"""

# returns the number of occurrences of element e in list l
def get_occurrences (l, e):
        count = 0
        for x in l:
                if x == e: count += 1
        return count

# remove elements according to exam question
# complexity: O(n^2)
def fun1 (l):
        lc = list(l) # copy of the input list (is this really needed? we can re-use the same list for occurrences count)
        i = 0 # pointer to the first element of the list l
        while i < len(l):
                occurrences = get_occurrences(lc, l[i]) # get number of occurrences of current element
                if occurrences == 1: del(l[i]) # if it appears only one time in the list delete it
                else: i += 1 # consider the next element otherwise

        return l

# remove elements according to exam question
# complexity: O(n)
def fun2 (l):
        d = dict() # create an empty dictionary (hash table)

        # for each element in input list
        for e in l:
                if d.has_key(e): d[e] += 1 # if this element has been already added to the dictonary increment the related number
                else: d[e] = 1 # set up the related number of this element to 1 (occurrence) otherwise

        # return the list in which all elements that appears only one time have been deleted
        return [ e for e in l if d[e] > 1 ]

# like fun1 but much more simple!
# complexity: O(n^2)
def fun3 (l):
        return [ e for e in l if l.count(e) > 1 ]

# main
l = [6, 1, 6, 1, 4, 1, 1, 3, 5]
print fun1(l)
print fun2(l)
print fun3(l)