How do I return a dictionary of values from a list?

Question

How can I edit the commented-off piece of code so that it returns a dictionary of the duplicate numbers rather than returning "{}"?

listOfElems = ["a", "c", "c"]

def checkIfDuplicates_3(listOfElems):
  duplicates = {}
  for x in listOfElems:
#    duplicates = ??
    if listOfElems.count(x) > 1:
      return duplicates
  return duplicates

#test 
test = [listOfElems]

for t in test:
  output = checkIfDuplicates_3(t)
  print("The duplicate in", t, "is", output)

Answer 1

I don't see why you would need to use a dictionary here. You don't have a key or a value. All you want to do is return a letter which is repeated in a list.

The below code uses list comprehension to solve your problem. First we call this function passing in listOfElems . Then we will iterate through the given list and check the occurrences of that each element. We can do this using count . It appears you have used that too. If a letter appears more than once we will append that letter the a new list.

In the example you have provided, we will end up with something like this, ['c','c'] . It would be nice if c only appeared once. To solve that we can use set which will give us {'c'} which is quite nice. To finish it all up, we return this set as a list.

listOfElems = ["a", "c", "c"]

def checkIfDuplicates_3(listOfElems):
    dups = [x for x in listOfElems if listOfElems.count(x) > 1]
    return list(set(dups))
print("The duplicates in", listOfElems, "are", checkIfDuplicates_3(listOfElems))

Answer 2

Here's a solution.

If there is more than one of the element in the list, listOfElems.count(x) > 1 , and our list of duplicates doesn't contain the element already, duplicates.count(x) == 0 , then we can add the element to the list of duplicates.

listOfElems = ["a", "c", "c"]


def checkIfDuplicates_3(listOfElems):
    duplicates = []
    for x in listOfElems:
        if listOfElems.count(x) > 1 and duplicates.count(x) == 0:
            duplicates.append(x)
    return duplicates


# test
test = [listOfElems]

for t in test:
    output = checkIfDuplicates_3(t)
    print("The duplicate in", t, "is", output)

Answer 3

Using set and list comprehension a one-liner function could do:

listOfElems = ["a", "c", "c","e","c","e","f"]

def checkIfDuplicates_3(listOfElems): return set([x for x in listOfElems if listOfElems.count(x)>1])

#test
test = [listOfElems]

for t in test:
    output = checkIfDuplicates_3(t)
    print("The duplicate in", t, "is", sorted(output))

Note that order of insertion may not be preserved by set. To get ordered output, use sorted (as shown in example above).

To get a dictionary counting the number of occurences>1 you may use:

def checkIfDuplicates_3(listOfElems):
    return {x:listOfElems.count(x) for x in set(listOfElems) if listOfElems.count(x)>1}

eg returning for checkIfDuplicates_3(listOfElems) the result {'e': 2, 'c': 3} (see note for ordering above).