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)

Does this answer your question? [How to print only the duplicate elements in python list](https://stackoverflow.com/questions/52072381/how-to-print-only-the-duplicate-elements-in-python-list) — sagar1025, May 05 '21 at 01:39
@sagar1025 While it does solve his overall issue, it doesn't answer his specific question, which was specifically the condition needed to determine duplicates. — lionrocker221, May 05 '21 at 01:41
you will never get a dictionary instead, you always will get a list as you assigned duplicates to empty dict and when you create for loop you called return statement so when it searches for the number of count and return runs will break the loop and return the end of code which will be empty dict so, no need to return dict here — Medo Abdin, May 05 '21 at 02:03

score 1 · Answer 1 · answered May 05 '21 at 01:51

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))

score 0 · Accepted Answer · answered May 05 '21 at 01:38

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)

MPuc · Answer 3 · 2021-05-05T02:20:29.710

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}

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

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

3 Answers3