Append to list in a dictionary after setdefault [duplicate]

Question

I have the below code where I am trying to append a 1 to the hash of an element, on every occurence of it in input.

def test(Ar):
    hash_table = {}
    for elem in Ar:
        if elem not in hash_table:
            hash_table.setdefault(elem,[]).append(1)
        else:
            hash_table[elem] = hash_table[elem].append(1)
    print(hash_table)

Ar = (1,2,3,4,5,1,2)
test(Ar)

Output:

{1: None, 2: None, 3: [1], 4: [1], 5: [1]}

Expected Output:

{1: [1,1], 2: [1,1], 3: [1], 4: [1], 5: [1]}

I am puzzled why None gets in on doing an append. Kindly explain what is happening.

Note:

On typing the else portion,

hash_table[elem] = hash_table[elem].append(1) # the append() was not suggested at all by the IDE. I forcibly put it, hoping things will work.

Answer 1

The list.append is an in-place operation. So it just modifies the list object and doesn't return anything. That is why, by default, None is getting returned from list.append and you are storing that corresponding to the key in this line

hash_table[elem] = hash_table[elem].append(1)

---

In your case, you don't need the if condition at all.

def test(Ar):
    hash_table = {}
    for elem in Ar:
        hash_table.setdefault(elem, []).append(1)
    print(hash_table)

because, setdefault will first look for the key elem in it and it found something, then it will return the value corresponding to it. And if it doesn't then it will create key elem and use the second argument passed to it as the value and then return then value.

---

Instead of using this, you can use collections.defaultdict, like this

from collections import defaultdict
def test(Ar):
    hash_table = defaultdict(list)
    for elem in Ar:
        hash_table[elem].append(1)
    print(hash_table)

This would do almost the same thing as the setdefault version

---

It looks like you are trying to find the frequency of elements. In that case you can simply use collections.Counter

from collections import Counter
Ar = (1, 2, 3, 4, 5, 1, 2)
print Counter(Ar)
# Counter({1: 2, 2: 2, 3: 1, 4: 1, 5: 1})

This will give the number of times each and every element occurred in the iterable passed.