Using dict.get

How often have you aggregated an item by its group like this:

ar= [1,1,3,1,2,1,3,3,3,3]
pairs = {}
for i in ar:
    if i not in pairs:
         pairs[i] = 0
    pairs[i] = pairs[i] + 1


print(pairs)
{1: 4, 3: 5, 2: 1}

The above code shows how to get the frequency of elements in a list.
While this code is good, the time complexity of in(membership operator) is O(n) worse case.
We can make this code better and more pythonic by using python's dict.get(k,d) method.

Using the same list, we have:

pairs = {}
for i in ar:
    pairs[i] = pairs.get(i, 0) + 1
print(pairs)
{1: 4, 3: 5, 2: 1}

the trick here is that dict.get(k,d) returns None by default or the specified default value if a key doesn't exist.

Using dict.setdefault

Similarly when grouping things together we usually do this:

items_by_type = {}
for item in items:
    if item.type not in items_by_type:
        items_by_type[item.type] = list()
    items_by_type[item.type].append(item)

Let's make it better and more pythonic:

items_by_type = {}
for item in items:
    items_by_type.setdefault(item.type, list()).append(item)

The trick here is that setdefault(k,d) only sets the item if it does not
exist. If it does exist then it simply returns the item.

Final Thoughts

You can use the Counter class from the collection module to get the frequency of elements in a list.
Just do this:

from collections import Counter
#using our above example:
print(Counter(ar))

Happy coding! ✌

Please drop your comments :)