Given a list of pairs, how to sort them by frequency of the second value?

Question

I am trying to sort a pair list such as [(1, 1), (2, 5), (2, 1), (3, 3), (5, 3)] by the frequency of the second term in each of the tuples. The resulting sorted list should be something like [(3,3),(5,3),(1,1),(2,1),(2,5)]. In this case, the frequency table would look like 1: 2, 3: 2, 5:1.

Answer 1

Using Counter to construct a key to pass to sort or sorted is a natural approach:

from collections import Counter
my_list = [(1, 1), (2, 5), (2, 1), (3, 3), (5, 3)]
counts = Counter(y for _,y in my_list)
sorted(my_list,key = lambda p:counts[p[1]],reverse = True)

Which evaluates to:

[(1, 1), (2, 1), (3, 3), (5, 3), (2, 5)]

Answer 2

Similar to @John's answer, we can build a frequency map using a collection.defaultdict, then sort by the second value of each tuple using the frequencies from the dictionary, making sure we sort in reverse:

from collections import defaultdict

lst = [(1, 1), (2, 5), (2, 1), (3, 3), (5, 3)]

frequency_map = defaultdict(int)
for _, snd in lst:
    frequency_map[snd] += 1

print(sorted(lst, key=lambda x: frequency_map[x[1]], reverse=True))

Output:

[(1, 1), (2, 1), (3, 3), (5, 3), (2, 5)]