Min value from a counter result in python

Question

I'm new to programming and I find myself in some trouble. I have a list, and I want to know how many times an item shows up and then print the minimum value that shows up. So if I have A=[1e, 2b, 3u, 2b, 1e, 1e, 3u, 3u], I want to show something like "What you want is a 2", where 2 is the least amount of times something shows up, in this case 2b is the one that shows up the least amount of times. This is my code so far:

import collections

collections.Counter(A)
B = {key: value for (key, value) in A}
result = []
min_value = None
minimum = min(B, key=B.get)
print(minimum, B[minimum])

The output for this is 2b, but what I want the amount of times 2b shows up, since it is the one that shows up the least. I'm having some difficulty with this. To clarify, I want the minimum number in a counter result.

Any help would be appreciated, I'm sorry if my question is confusing English is not my first language and it's my first time doing something like this.

Answer 1

Just use min on dict.items, where dict is a Counter object:

from collections import Counter
from operator import itemgetter

c = Counter(A)
min_key, min_count = min(c.items(), key=itemgetter(1))

Since dict.items returns a view of key-value pairs, you can unpack directly to min_key, min_count variables.