按出现次数对列表进行排序并删除重复项

Question

例如，我有以下列表，其中每个元素的出现次数是：apple - 3banana - 4orange - 2列表：["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"]我需要按流行度对列表进行排序而没有重复，因此预期结果将是：["banana", "apple", "orange"]我想创建一个以每个元素为键的字典，遍历列表，然后在每次找到键时添加 +1，所以我最终会得到一个示例字典：dic = {"apple": 3, "banana": 4, "orange":2}但是有点卡住了如何在没有欺骗的情况下对列表本身进行排序..提前致谢。编辑：谢谢大家，我不知道计数器。节日快乐！

Answer 1

您可以使用 acollections.Counter及其most_common方法：

from collections import Counter

lst = ["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"]

res = [k for k, _ in Counter(lst).most_common()]

# ['banana', 'apple', 'orange']

Answer 2

使用计数器：

from collections import Counter

data = ["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"]

counts = Counter(data)

result = sorted(counts, key=counts.get, reverse=True)

print(result)

输出

['banana', 'apple', 'orange']

Answer 3

根据原始列表的计数对集合进行排序。编辑：正如评论中所指出的，如果您有很多候选人，您可能想要使用其他解决方案，多次调用列表的计数方法不是最佳选择。

a = ["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"]

sorted(set(a), key = lambda x: a.count(x), reverse = True) #reverse for descending

结果：

['banana', 'apple', 'orange']