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 - 3

banana - 4

orange - 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 Counterlst = ["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 Counterdata = ["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']