python - 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为：

apple - 3
banana - 4
orange - 2

名单：

["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"]

我需要按流行度对列表进行排序，没有重复，所以预期的结果是：

["banana", "apple", "orange"]

我考虑过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}

但是有点坚持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：正如评论中指出的那样，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并不是最佳的。

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']

Answer 4

from itertools import groupby

L = ["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"] # Input list

counts = [(i, len(list(c))) for i,c in groupby(sorted(L))]      # Create value-count pairs as list of tuples 
counts =  sorted(counts, key = lambda i: i[1] , reverse=True)    #sort value-count list
out = [key for key, value in counts]   #extract key
print (out)

Answer 5

使用list和set

a = ["apple", "apple", "banana", "orange", "orange", "banana", "banana", "apple", "banana"]
result =sorted(list(set(a)))

输出 ：

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