Python 3.x 根据频率按降序对列表进行排序

我有一个数字列表： [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]. 我想按元素的频率对其进行排序，以得到以下结果： [1, 1, 1, 1, 3, 3, 3, 2, 2, 2].

如果多个元素具有相同的频率，请按降序值对它们进行排序。你能找到办法吗。我使用上述方法，但得到的输出为：[1，1，1，2，2，2，3，3，3]

from collections import Counter 
list = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(list)
x = sorted(c.most_common(), key=lambda x: (-x[1], x[0])) 
y = [([v] * n) for (v, n) in x]
z = sum(y, [])
print(z)

---

看起来您需要使用

reverse=True

Ex:

from collections import Counter 

data = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(data)

data.sort(key=lambda x: (c[x], x), reverse=True)
print(data)

[1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

输出：

from collections import Counter 

data = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(data)

data.sort(key=lambda x: (c[x], x), reverse=True)
print(data)

[1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

---

看起来您需要使用

reverse=True

Ex:

from collections import Counter 

data = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(data)

data.sort(key=lambda x: (c[x], x), reverse=True)
print(data)

[1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

输出：

from collections import Counter 

data = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(data)

data.sort(key=lambda x: (c[x], x), reverse=True)
print(data)

[1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

---

如果列表很长，您可能只需要对出现次数相同的项目进行排序：

from collections import Counter
from itertools import groupby

lst = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(lst)

ret = []
for key, group in groupby(c.most_common(), key=lambda x: x[1]):
    items = sorted((item for item, _ in group), reverse=True)
    for item in items:
        ret.extend(key * [item])
print(ret)
# [1, 1, 1, 1, 3, 3, 3, 2, 2, 2]

---

如果列表很长，您可能只需要对出现次数相同的项目进行排序：

from collections import Counter
from itertools import groupby

lst = [1, 2, 3, 1, 1, 2, 3, 1, 2, 3]
c = Counter(lst)

ret = []
for key, group in groupby(c.most_common(), key=lambda x: x[1]):
    items = sorted((item for item, _ in group), reverse=True)
    for item in items:
        ret.extend(key * [item])
print(ret)
# [1, 1, 1, 1, 3, 3, 3, 2, 2, 2]