按频率排序列表

Question

在 Python 中有什么方法可以按频率对列表进行排序吗？

例如，

[1,2,3,4,3,3,3,6,7,1,1,9,3,2]

上面的列表将按照其值出现的频率顺序进行排序以创建以下列表，其中频率最高的项目位于最前面：

[3,3,3,3,3,1,1,1,2,2,4,6,7,9]

Answer 1

def orderByFrequency(list):

    listUniqueValues = np.unique(list)
    listQty = []
    listOrderedByFrequency = []
    
    for i in range(len(listUniqueValues)):
        listQty.append(list.count(listUniqueValues[i]))
    for i in range(len(listQty)):
        index_bigger = np.argmax(listQty)
        for j in range(listQty[index_bigger]):
            listOrderedByFrequency.append(listUniqueValues[index_bigger])
        listQty[index_bigger] = -1
    
    return listOrderedByFrequency

#tests:
print(orderByFrequency([1,2,3,4,3,3,3,6,7,1,1,9,3,2]))
print(orderByFrequency([1,2,2]))
print(orderByFrequency([1,2,1,2]))
print(orderByFrequency([2,1,2,1]))
print(orderByFrequency([3,3,3,4,4,4,4,1,5,5,5,5,5,2,2]))
print(orderByFrequency([3,3,3,6,6,6,4,4,4,4,1,6,6,5,5,5,5,5,2,2]))
print(orderByFrequency([10,20,30,30,30,40,40,50,50,50]))

结果：

[3, 3, 3, 3, 3, 1, 1, 1, 2, 2, 4, 6, 7, 9]
[2, 2, 1]
[1, 1, 2, 2]
[1, 1, 2, 2]
[5, 5, 5, 5, 5, 4, 4, 4, 4, 3, 3, 3, 2, 2, 1]
[5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 4, 4, 4, 4, 3, 3, 3, 2, 2, 1]
[30, 30, 30, 50, 50, 50, 40, 40, 10, 20]

Answer 2

如果您想使用双比较器。

例如：按频率降序对列表进行排序，如果发生冲突，则较小的排在第一位。

import collections 

def frequency_sort(a):
    f = collections.Counter(a)
    a.sort(key = lambda x:(-f[x], x))
    return a

Answer 3

您可以使用Counter 获取每个项目的计数，使用其most_common 方法按排序顺序获取它，然后使用列表推导再次展开

>>> lst = [1,2,3,4,3,3,3,6,7,1,1,9,3,2]
>>> 
>>> from collections import Counter
>>> [n for n,count in Counter(lst).most_common() for i in range(count)]
[3, 3, 3, 3, 3, 1, 1, 1, 2, 2, 4, 6, 7, 9]

Answer 4

练习这个是为了好玩。此解决方案使用较少的时间复杂度。

from collections import defaultdict

lis = [1,2,3,4,3,3,3,6,7,1,1,9,3,2]

dic = defaultdict(int)
for num in lis:
    dic[num] += 1

s_list = sorted(dic, key=dic.__getitem__, reverse=True)

new_list = []
for num in s_list:
    for rep in range(dic[num]):
        new_list.append(num)

print(new_list)

Answer 5

我认为这对collections.Counter 来说是一份不错的工作：

counts = collections.Counter(lst)
new_list = sorted(lst, key=lambda x: -counts[x])

---

或者，您可以在没有 lambda 的情况下编写第二行：

counts = collections.Counter(lst)
new_list = sorted(lst, key=counts.get, reverse=True)

如果您有多个具有相同频率的元素并且您关心这些元素保持分组，我们可以通过更改排序键来做到这一点，不仅包括计数，还包括 值：

counts = collections.Counter(lst)
new_list = sorted(lst, key=lambda x: (counts[x], x), reverse=True)

Answer 6

您可以使用以下方法。它是用简单的python编写的。

def frequencyIdentification(numArray):
frequency = dict({});
for i in numArray:
    if i in frequency.keys():
            frequency[i]=frequency[i]+1;
    else:
            frequency[i]=1;         
return frequency;

def sortArrayBasedOnFrequency(numArray):
    sortedNumArray = []
    frequency = frequencyIdentification(numArray);
    frequencyOrder = sorted(frequency, key=frequency.get);
    loop = 0;
    while len(frequencyOrder) > 0:
        num = frequencyOrder.pop()
        count = frequency[num];
        loop = loop+1;
        while count>0:
            loop = loop+1;
            sortedNumArray.append(num);
            count=count-1;
    print("loop count");
    print(loop);
    return sortedNumArray;  

a=[1, 2, 3, 4, 3, 3, 3, 6, 7, 1, 1, 9, 3, 2]
print(a);
print("sorted array based on frequency of the number"); 
print(sortArrayBasedOnFrequency(a));

Answer 7

from collections import Counter
a = [2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8]
count = Counter(a)
a = []
while len(count) > 0:
    c = count.most_common(1)
    for i in range(c[0][1]):
        a.append(c[0][0])
    del count[c[0][0]]
print(a)

Answer 8

l = [1,2,3,4,3,3,3,6,7,1,1,9,3,2]
print sorted(l,key=l.count,reverse=True)

[3, 3, 3, 3, 3, 1, 1, 1, 2, 2, 4, 6, 7, 9]