python实现：重复列表按重复次数排序

#重复列表按重复次数排序方法1
s = 'aacbddbcdadb'
lists1=list(s)
uniques1 = set(lists1)
dict_str = {}
for unique1 in uniques1:
    i=0
    for list1 in lists1:
        if list1 == unique1:
            i += 1
    dict_str[unique1]=i

print(dict_str)
list2=[]

while dict_str:
    max_count = max(dict_str.values())
    for key,value in dict_str.items():
        if value == max_count:
            for i in range(0,value):
                list2.append(key)
            break
    del dict_str[key]
    print("after del dict_str-"+str(dict_str))
print(list2)

#重复列表按重复次数排序方法2
s = 'aacbddbcdadb'
lists = list(s)
uniquelist = set(lists)
dict={}
result_list = []
for unique in uniquelist:
    dict[unique] = lists.count(unique)
print(dict)
while dict:
    max_count = max(dict.values())
    for key,value in dict.items():
        if value == max_count:
            for i in range(1,value+1):
                result_list.append(key)
            break
    del dict[key]

print(result_list)

#重复列表按重复次数排序方法3  ------------这个方案比较好，对字典按照value进行了倒序排列，使用到了lambda匿名函数。
s = 'aacbddbcdadb'
lists = list(s)
uniquelist = set(lists)
dict={}
ordered_dict={}
result_list = []
for unique in uniquelist:
    dict[unique] = lists.count(unique)
print(dict)
ordered_dict = sorted(dict.items(),key=lambda x:x[1],reverse=True)  ##对字典按照value进行倒序排序，使用了匿名函数lambda
print(ordered_dict)
for key,value in ordered_dict:
    for i in range(1,value+1):
        result_list.append(key)
print(result_list)

# ordered_dict = sorted(dict.items(),key=lambda x:x[1],reverse=True)