判断是否有重复的元素，按元素出现的次数从高到低的顺序输出

第一种用集合法判断 普通

def isrepeat(list):#判断是否有重复数据,输出重复元素的个数
    a={}
    reelmt_l=[]
    e_index=len(list)
    re_len=len(set(list))#判断是否有重复元素
    if e_index==re_len:
        print('no doubt data')
    else:
        relist = set(list)
        for i in relist:
            if list.count(i) > 1:
                a[i]=list.count(i)#
                reelmt_l.append(i)
                print("the %d has found count %d" % (i, list.count(i)))#输出重复元素的个数
        a = sorted(a.items(), key=lambda item: item[1], reverse=True)#重复元素倒序排列
        print(a)
        print(find_index(reelmt_l,list))

#查找重复元素在list位置index
def find_index(reelemtl,list):
    re_dict={}
    e_index = len(list)
    for i in reelemtl :
        s_index = 0
        re_index = []
        while (s_index < e_index):
            try:
                temp = list.index(i, s_index, e_index)
                re_index.append(temp)
                s_index = temp + 1
            except ValueError:
                break
        re_dict[i]=re_index
    return re_dict

第二种方法 快

def setpreatlist(list):
    setlist=Counter(list)#显示重复的元素
    if len(setlist)!=0:
        # dic = sorted(setlist.items(), key=lambda item: item[1])#升序排列
        dic=sorted(setlist.items(),key=lambda  item:item[1],reverse=True)#降序
        print(dic)
    else:
        print('no doubt data！')