我认为有predict.py（202~235）是否有不充分之处？期望回复.

MaxScores 列表与 MaxScoreFaceProposal列表同时生成，索引对齐，但是DiffentGroups列表是MaxScoreFaceProposal 去重后的列表，如果MaxScoreFaceProposal列表中确实有重复元素，那么在后续: IntersectScores.append(MaxScores[kk]) 这里将有问题

          # 上次循环
            maxscore = max(scoreinonepropose)
            MaxScores.append(maxscore)
            indexofMaxScore = scoreinonepropose.index(maxscore)
            MaxScoreFaceProposal.append(myindexofallindexsim[indexofMaxScore])
            allscoreofproposal.append(scoreinonepropose)
          # 循环结束

        DiffentGroups = []
        for j in range(len(MaxScoreFaceProposal)):
            if j == 0:
                DiffentGroups.append(MaxScoreFaceProposal[j])
            else:
                if MaxScoreFaceProposal[j] not in DiffentGroups:
                    DiffentGroups.append(MaxScoreFaceProposal[j])
            for diffentgroup in DiffentGroups:
            num = 0
            for element in diffentgroup:
                for otherdiffentgroup in DiffentGroups:
                    if diffentgroup != otherdiffentgroup:
                        if element in otherdiffentgroup:
                            num += 1
                            break
            if num == 0:
                IsolatedProposals.append(diffentgroup)
            else:
                IntersectProposals.append(diffentgroup)
                IntersectScores.append(MaxScores[kk])   # 这里有问题

还有一处，万一下面代码中的score值有相同的，但是list.index()只会返回第一个索引：

    IntersectScoresCopy = IntersectScores.copy()
    IntersectScoresCopy.sort(reverse=True)
    indexesofIntersectScores = []
    for IntersectScore in IntersectScoresCopy:
        index = IntersectScores.index(IntersectScore) #万一有相同的score
        indexesofIntersectScores.append(index)

3.我的修正如下：

DiffentGroups = []
New_MaxScores=[] #-----------------> 增加
for j in range(len(MaxScoreFaceProposal)):
    if j == 0:
        DiffentGroups.append(MaxScoreFaceProposal[j])
        New_MaxScores.append(MaxScores[j])

    else:
        if MaxScoreFaceProposal[j] not in DiffentGroups:
            DiffentGroups.append(MaxScoreFaceProposal[j])
            New_MaxScores.append(MaxScores[j]) #-----------------> 增加

IsolatedProposals = []
IntersectProposals = []
IntersectScores = []
kk = 0 #记录DiffentGroups的索引
for diffentgroup in DiffentGroups: #对于每一个group 计算与其他group的交叉情况，如果没有交叉，正好，他就是一个实例
    num = 0
    for element in diffentgroup:
        for otherdiffentgroup in DiffentGroups:
            if diffentgroup != otherdiffentgroup:
                if element in otherdiffentgroup:
                    num += 1
                    break
    if num == 0:
        IsolatedProposals.append(diffentgroup)
    else:
        IntersectProposals.append(diffentgroup)
        #IntersectScores.append(MaxScores[kk]) #感觉有问题,因为diffentgroup那一步真的可能过滤调一些东西，造成索引错误
        IntersectScores.append(New_MaxScores[kk])
    kk += 1

# 对于存在交叉的组，按照分去重比如【0，1，2】与【1，2】
# 首先先整理一下得分---------------------------------
# IntersectScoresCopy = IntersectScores.copy()
# IntersectScoresCopy.sort(reverse=True)# 降序拍，第一个最大
# indexesofIntersectScores = []
# 找降序排后的索引，因为index只返回以一个索引，有问题！！！
# for IntersectScore in IntersectScoresCopy:
#     index = IntersectScores.index(IntersectScore)
#     indexesofIntersectScores.append(index) # 万一有得分一样的？

# 上方注释，只用下面的一行即可
indexesofIntersectScores = sorted(range(len(IntersectScores)), key=lambda k: IntersectScores[k], reverse=True)
# 得分整理完毕------------------------------------

HARRIXJANG / ASIN-master

我认为有predict.py（202~235）是否有不充分之处？期望回复. #4