根据显著性标记来自Tueky测试结果的组

Question

我有一个Tukey的测试表来自python statsmodels.stats.multicomp的{}。在

    group1  group2  meandiff    lower   upper   reject
0   101     102     0.2917    -0.0425   0.6259  False
1   101     103     0.1571    -0.1649   0.4792  False
2   101     104     -0.1333   -0.4675   0.2009  False
3   101     105     0.0833    -0.2509   0.4175  False
4   101     106     -0.0500   -0.3626   0.2626  False
5   102     103     -0.1345   -0.4566   0.1875  False
6   102     104     -0.4250   -0.7592  -0.0908  True
7   102     105     -0.2083   -0.5425   0.1259  False
8   102     106     -0.3417   -0.6543  -0.0290  True
9   103     104     -0.2905   -0.6125   0.0316  False
10  103     105     -0.0738   -0.3959   0.2482  False
11  103     106     -0.2071   -0.5067   0.0924  False
12  104     105     0.2167    -0.1175   0.5509  False
13  104     106     0.0833    -0.2293   0.3960  False
14  105     106     -0.1333   -0.4460   0.1793  False

我把这个表命名为pandasdf。我想（用字母）标记表示统计关系的组（101-106）。对于这个特定的例子，期望的结果是：（我不介意结果是df、一个列表、一个字典）

^{pr2}$

如您所见，共享相似字母的所有组都具有相同的平均值（reject column=False），具有不同字母（reject column=True）的组具有不同的平均值。例如，第101组的平均值等于所有其他组的平均值，因为第101组的字母为ab，而所有其他组的字母都是a或b或ab。另一方面，106组只有字母b，这表明它与所有组相似，除了组102只有字母a。在

我找不到一个自动的python解决方案。我sawR有一个名为multcompLetters的包，在python中有类似的东西吗？在

Answer 1

所以，经过几天的研究，没有其他用户的建议答案/评论，我想我已经明白了。假设我问题中的表名为df。下面的脚本是为我的需要，但我希望它可以帮助其他人。我添加了一些评论以便于理解。在

    df_True = df.loc[df.reject==True,:]

    letters = list(string.ascii_lowercase)
    n = 0

    group1_list = df_True.group1.tolist() #get the groups from the df with only True (True df) to a list
    group2_list = df_True.group2.tolist()
    group3 = group1_list+group2_list #concat both lists
    group4 = list(set(group3)) #get unique items from the list
    group5 = [str(i) for i in group4 ] #convert unicode to a str
    group5.sort() #sort the list

    gen = ((i, 0) for i in group5) #create dict with 0 so the dict won't be empty when starts
    dictionary = dict(gen)

    group6 = [(group5[i],group5[j]) for i in range(len(group5)) for j in range(i+1, len(group5))] #get all combination pairs
    for pairs in group6: #check for each combination if it is present in df_True

        print n
        print dictionary

        try:        
            a = df_True.loc[(df_True.group1==pairs[0])&(df_True.group2==pairs[1]),:] #check if the pair exists in the df

        except:
            a.shape[0] == 0 

        if a.shape[0] == 0: #it mean that the df is empty as it does not appear in df_True so this pair is equal

            print 'equal'

            if dictionary[pairs[0]] != 0 and dictionary[pairs[1]] == 0: #if the 1st is populated but the 2nd in not populated
                print "1st is populated and 2nd is empty"      
                dictionary[pairs[1]] = dictionary[pairs[0]]

            elif dictionary[pairs[0]] != 0 and dictionary[pairs[1]] != 0: #if both are populated, check matching labeles
                print "both are populated"
                if len(list(set([c for c in dictionary[pairs[0]] if c in dictionary[pairs[1]]]))) >0: #check if they have a common label
                        print "they have a shared character"
                else:
                    print "equal but have different labels"
                 #check if the 1st group label doesn't appear in anyother labels, if it is unique then the 2nd group can have the first group label
                    m = 0 #count the number of groups that have a shared char with 1st group
                    j = 0 #count the number of groups that have a shared char with 2nd group
                    for key, value in dictionary.iteritems():
                        if key != pairs[0] and len(list(set([c for c in dictionary[pairs[0]] if c in value])))==0:
                            m+=1
                    for key, value in dictionary.iteritems():
                        if key != pairs[1] and len(list(set([c for c in dictionary[pairs[1]] if c in value])))==0:
                            j+=1
                    if m == len(dictionary)-1 and j == len(dictionary)-1: #it means that this value is unique because it has no shared char with another group
                        print "unique"
                        dictionary[pairs[1]] = dictionary[pairs[0]][0]

                    else:
                        print "there is at least one group in the dict that shares a char with the 1st group"                                 
                        dictionary[pairs[1]] = dictionary[pairs[1]] + dictionary[pairs[0]][0]


            else:  # if it equals 0, meaning if the 1st is empty (which means that the 2nd must be also empty)
                print "both are empty"
                dictionary[pairs[0]] = letters[n]
                dictionary[pairs[1]] = letters[n]

        else:

            print "not equal"

            if dictionary[pairs[0]] != 0: # if the first one is populated (has a value) then give a value only to the second 

                print '1st is populated'            
                # if the 2nd is not empty and they don't share a charcter then no change is needed as they already have different labels
                if dictionary[pairs[1]] != 0 and len(list(set([c for c in dictionary[pairs[0]] if c in dictionary[pairs[1]]]))) == 0: 
                    print "no change"

                elif dictionary[pairs[1]] == 0: #if the 2nd is not populated give it a new letter
                    dictionary[pairs[1]] = letters[n+1]

                #if the 2nd is populated and equal to the 1st, then change the letter of the 2nd to a new one and assign its original letter to all the others that had the same original letter       
                elif  dictionary[pairs[1]] != 0 and len(list(set([c for c in dictionary[pairs[0]] if c in dictionary[pairs[1]]]))) > 0:
                    #need to check that they don't share a charcter
                    print "need to add a letter"
                    original_value = dictionary[pairs[1]]
                    dictionary[pairs[1]] = letters[n]
                    for key, value in dictionary.iteritems():
                        if key != pairs[0] and len(list(set([c for c in original_value if c in value])))>0: #for any given value, check if it had a character from the group that will get a new letter, if so, it means  that they are equal and thus the new letter should also appear in the value of the "old" group 
                            dictionary[key] = original_value + letters[n]  #add the original letter of the group to all the other groups it was similar to               

            else:

                print '1st is empty'
                dictionary[pairs[0]] = letters[n]
                dictionary[pairs[1]] = letters[n+1]
                print dictionary
            n+=1

    # get the letter out the dictionary

    labels = list(dictionary.values())
    labels1 = list(set(labels))
    labels1.sort()
    final_label = ''.join(labels1)

    for GroupName in group_names:
        if GroupName in dictionary:
            print "already exists"
        else:
            dictionary[GroupName] = final_label

    for key, value in dictionary.iteritems(): #this keeps only the unique char per group and sort it by group
        dictionary[key] =  ''.join(set(value))