我有来自py pairwise_tukeyhsd的{{1}}的Tukey测试表。
statsmodels.stats.multicomp我将此表格设为 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
pandas。我想（通过字母）标记表示统计关系的组（101-106）。对于这个具体的例子，期望的结果是:(我不介意结果将是df，列表，字典）
df如您所见，所有共享相似字母的组具有相等的平均值（拒绝列= False），具有不同字母的组（拒绝列= True）具有不同的均值。例如，组101的平均值等于所有其他组的平均值，因为组101具有字母ab，并且所有其他组具有 a 或 b 或 ab 。另一方面，组106只有字母 b ，表示它与的所有组相似，除了，组102只有字母 a < / em>的
我找不到自动python解决方案。我saw R有一个名为group label
101 ab
102 a
103 ab
104 b
105 ab
106 b
的包，在python中有类似的东西吗？
答案 0 :(得分：1)
感谢您的贡献。我不得不稍微修改一下您的代码，以更正一些缺失的内容并适应python3。主要变化是
您的原始数据位于现在名为input2.csv的csv文件中
,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
import pandas as pd
import numpy as np
import math
import itertools
import string
df = pd.read_csv('input2.csv', index_col=0)
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.items():
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.items():
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.items():
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)
df2=pd.concat([df.group1,df.group2])
group_names=df2.unique()
for GroupName in group_names:
if GroupName in dictionary:
print ("already exists")
else:
dictionary[str(GroupName)] = final_label
for key, value in dictionary.items(): #this keeps only the unique char per group and sort it by group
dictionary[key] = ''.join(set(value))
dict2 = dict(sorted(dictionary.items())) # the final output
答案 1 :(得分：0)
所以，经过几天的讨论，没有其他用户的建议答案/评论，我想我已经明白了。假设我的问题中的表名为df。以下脚本特定于我的需求，但我希望它可以帮助其他人。我添加了评论以促进理解。
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))