Question

我有两个数据框。第一个：

import pandas as pd
a = [['xxx', 'admin'], ['yyy', 'admin,super admin'], ['zzz', 'guest,admin,superadmin']]
df1 = pd.DataFrame(a, columns=['user', 'groups'])

第二个：

b = [['xxx', 'admin,super admin'], ['www', 'admin,super admin'], ['zzz', 'guest,superadmin']]
df2 = pd.DataFrame(b, columns=['user', 'groups'])

这是第一个：

 user  groups
0  xxx  admin
1  yyy  admin,super admin
2  zzz  guest,admin,superadmin

这是第二个：

 user  groups
0  xxx  admin,super admin
1  www  admin,super admin
2  zzz  guest,superadmin

我想做两件事：

如果第二个用户不在第一个用户中，则打印出来。像：www不在列表中
如果用户在列表中，但组不相等，则打印出：
像xxx这样的用户拥有更多：super admin比列表更多
zzz用户的数量：admin比列表少。

Answer 1

如果两个DataFrame中都有相同的索引值且长度不同，则需要每行比较值：

print (df1.index.equals(df2.index))
True

#compare rows for not equal
mask = df1['user'].ne(df2['user'])
#filter rows by mask and column user in df2
a = df2.loc[mask, 'user'].tolist()
print (a)
['www']

#join both DataFrames together
df1 = pd.concat([df1, df2], axis=1, keys=('a','b'))
df1.columns  = df1.columns.map('_'.join)
#filter only same user rows
df1 = df1[~mask]
#split columns by , ans convert to sets
df1['a'] = df1['a_groups'].apply(lambda x: set(x.split(',')))
df1['b'] = df1['b_groups'].apply(lambda x: set(x.split(',')))
#get difference of sets, join to strings with separator ,
df1['a_diff'] = [', '.join(x.difference(y)) for x, y in zip(df1['b'],df1['a'] )]
df1['b_diff'] = [', '.join(x.difference(y)) for x, y in zip(df1['a'],df1['b'] )]
print (df1)
  a_user                a_groups b_user           b_groups  \
0    xxx                   admin    xxx  admin,super admin   
2    zzz  guest,admin,superadmin    zzz   guest,superadmin   

                            a                     b       a_diff b_diff  
0                     {admin}  {admin, super admin}  super admin         
2  {admin, superadmin, guest}   {superadmin, guest}               admin

#filter by casting set columns to boolean, empty sets are converted to False
b = df1.loc[df1['a_diff'].astype(bool), ['a_user','a_diff']]
print (b)
  a_user       a_diff
0    xxx  super admin

c = df1.loc[df1['b_diff'].astype(bool), ['a_user','b_diff']]
print (c)
  a_user b_diff
2    zzz  admin

我想比较熊猫的价值

1 个答案: