Python pandas 数据框，检查列值是否与每一行的其他列值匹配

Question

Imagine having the columns username and userid.

Username UserId
user1    1
user1    1
user2    2
user3    1    <- this is wrong for example
user1    1

User3 has the same userid as user1, which is not supposed to be possible. How can I check if any of these occurences exist?

Answer 1

First remove all duplicates by both columns:

df1  = df.drop_duplicates(['Username','UserId'])
print (df1)
  Username  UserId
0    user1       1
2    user2       2
3    user3       1

And then get all duplicates by UserId - but here still more logic for ditingush if wrong value is for user1 or user3 :

dups = df1[df1['UserId'].duplicated(keep=False)]
print (dups)
  Username  UserId
0    user1       1
3    user3       1

---

Sample data - added next user4 for I hope better data:

print (df)
  Username  UserId
0    user1       1
1    user1       1
2    user2       2
3    user3       1
4    user4       1
5    user4       1
6    user1       1

One idea is get counts per groups by both columns by GroupBy.transform :

df['count'] = df.groupby(['Username','UserId'])['UserId'].transform('size')
print (df)
  Username  UserId  count
0    user1       1      3
1    user1       1      3
2    user2       2      1
3    user3       1      1
4    user4       1      2
5    user4       1      2
6    user1       1      3

Then remove duplicates and sorting by DataFrame.sort_values :

df1  = df.drop_duplicates(['Username','UserId']).sort_values(['UserId','count'])
print (df1)
  Username  UserId  count
3    user3       1      1
4    user4       1      2
0    user1       1      3
2    user2       2      1

Get all dupes:

mask1 = df1['UserId'].duplicated(keep=False)
dups = df1[mask1]
print (dups)
  Username  UserId  count
3    user3       1      1
4    user4       1      2
0    user1       1      3

Remove dupe with maximal count by chain Series.duplicated with keep='last' :

dups_without_max_count = df1[df1['UserId'].duplicated(keep='last') & mask1]
print (dups_without_max_count)
  Username  UserId  count
3    user3       1      1
4    user4       1      2