如何删除数据集中的重复值:python
我希望通过保留具有最高值的项来删除数据集中的重复项。现在我正在使用熊猫:如何删除数据集中的重复值:python,python,pandas,Python,Pandas,我希望通过保留具有最高值的项来删除数据集中的重复项。现在我正在使用熊猫: c_maxes = hospProfiling.groupby(['Hospital_ID', 'District_ID'], group_keys=False)\ .apply(lambda x: x.ix[x['Hospital_employees'].idxmax()]) print c_maxes c_maxes.to_csv('data/external/HospitalProf
c_maxes = hospProfiling.groupby(['Hospital_ID', 'District_ID'], group_keys=False)\
.apply(lambda x: x.ix[x['Hospital_employees'].idxmax()])
print c_maxes
c_maxes.to_csv('data/external/HospitalProfilingMaxes.csv')
这样做会导致初始数据集:Hospital\u ID,District\u ID,Hospital\u employees
变成Hospital\u ID,District\u ID,Hospital\u ID,Hospital\u employees
正在复制用于分组的列。这里有什么错误
编辑:
在使用groupby()函数时,会在数据的开头添加一个额外的列。列没有名称,只是所有行的序列号。这在此处问题的输出第二个答案中显示。我想删除这个额外的列,因为我不需要它。我试过这个:
hosprofiling.drop(hosprofiling.columns[0],axis=1)
此代码不会删除该列。如何将其删除?我认为您需要:
hospProfiling.loc[hospProfiling.groupby(['Hospital_ID', 'District_ID'])['Hospital_employees']
.idxmax()]
我对另一个答案感到非常惊讶,我做了一些研究,看看函数是否无用:
样本:
hospProfiling = pd.DataFrame({'Hospital_ID': {0: 'A', 1: 'A', 2: 'B', 3: 'A', 4: 'A', 5: 'B', 6: 'A', 7: 'A', 8: 'B', 9: 'B', 10: 'A', 11: 'B', 12: 'A'}, 'Name': {0: 'Sam', 1: 'Annie', 2: 'Fred', 3: 'Sam', 4: 'Annie', 5: 'Fred', 6: 'Sam', 7: 'Annie', 8: 'Fred', 9: 'James', 10: 'Alan', 11: 'Julie', 12: 'Greg'}, 'District_ID': {0: 'M', 1: 'F', 2: 'M', 3: 'M', 4: 'F', 5: 'M', 6: 'M', 7: 'F', 8: 'M', 9: 'M', 10: 'M', 11: 'F', 12: 'M'}, 'Hospital_employees': {0: 25, 1: 41, 2: 70, 3: 44, 4: 12, 5: 14, 6: 20, 7: 10, 8: 30, 9: 18, 10: 56, 11: 28, 12: 33}, 'Val': {0: 100, 1: 7, 2: 14, 3: 200, 4: 5, 5: 20, 6: 1, 7: 0, 8: 7, 9: 9, 10: 6, 11: 9, 12: 47}})
hospProfiling = hospProfiling[['Hospital_ID','District_ID','Hospital_employees','Val','Name']]
hospProfiling.sort_values(by=['Hospital_ID','District_ID'], inplace=True)
print (hospProfiling)
Hospital_ID District_ID Hospital_employees Val Name
1 A F 41 7 Annie
4 A F 12 5 Annie
7 A F 10 0 Annie
0 A M 25 100 Sam
3 A M 44 200 Sam
6 A M 20 1 Sam
10 A M 56 6 Alan
12 A M 33 47 Greg
11 B F 28 9 Julie
2 B M 70 14 Fred
5 B M 14 20 Fred
8 B M 30 7 Fred
9 B M 18 9 James
主要区别在于如何处理其他列,如果使用max
它将从每个列返回最大值-这里是Hospital_employees
和Val
:
c_maxes = hospProfiling.groupby(['Hospital_ID','District_ID'],as_index = False).max()
print (c_maxes)
Hospital_ID District_ID Hospital_employees Name Val
0 A F 41 Annie 7
1 A M 56 Sam 200
2 B F 28 Julie 9
3 B M 70 James 20
c_maxes = hospProfiling.groupby(['Hospital_ID','District_ID'],as_index = False)
.agg({'Hospital_employees': max})
print (c_maxes)
Hospital_ID District_ID Hospital_employees
0 A F 41
1 A M 56
2 B F 28
3 B M 70
函数idxmax
返回另一列中最大值的索引:
print (hospProfiling.groupby(['Hospital_ID', 'District_ID'])['Hospital_employees'].idxmax())
A F 1
M 10
B F 11
M 2
Name: Hospital_employees, dtype: int64
然后您只能通过以下方式选择DataFrame
:
我认为你需要:
hospProfiling.loc[hospProfiling.groupby(['Hospital_ID', 'District_ID'])['Hospital_employees']
.idxmax()]
我对另一个答案感到非常惊讶,我做了一些研究,看看函数是否无用:
样本:
hospProfiling = pd.DataFrame({'Hospital_ID': {0: 'A', 1: 'A', 2: 'B', 3: 'A', 4: 'A', 5: 'B', 6: 'A', 7: 'A', 8: 'B', 9: 'B', 10: 'A', 11: 'B', 12: 'A'}, 'Name': {0: 'Sam', 1: 'Annie', 2: 'Fred', 3: 'Sam', 4: 'Annie', 5: 'Fred', 6: 'Sam', 7: 'Annie', 8: 'Fred', 9: 'James', 10: 'Alan', 11: 'Julie', 12: 'Greg'}, 'District_ID': {0: 'M', 1: 'F', 2: 'M', 3: 'M', 4: 'F', 5: 'M', 6: 'M', 7: 'F', 8: 'M', 9: 'M', 10: 'M', 11: 'F', 12: 'M'}, 'Hospital_employees': {0: 25, 1: 41, 2: 70, 3: 44, 4: 12, 5: 14, 6: 20, 7: 10, 8: 30, 9: 18, 10: 56, 11: 28, 12: 33}, 'Val': {0: 100, 1: 7, 2: 14, 3: 200, 4: 5, 5: 20, 6: 1, 7: 0, 8: 7, 9: 9, 10: 6, 11: 9, 12: 47}})
hospProfiling = hospProfiling[['Hospital_ID','District_ID','Hospital_employees','Val','Name']]
hospProfiling.sort_values(by=['Hospital_ID','District_ID'], inplace=True)
print (hospProfiling)
Hospital_ID District_ID Hospital_employees Val Name
1 A F 41 7 Annie
4 A F 12 5 Annie
7 A F 10 0 Annie
0 A M 25 100 Sam
3 A M 44 200 Sam
6 A M 20 1 Sam
10 A M 56 6 Alan
12 A M 33 47 Greg
11 B F 28 9 Julie
2 B M 70 14 Fred
5 B M 14 20 Fred
8 B M 30 7 Fred
9 B M 18 9 James
主要区别在于如何处理其他列,如果使用max
它将从每个列返回最大值-这里是Hospital_employees
和Val
:
c_maxes = hospProfiling.groupby(['Hospital_ID','District_ID'],as_index = False).max()
print (c_maxes)
Hospital_ID District_ID Hospital_employees Name Val
0 A F 41 Annie 7
1 A M 56 Sam 200
2 B F 28 Julie 9
3 B M 70 James 20
c_maxes = hospProfiling.groupby(['Hospital_ID','District_ID'],as_index = False)
.agg({'Hospital_employees': max})
print (c_maxes)
Hospital_ID District_ID Hospital_employees
0 A F 41
1 A M 56
2 B F 28
3 B M 70
函数idxmax
返回另一列中最大值的索引:
print (hospProfiling.groupby(['Hospital_ID', 'District_ID'])['Hospital_employees'].idxmax())
A F 1
M 10
B F 11
M 2
Name: Hospital_employees, dtype: int64
然后您只能通过以下方式选择DataFrame
:
为什么不使用groupby
max
方法
hopsProfiling.groupby(['Hospital_ID','District_ID'],as_index = False).max()
如果恰好有三列以上,请将max替换为agg:
hopsProfiling.groupby(['Hospital_ID','District_ID'],as_index = False).agg({'Hospital employees': max})
为什么不使用groupby
max
方法
hopsProfiling.groupby(['Hospital_ID','District_ID'],as_index = False).max()
如果恰好有三列以上,请将max替换为agg:
hopsProfiling.groupby(['Hospital_ID','District_ID'],as_index = False).agg({'Hospital employees': max})