使用此代码
a = pd.DataFrame([1,2,3])
如果我这样做
a.loc[0,'A']=3
我明白了:
0 A
0 1 3
1 2 NaN
2 3 NaN
现在假设:
a = pd.DataFrame([1,2,3])
b=pd.DataFrame([[5,4],[6,3],[7,2]],columns=['B','C'])
有没有办法做[1,'A','B'] = b [2]得到的结果
0 B C
0 1 NaN NaN
1 2 7 2
2 3 NaN NaN
更新:我已经改变了一点问题,因为答案说明a和b有相同的索引,而不是。
答案 0 :(得分:4)
您可以在join
上使用a
所需的b
行
In [150]: a.join(b.loc[0:0])
Out[150]:
0 B C
0 1 5.0 4.0
1 2 NaN NaN
2 3 NaN NaN
答案 1 :(得分:1)
一种可能性:
public Details(String abc) {
initComponents();
jLabel5.setText(abc);
}
//To change body of generated methods, choose Tools | Templates.
Details() {
//To change body of generated methods, choose Tools | Templates.
}
private void jButton2ActionPerformed(java.awt.event.ActionEvent evt) {
try{
Class.forName("java.sql.Driver");
Connection con=(Connection) DriverManager.getConnection("jdbc:mysql://localhost/foodframe", "root", "valli99");
Statement stmt=(Statement) con.createStatement();
String query="Update user set Name='"+jTextField1.getText()+"', Age='"+age+"', Restrictions='"+Restriction+"', Dislikes_Allergies='"+jTextArea1.getText()+"' where Username='"+jLabel5.getText()+"';";
stmt.executeUpdate(query);
rs.close();
stmt.close();
con.close();
}
catch(Exception e){
JOptionPane.showMessageDialog(null, e.getMessage());
}
}
给出import pandas as pd
a = pd.DataFrame([1, 2, 3])
b = pd.DataFrame([[5, 4], [6, 3], [7, 2]], columns=['B', 'C'])
a.loc[0, 'B'] = b.loc[0, 'B']
a.loc[0, 'C'] = b.loc[0, 'C']
:
a
答案 2 :(得分:1)
pd.concat
:
pd.concat([a,b.loc[b.index==0,:]],axis=1)
Out[53]:
0 B C
0 1 5.0 4.0
1 2 NaN NaN
2 3 NaN NaN