我是python的新手并且有一个问题。我已经想出了如何绘制函数图,但是如何绘制一个指示最大值和最小值的点?这是我的代码,它相信我正确的图表。谢谢。
import numpy as np
import matplotlib.pyplot as plt
def graph(formula, x_range):
x = np.array(x_range)
y = eval(formula)
plt.plot(x, y)
plt.show()
graph('-x**4 + 508 * x + 40', range(-10, 200))
答案 0 :(得分:2)
import java.sql.Array;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.Struct;
import oracle.jdbc.OracleCallableStatement;
import oracle.jdbc.OracleConnection;
public class ArrayExampleMain {
public static void main(String[] args) throws Exception {
OracleConnection conn = getOracleConnection().unwrap(OracleConnection.class);
System.out.println("Got Connection.");
OracleCallableStatement callStmt = null;
try {
callStmt = (OracleCallableStatement)conn.prepareCall("{call add_projects(?)}");
// create array holding values for ProjectType object's properties
Object[] project1 = new Object[] {"1", "Title 1"};
Object[] project2 = new Object[] {"2", "Title 2"};
// each struct is one ProjectType object
Struct structProject1 = conn.createStruct("PROJECT_TYPE", project1);
Struct structProject2 = conn.createStruct("PROJECT_TYPE", project2);
Struct[] structArrayOfProjects = {structProject1, structProject2};
// array holding two ProjectType objects
Array arrayOfProjects = conn.createOracleArray("MY_ARRAY", structArrayOfProjects);
callStmt.setArray(1, arrayOfProjects);
callStmt.execute();
conn.commit();
System.out.println("Committed.");
} catch (Exception e) {
if (conn != null) try { conn.rollback(); } catch (Exception ex) { System.out.println("Rollback failed."); }
throw e;
} finally {
callStmt.close();
conn.close();
}
}
public static Connection getOracleConnection() throws Exception {
String driver = "oracle.jdbc.driver.OracleDriver";
String url = "jdbc:oracle:thin:@YOUR_HOST:orcl";
String username = "hr";
String password = "password";
Class.forName(driver); // load Oracle driver
Connection conn = DriverManager.getConnection(url, username, password);
return conn;
}
}
这样的事情?