2).words = ["apple", "banana", "cherry", "date", "elephant", "grape"]，编写一个列表生成式，生成一个新的列表，其中包含words中长度超过5个字符的单词

答案:
你可以使用列表生成式来生成一个新的列表，其中包含words中长度超过5个字符的单词。以下是相应的代码：

words = ["apple", "banana", "cherry", "date", "elephant", "grape"]
new_list = [word for word in words if len(word) > 5]
print(new_list)

输出结果为：['banana', 'cherry', 'elephant']

相关问题

续写代码public abstract class Animal { private String name; public Animal(String name) { this.name = name; } public String getName() { return name; } public void setName(String name) { this.name = name; } public abstract void makeSound(); } // 包 animal package animal; public class Cat extends Animal { private int age; public Cat(String name, int age) { super(name); this.age = age; } public int getAge() { return age; } public void setAge(int age) { this.age = age; } @Override public void makeSound() { System.out.println("Meow"); } } public class Dog extends Animal { private String breed; public Dog(String name, String breed) { super(name); this.breed = breed; } public String getBreed() { return breed; } public void setBreed(String breed) { this.breed = breed; } @Override public void makeSound() { System.out.println("Woof"); } } public class Elephant extends Animal { private int weight; public Elephant(String name, int weight) { super(name); this.weight = weight; } public int getWeight() { return weight; } public void setWeight(int weight) { this.weight = weight; } @Override要求0.必须有抽象父类，且被继承 1.必须包含 >=2 个包2.必须创建 >=5 个类(其中包含抽象父类)3.类中属性皆为私有属性，需定义读写方法 4.子类需要重写个别父类方法

续写代码：

// 包 animal
package animal;

public abstract class Animal {
private String name;

public Animal(String name) {
this.name = name;
}

public String getName() {
return name;
}

public void setName(String name) {
this.name = name;
}

public abstract void makeSound();
}

public class Cat extends Animal {
private int age;

public Cat(String name, int age) {
super(name);
this.age = age;
}

public int getAge() {
return age;
}

public void setAge(int age) {
this.age = age;
}

@Override
public void makeSound() {
System.out.println("Meow");
}
}

public class Dog extends Animal {
private String breed;

public Dog(String name, String breed) {
super(name);
this.breed = breed;
}

public String getBreed() {
return breed;
}

public void setBreed(String breed) {
this.breed = breed;
}

@Override
public void makeSound() {
System.out.println("Woof");
}
}

public class Elephant extends Animal {
private int weight;

public Elephant(String name, int weight) {
super(name);
this.weight = weight;
}

public int getWeight() {
return weight;
}

public void setWeight(int weight) {
this.weight = weight;
}

@Override
＊＊＊

图片有Apple Grape Banana Elephant Pigeon Leopard

Thank you for providing more information. Based on the given objects, I would design the following class hierarchy:

Animal
├── Elephant
└── Leopard

Fruit
├── Apple
├── Grape
└── Banana

Bird
└── Pigeon

In this hierarchy, `Animal`, `Fruit`, and `Bird` are abstract base classes. `Elephant` and `Leopard` are concrete subclasses of `Animal`, `Apple`, `Grape`, and `Banana` are concrete subclasses of `Fruit`, and `Pigeon` is a concrete subclass of `Bird`.

This hierarchy reflects the common characteristics and behaviors of each group of objects. `Animal` has common properties such as habitat, diet, and behavior, and it makes sense to group `Elephant` and `Leopard` together as they share those characteristics. Similarly, `Fruit` has common properties such as taste, nutrients, and usage, and it makes sense to group `Apple`, `Grape`, and `Banana` together. `Bird`, on the other hand, only has one concrete subclass `Pigeon`, but it still makes sense to group it separately from other objects.