java代码实现:从一个list中取出属性A的值不大于5且不小于4对应的对象,用lumbda表达式实现
时间: 2024-03-27 18:34:51 浏览: 40
使用Lambda表达式实现代码如下:
```java
List<Obj> list = new ArrayList<>(); // 假设已经有了一个Obj对象的List
List<Obj> result = list.stream()
.filter(obj -> obj.getA() >= 4 && obj.getA() <= 5)
.collect(Collectors.toList());
```
其中,`.stream()`将List转为Stream,`.filter()`用于过滤符合条件的对象,`.collect(Collectors.toList())`将过滤后的结果转为List。`obj -> obj.getA() >= 4 && obj.getA() <= 5`为Lambda表达式,表示筛选出属性A的值不大于5且不小于4的对象。
相关问题
write a lumbda function to formate a name by python
Here's an example of a Lambda function written in Python that formats a name:
```
import json
def lambda_handler(event, context):
first_name = event['first_name'].capitalize()
last_name = event['last_name'].capitalize()
full_name = f"{first_name} {last_name}"
return {
'statusCode': 200,
'body': json.dumps(full_name)
}
```
This function takes in two parameters, `first_name` and `last_name`, which are passed in as part of an event object. It then capitalizes the first letter of each name and combines them into a full name. Finally, the function returns the formatted name as a JSON object with a status code of 200.