使用Java流将对象列表拆分为多个字段值列表 [英] Split list of objects into multiple lists of fields values using Java streams

问题描述

假设我有这样的对象:

public class Customer {

    private Integer id;
    private String country;
    private Integer customerId;
    private String name;
    private String surname;
    private Date dateOfBirth;
}

，我有一个List<Customer>.我想用Java流拆分这样的列表，以便获得IDs List<Integer>，国家/地区List<String>，customerIds List<Integer>等的列表.

and I have a List<Customer>. I would like to split such list with Java streams so that I would get a list of ids List<Integer>, countries List<String>, customerIds List<Integer> etc.

我知道我可以像制作6个流一样简单:

I know that I could do it as simple as making 6 streams such as:

List<Integer> idsList = customerList.stream()
        .map(Customer::getId)
        .collect(Collectors.toList());

但是这样做很多次我都觉得很乏味.我当时在考虑自定义收集器，但是我想不出任何既实用又有效的有用的东西.

but doing it that many times that I have fields seems pretty dull. I was thinking about custom Collector but I could not come up with anything useful that would be both neat and efficient.

推荐答案

对于类型安全的解决方案，您需要定义一个包含所需结果的类.此类型还可能提供添加其他Customer或部分结果的必要方法:

For a type safe solution, you’d need to define a class holding the desired results. This type may also provide the necessary methods for adding another Customer or partial result:

public class CustomerProperties {
    private List<Integer> id = new ArrayList<>();
    private List<String> country = new ArrayList<>();
    private List<Integer> customerId = new ArrayList<>();
    private List<String> name = new ArrayList<>();
    private List<String> surname = new ArrayList<>();
    private List<Date> dateOfBirth = new ArrayList<>();

    public void add(Customer c) {
        id.add(c.getId());
        country.add(c.getCountry());
        customerId.add(c.getCustomerId());
        name.add(c.getName());
        surname.add(c.getSurname());
        dateOfBirth.add(c.getDateOfBirth());
    }
    public void add(CustomerProperties c) {
        id.addAll(c.id);
        country.addAll(c.country);
        customerId.addAll(c.customerId);
        name.addAll(c.name);
        surname.addAll(c.surname);
        dateOfBirth.addAll(c.dateOfBirth);
    }
}

然后，您可以收集所有结果，例如

Then, you can collect all results like

CustomerProperties all = customers.stream()
    .collect(CustomerProperties::new, CustomerProperties::add, CustomerProperties::add);

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