问题
我有一个BigDecimals的集合(在这个例子中,aLinkedList
),我想加在一起。是否可以使用流?
我注意到Stream
class有几种方法
Stream::mapToInt
Stream::mapToDouble
Stream::mapToLong
每个都有方便的sum()
方法。但是,正如我们所知,float
和double
算术几乎总是一个坏主意。
那么,有没有一种方便的方法来总结BigDecimals?
这是我到目前为止的代码。
public static void main(String[] args) {
LinkedList<BigDecimal> values = new LinkedList<>();
values.add(BigDecimal.valueOf(.1));
values.add(BigDecimal.valueOf(1.1));
values.add(BigDecimal.valueOf(2.1));
values.add(BigDecimal.valueOf(.1));
// Classical Java approach
BigDecimal sum = BigDecimal.ZERO;
for(BigDecimal value : values) {
System.out.println(value);
sum = sum.add(value);
}
System.out.println("Sum = " + sum);
// Java 8 approach
values.forEach((value) -> System.out.println(value));
System.out.println("Sum = " + values.stream().mapToDouble(BigDecimal::doubleValue).sum());
System.out.println(values.stream().mapToDouble(BigDecimal::doubleValue).summaryStatistics().toString());
}
正如你所看到的,我正在使用BigDecimal::doubleValue()
总结BigDecimals,但这(正如预期的那样)并不准确。
后代的编辑后编辑:
这两个答案都非常有帮助。我想补充一点:我的现实场景不涉及rawBigDecimal
s的集合,它们包含在发票中。但是,通过使用流的map()
函数,我能够修改Aman Agnihotri的答案以解决这个问题:
public static void main(String[] args) {
LinkedList<Invoice> invoices = new LinkedList<>();
invoices.add(new Invoice("C1", "I-001", BigDecimal.valueOf(.1), BigDecimal.valueOf(10)));
invoices.add(new Invoice("C2", "I-002", BigDecimal.valueOf(.7), BigDecimal.valueOf(13)));
invoices.add(new Invoice("C3", "I-003", BigDecimal.valueOf(2.3), BigDecimal.valueOf(8)));
invoices.add(new Invoice("C4", "I-004", BigDecimal.valueOf(1.2), BigDecimal.valueOf(7)));
// Classical Java approach
BigDecimal sum = BigDecimal.ZERO;
for(Invoice invoice : invoices) {
BigDecimal total = invoice.unit_price.multiply(invoice.quantity);
System.out.println(total);
sum = sum.add(total);
}
System.out.println("Sum = " + sum);
// Java 8 approach
invoices.forEach((invoice) -> System.out.println(invoice.total()));
System.out.println("Sum = " + invoices.stream().map((x) -> x.total()).reduce((x, y) -> x.add(y)).get());
}
static class Invoice {
String company;
String invoice_number;
BigDecimal unit_price;
BigDecimal quantity;
public Invoice() {
unit_price = BigDecimal.ZERO;
quantity = BigDecimal.ZERO;
}
public Invoice(String company, String invoice_number, BigDecimal unit_price, BigDecimal quantity) {
this.company = company;
this.invoice_number = invoice_number;
this.unit_price = unit_price;
this.quantity = quantity;
}
public BigDecimal total() {
return unit_price.multiply(quantity);
}
public void setUnit_price(BigDecimal unit_price) {
this.unit_price = unit_price;
}
public void setQuantity(BigDecimal quantity) {
this.quantity = quantity;
}
public void setInvoice_number(String invoice_number) {
this.invoice_number = invoice_number;
}
public void setCompany(String company) {
this.company = company;
}
public BigDecimal getUnit_price() {
return unit_price;
}
public BigDecimal getQuantity() {
return quantity;
}
public String getInvoice_number() {
return invoice_number;
}
public String getCompany() {
return company;
}
}
#1 热门回答(209 赞)
##原始答案
是的,这是可能的:
648052058
它的作用是:
- 获取List <BigDecimal>。
- 将其转换为Stream <BigDecimal>
- 调用reduce方法。 3.1。我们提供添加的标识值,即BigDecimal.ZERO。 3.2。我们指定BinaryOperator <BigDecimal>,它通过方法引用BigDecimal :: add添加两个BigDecimal。
##在编辑后更新了答案
我看到你添加了新数据,因此新答案将变为:
List<Invoice> invoiceList = new ArrayList<>();
//populate
Function<Invoice, BigDecimal> totalMapper = invoice -> invoice.getUnit_price().multiply(invoice.getQuantity());
BigDecimal result = invoiceList.stream()
.map(totalMapper)
.reduce(BigDecimal.ZERO, BigDecimal::add);
它大致相同,只是我添加了atotalMapper
变量,其功能从Invoice
到BigDecimal
并返回该发票的总价。
然后我获得aStream<Invoice>
,将其映射到aStream<BigDecimal>
然后将其减少到aBigDecimal
。
现在,从OOP设计的角度来看,我建议你实际使用已经定义的total()
方法,然后它变得更容易:
List<Invoice> invoiceList = new ArrayList<>();
//populate
BigDecimal result = invoiceList.stream()
.map(Invoice::total)
.reduce(BigDecimal.ZERO, BigDecimal::add);
这里我们直接使用map
方法中的方法参考。
#2 热门回答(5 赞)
使用此方法对BigDecimal列表求和:
List<BigDecimal> values = ... // List of BigDecimal objects
BigDecimal sum = values.stream().reduce((x, y) -> x.add(y)).get();
此方法仅将每个BigDecimal映射为BigDecimal,并通过对它们求和来减少它们,然后使用get()
方法返回它们。
这是进行相同求和的另一种简单方法:
List<BigDecimal> values = ... // List of BigDecimal objects
BigDecimal sum = values.stream().reduce(BigDecimal::add).get();
更新
如果我在编辑的问题中编写类和lambda表达式,我会写如下:
import java.math.BigDecimal;
import java.util.LinkedList;
public class Demo
{
public static void main(String[] args)
{
LinkedList<Invoice> invoices = new LinkedList<>();
invoices.add(new Invoice("C1", "I-001", BigDecimal.valueOf(.1), BigDecimal.valueOf(10)));
invoices.add(new Invoice("C2", "I-002", BigDecimal.valueOf(.7), BigDecimal.valueOf(13)));
invoices.add(new Invoice("C3", "I-003", BigDecimal.valueOf(2.3), BigDecimal.valueOf(8)));
invoices.add(new Invoice("C4", "I-004", BigDecimal.valueOf(1.2), BigDecimal.valueOf(7)));
// Java 8 approach, using Method Reference for mapping purposes.
invoices.stream().map(Invoice::total).forEach(System.out::println);
System.out.println("Sum = " + invoices.stream().map(Invoice::total).reduce((x, y) -> x.add(y)).get());
}
// This is just my style of writing classes. Yours can differ.
static class Invoice
{
private String company;
private String number;
private BigDecimal unitPrice;
private BigDecimal quantity;
public Invoice()
{
unitPrice = quantity = BigDecimal.ZERO;
}
public Invoice(String company, String number, BigDecimal unitPrice, BigDecimal quantity)
{
setCompany(company);
setNumber(number);
setUnitPrice(unitPrice);
setQuantity(quantity);
}
public BigDecimal total()
{
return unitPrice.multiply(quantity);
}
public String getCompany()
{
return company;
}
public void setCompany(String company)
{
this.company = company;
}
public String getNumber()
{
return number;
}
public void setNumber(String number)
{
this.number = number;
}
public BigDecimal getUnitPrice()
{
return unitPrice;
}
public void setUnitPrice(BigDecimal unitPrice)
{
this.unitPrice = unitPrice;
}
public BigDecimal getQuantity()
{
return quantity;
}
public void setQuantity(BigDecimal quantity)
{
this.quantity = quantity;
}
}
}
#3 热门回答(3 赞)
你可以使用a可重复使用CollectornamedsummingUp
来总结aBigDecimal
流的值:
BigDecimal sum = bigDecimalStream.collect(summingUp());
Collector
可以像这样实现:
public static Collector<BigDecimal, ?, BigDecimal> summingUp() {
return Collectors.reducing(BigDecimal.ZERO, BigDecimal::add);
}