我有一个对象:
public class CurrencyItem {
private CurrencyName name;
private BigDecimal buy;
private BigDecimal sale;
private Date date;
//...
}
其中CurrencyName
之一是:EUR,USD,RUR等。
两个清单
List<CurrencyItem> currenciesByCommercialBank = ...
List<CurrencyItem> currenciesByCentralBank = ...
如何将此列表合并到Map<CurrencyItem, CurrencyItem>
,其中键为currenciesByCommercialBank
,值为currenciesByCentralBank
且条件为
currenciesByCommercialBank.CurrencyName == currenciesByCentralBank.CurrencyName
答案 0 :(得分:6)
这应该是最佳的。您首先要建立一个从货币到商业银行的地图。然后你穿过你的中心建立一个从商业到中心的地图(在第一张地图中查找)。
List<CurrencyItem> currenciesByCommercialBank = new ArrayList<>();
List<CurrencyItem> currenciesByCentralBank = new ArrayList<>();
// Build my lookup from CurrencyName to CommercialBank.
Map<CurrencyName, CurrencyItem> commercials = currenciesByCommercialBank
.stream()
.collect(
Collectors.toMap(
// Map from currency name.
ci -> ci.getName(),
// To the commercial bank itself.
ci -> ci));
Map<CurrencyItem, CurrencyItem> commercialToCentral = currenciesByCentralBank
.stream()
.collect(
Collectors.toMap(
// Map from the equivalent commercial
ci -> commercials.get(ci.getName()),
// To this central.
ci -> ci
));
答案 1 :(得分:2)
以下代码是O(n 2 ),但对于小型集合(可能是您的列表)应该没问题:
return currenciesByCommercialBank
.stream()
.map(c ->
new AbstractMap.SimpleImmutableEntry<>(
c, currenciesByCentralBank.stream()
.filter(c2 -> c.currencyName == c2.currencyName)
.findFirst()
.get()))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
}
如果您想声明currenciesByCentralBank
包含currenciesByCommercialBank
中每个项目的匹配项,则上述内容是合适的。如果这两个列表可能不匹配,那么以下内容是合适的:
currenciesByCommercialBank
.stream()
.flatMap(c ->
currenciesByCentralBank.stream()
.filter(c2 -> c.currencyName == c2.currencyName)
.map(c2 -> new AbstractMap.SimpleImmutableEntry<>(c, c2)))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
在这种情况下,地图将包含所有匹配项,并且不会抱怨缺少条目。