带有Map的Java8流？

Question

我有以下地图（每个键是一个String，每个值是一个List<Message>）

我map的是这样的：

1st entry :"MONDAY" -> [message1, message2, message3]

2nd entry : "TUESDAY" -> [message4, message5]

...

我的目标是更改每条消息的内容：

我在想这个：

map.entrySet().stream().peek(entry -> {

    entry.getValue().stream().peek(m -> m.setMessage(changeMessage()))

})

但是不知道如何完成并正确地完成它。

Answer 1

很遗憾，java流没有提供在不违反副作用原则的情况下更改值的直接方法：Map通常，不鼓励对流操作的行为参数产生副作用，因为它们通常会导致无意中违反无状态要求，以及其他线程安全隐患。这是一个可能的解决方案：Map<String, List<Message>> = map.entrySet().stream()&nbsp; &nbsp; .map(e -> {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // iterate entries&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; e.setValue(e.getValue().stream()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;// set a new value&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .map(message -> {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; message&nbsp; -> message.setMessage(changeMessage()); // .. update message&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return message;})&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // .. use it&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.toList()));&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // return as a List&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return e;})&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; // return an updated value&nbsp; &nbsp; .collect(Collectors.toMap(Entry::getKey, Entry::getValue));&nbsp; &nbsp; &nbsp; // collec to a Map但是，Java 提供了一个众所周知的for-each功能，可以以更直接、更易读的方式实现您的目标：for (List<Message> list: map.values()) {&nbsp;&nbsp; &nbsp; for (Message message: list) {&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; message.setMessage(changeMessage());&nbsp;&nbsp; &nbsp; }&nbsp;}

Answer 2

迭代地图，再次更改列表的每个元素，将收集的列表放在地图的相同键上。map.forEach((k,v)->{&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; map.put(k, v.stream().map(i->i+"-changed").collect(Collectors.toList()));&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; });

Answer 3

如果您只想更新message所有消息的，则无需使用整个条目集。您可以只流式传输地图的值并映射项目。forEach()更新它们的用途：map.values().stream().flatMap(List::stream) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.forEach(m&nbsp;->&nbsp;m.setMessage(changeMessage(m.getMessage())));如果您需要密钥来更改消息，您可以使用：map.forEach((key,&nbsp;messages)&nbsp;->&nbsp;messages.forEach(m&nbsp;->&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;m.setMessage(changeMessage(key,&nbsp;m.getMessage()))));