使用流转换和过滤 Java Map

2022-01-22 00:00:00 java-8 java java-stream collectors

我有一个想要转换和过滤的 Java 地图.作为一个简单的例子，假设我想将所有值转换为整数，然后删除奇数项.

I have a Java Map that I'd like to transform and filter. As a trivial example, suppose I want to convert all values to Integers then remove the odd entries.

Map<String, String> input = new HashMap<>();
input.put("a", "1234");
input.put("b", "2345");
input.put("c", "3456");
input.put("d", "4567");

Map<String, Integer> output = input.entrySet().stream()
        .collect(Collectors.toMap(
                Map.Entry::getKey,
                e -> Integer.parseInt(e.getValue())
        ))
        .entrySet().stream()
        .filter(e -> e.getValue() % 2 == 0)
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));


System.out.println(output.toString());

这是正确的并产生:{a=1234, c=3456}

但是，我不禁想知道是否有办法避免两次调用 .entrySet().stream().

However, I can't help but wonder if there's a way to avoid calling .entrySet().stream() twice.

有没有一种方法可以同时执行转换和过滤操作，并且在最后只调用一次 .collect()?

Is there a way I can perform both transform and filter operations and call .collect() only once at the end?

推荐答案

是的，您可以将每个条目映射到另一个临时条目，该条目将保存键和解析的整数值.然后您可以根据每个条目的值过滤它们.

Yes, you can map each entry to another temporary entry that will hold the key and the parsed integer value. Then you can filter each entry based on their value.

Map<String, Integer> output =
    input.entrySet()
         .stream()
         .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(), Integer.valueOf(e.getValue())))
         .filter(e -> e.getValue() % 2 == 0)
         .collect(Collectors.toMap(
             Map.Entry::getKey,
             Map.Entry::getValue
         ));

请注意，我使用 Integer.valueOf 而不是 parseInt 因为我们实际上想要一个装箱的 int.

Note that I used Integer.valueOf instead of parseInt since we actually want a boxed int.

如果你有幸使用 StreamEx 库，你可以很简单地做到这一点:

If you have the luxury to use the StreamEx library, you can do it quite simply:

Map<String, Integer> output =
    EntryStream.of(input).mapValues(Integer::valueOf).filterValues(v -> v % 2 == 0).toMap();

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