基于布尔值的 Java 8 过滤器
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Java 8 filter based on a boolean
提问by Nick01
I want to be able to apply a filter based on the boolean passed in.
我希望能够根据传入的布尔值应用过滤器。
public static List<Integer> multiplyNumbers(List<Integer> input, boolean ignoreEven){
return input.stream()
.filter(number -> !(number%2==0))
.map(number -> number*2)
.collect(Collectors.toList());
}
I want to make the filter step based on the ignoreEven flag. If its true, ignore even numbers. How do I go about doing it? I am doing this to avoid code duplication
我想根据 ignoreEven 标志进行过滤步骤。如果为真，则忽略偶数。我该怎么做？我这样做是为了避免代码重复
采纳答案by Joe C
Sounds like a straightforward or condition to me.
对我来说，这听起来像是一个简单的条件。
.filter(number -> !ignoreEven || (number % 2 != 0))
回答by shmosel
If you don't want to check ignoreEvenfor each element, you can define the predicate itself conditionally:
如果不想检查ignoreEven每个元素，可以有条件地定义谓词本身：
.filter(ignoreEven ? (n -> n % 2 != 0) : (n -> true))
回答by Jagannath
You can pass the condition to your multiply(...)method and pass it in filteras shown below.
您可以将条件传递给您的multiply(...)方法并将其传入filter，如下所示。
multiplyNumbers(Arrays.asList(1, 2, 3, 4, 5), (x) -> x % 2 != 0).forEach(System.out::println);
public static List<Integer> multiplyNumbers(List<Integer> input,
Predicate<Integer> filterCondition){
return input.stream()
.filter(filterCondition)
.map(number -> number*2)
.collect(Collectors.toList());
}
回答by Grzegorz Piwowarek
You can define two additional helper methods and compose them using or()method:
您可以定义两个额外的辅助方法并使用or()方法组合它们：
private static Predicate<Integer> ignoringEven(boolean ignoreEven) {
return number -> !ignoreEven;
}
private static Predicate<Integer> ignoringOdd() {
return number -> (number % 2 != 0);
}
Final result:
最后结果：
return input.stream()
.filter(ignoringEven(ignoreEven).or(ignoringOdd()))
.map(number -> number * 2)
.collect(Collectors.toList());
回答by David Leonardo Bernal
You can use the anyMatch method:
您可以使用 anyMatch 方法：
return swiftTemplateList.stream()
.map(swiftTemplate -> swiftTemplate.getName())
.anyMatch(name -> name.equalsIgnoreCase(messageName));
