使用流和减少将 Java 字符串中的第 n 个字符乘以 2？

Question

尝试使用 Java 流并减少这段代码

//get string as char array
char[] chars = ccNumber.toCharArray();
// build array list to store integer value of each char
List<Integer> multipliedChars = new ArrayList<>();
//get char values on even positions of the array, multiply them by two and add them to array list
for (int i = 1; i < chars.length; i+=2) {
    multipliedChars.add((chars[i] * 2) / 10);
}

Java 8版本

return IntStream.range(0, ccNumber.length())
            .filter(n -> n % 2 != 0)
            .map(ccNumber::charAt)
            .reduce(1 , (x) -> (x * 2) / 10)
            .sum();

但是它抱怨说我在 reduce 方法中的参数与我正在尝试做的不兼容？

有什么解决办法吗？

Answer 1

本来我在想这里需要用第二种map方法。但实际上我们仍然可以使用 reduce:

IntStream.range(0, str.length())
            .filter(n -> n % 2 != 0)
            .map(str::charAt)
            .reduce(0, (result,next) -> result + next / 5));

根据 javadocs reduction

reduce(
       0,
       (result, next) -> result + next / 5)

The reduce operation takes two arguments:

identity: The identity element is both the initial value of the reduction and the default result if there are no elements in the stream. In this example, the identity element is 0; this is the initial value of the sum and the default value if no members exist in the collection.

accumulator: The accumulator function takes two parameters: a partial result of the reduction (in this example, the sum of all processed integers so far) and the next element of the stream (in this example, an integer). It returns a new partial result. In this example, the accumulator function is a lambda expression that adds two Integer values and returns an Integer value:
(result, next) -> result + next * / 5)

现在我们可以使用第二个 map 而不是 reduce 来做同样的事情：

IntStream.range(0, str.length())
            .filter(n -> n % 2 != 0)
            .map(str::charAt)
            .map(x -> x  / 5)
            .sum();

结果将相同：

assert  IntStream.range(0, str.length())
                .filter(n -> n % 2 != 0)
                .map(str::charAt)
                .map(x -> x  / 5)
                .sum()
                ==
        IntStream.range(0, str.length())
                .filter(n -> n % 2 != 0)
                .map(str::charAt)
                .reduce(0, (result, next) -> result + next / 5);

但是如果需要采集值，我们需要使用map:

IntStream.range(0, str.length())
                .filter(n -> n % 2 != 0)
                .map(str::charAt)
                .map(x -> x / 5).boxed()
                .collect(Collectors.toList());

正如@Holger 在下面的评论中指出的那样“除以十仍然没有任何意义。但是无论如何，乘以二，然后除以十，与除以五相同”

使用流和减少将 Java 字符串中的第 n 个字符乘以 2？

Multiplying nth character in Java String by 2 using stream and reduce?

java

reduce

integer

java-8

java-stream