问题

我创建了一段代码，它读取一个字符串并创建出频率映射。我是在Java 8中提出的这个想法。

```java
final Map<Character, Integer> frequencyMap = new HashMap<>(26);
myString.chars()
    .forEach(ch -> frequencyMap.put((char)ch, frequencyMap.getOrDefault(ch, 0) + 1)); 

然而，我的输出结果如下 -

[1, 1, 1, 1, 1, 1, 1, 1]

我理解Lambda函数只能访问它作用域外部的final/实际上的final变量，但在这里我似乎也没有违反这个规则。
我尝试通过显式地使流变成顺序流来解决这个问题，但仍然不起作用。

str.chars()
    .sequential().forEach(ch -> frequencyMap.put((char)ch, frequencyMap.getOrDefault(ch, 0) + 1));

为了确保我在使用Map函数时没有问题，我尝试了相同的Map函数与数组，它可以工作 -

final Map<Character, Integer> frequencyMap = new HashMap<>(26);
for (char ch: str.toCharArray()) {
  frequencyMap.put((char)ch, frequencyMap.getOrDefault(ch, 0) + 1);
}

这里的输出结果 -

[1, 1, 1, 1, 1, 2, 3, 4]

我知道存在许多不同的解决方案来执行这个操作，但我对于我漏掉了哪个概念很感兴趣。

<details>
<summary>英文:</summary>

I created a piece of code which reads a String &amp; creates a frequency map out of it. I came up with this in Java 8.

final Map<Character, Integer> frequencyMap = new HashMap<>(26);
myString.chars()
.forEach(ch -> frequencyMap.put((char)ch, frequencyMap.getOrDefault(ch, 0) + 1));

However my output is as follows -

[1, 1, 1, 1, 1, 1, 1, 1]

I understand that the lambda functions can access only final/effectively final variables outside it's scope but here I don't seem to violate that rule as well.
I tried making the stream sequential explicitly by doing this but it still doesn't work

str.chars()
.sequential().forEach(ch -> frequencyMap.put((char)ch, frequencyMap.getOrDefault(ch, 0) + 1));

To be sure that there isn't a problem with the way I'm using Map functions I tried the same map functions with array & it works - 

final Map<Character, Integer> frequencyMap = new HashMap<>(26);
for (char ch: str.toCharArray()) {
frequencyMap.put((char)ch, frequencyMap.getOrDefault(ch, 0) + 1);
}

Output here - 

[1, 1, 1, 1, 1, 2, 3, 4]

I'm aware of the many different solutions that exist to do this but I'm interested in knowing what concept I'm missing here?

</details>


# 答案1
**得分**: 2

你的想法是正确的。如果你还将`ch`转换为`char`，你的初始代码就能够工作：

```java
final Map<Character, Integer> frequencyMap = new HashMap<>(26);
myString.chars()
    .forEach(ch -> frequencyMap.put((char)ch, frequencyMap.getOrDefault((char)ch, 0) + 1));

如果你省略这个步骤，映射的键会是java.lang.Character类型。搜索键是java.lang.Integer类型，因此它们不相等。这会导致计数始终保持为1。

final Map<Character, Integer> frequencyMap = new HashMap<>(26);
    myString.chars()
        .forEach(ch -> frequencyMap.put((char)ch, frequencyMap.getOrDefault((char)ch, 0) + 1));

答案2

得分: 0

试试这个。

Map<Character, Integer> frequencyMap = new HashMap<>(26);
myString.chars()
    .forEach(ch -> frequencyMap.compute((char)ch, (k, v) -> v == null ? 1 : v + 1));

或者

Map<Character, Integer> frequencyMap = myString.chars()
    .mapToObj(i -> (char)i)
    .collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(e -> 1)));

或者

Map<Character, Long> frequencyMap = myString.chars()
    .mapToObj(i -> (char)i)
    .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

Map<Character, Integer> frequencyMap = new HashMap<>(26);
myString.chars()
    .forEach(ch -> frequencyMap.compute((char)ch, (k, v) -> v == null ? 1 : v + 1));

Map<Character, Integer> frequencyMap = myString.chars()
    .mapToObj(i -> (char)i)
    .collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(e -> 1)));

Map<Character, Long> frequencyMap = myString.chars()
    .mapToObj(i -> (char)i)
    .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));