java 计算给定字符串在 ArrayList 中的出现次数

Count occurrences of a given string in an ArrayList

提问by lola

I have a list of strings , I browse it and count number of "x" strings as below but the count doesn't print me the expected value:

我有一个字符串列表，我浏览它并计算“x”字符串的数量，如下所示，但计数没有打印出预期值：

ArrayList<Integer> list = new ArrayList<Integer>();

List<String> strings = table.getValue(); //this gives  ["y","z","d","x","x","d"]

int count = 0;
for (int i = 0; i < strings.size(); i++) {
    if ((strings.get(i) == "x")) {
        count++;
        list.add(count);
    }
}

System.out.println(list);

this gives []it should be 2 as I have 2 occurrences of "x"

这[]使它应该是 2，因为我有 2 次出现“x”

回答by JRL

There already is an existing methodfor this:

已经有一个现有的方法：

Collections.frequency(collection, object);

In your case, use like this (replace all of your posted code with this):

在你的情况下，像这样使用（用这个替换你发布的所有代码）：

System.out.println(java.util.Collections.frequency(table.getValue(), "x"));

回答by aioobe

You should compare strings using equalsinstead of ==. I.e. change

您应该使用equals而不是比较字符串==。即改变

if ((list.get(i) == "x"))
                 ^^

to

到

if ((list.get(i).equals("x")))
                 ^^^^^^

==compares references, while .equalscompares actual content of strings.

==比较引用，而.equals比较字符串的实际内容。

---

---

Related questions:

相关问题：

• How do I compare strings in Java?
• Java Compare Strings
• Comparing strings in Java
• https://stackoverflow.com/questions/5116595/compare-two-strings-java
• What's the quickest way to compare strings in Java?
• String Comparison in Java
• Java String.equals versus ==
• How do I make my string comparison case insensitive?
• comparing strings in java
• Strings don't seem to be equal in Java on Android, even though they print the same
• Comparing java Strings with ==
• How can I compare String value with ArrayList of String type in Java?
• How can I compare two strings in java and define which of them is smaller than the other alphabetically?

• 如何比较 Java 中的字符串？
• Java 比较字符串
• 比较Java中的字符串
• https://stackoverflow.com/questions/5116595/compare-two-strings-java
• 在 Java 中比较字符串的最快方法是什么？
• Java 中的字符串比较
• Java String.equals 与 ==
• 如何使我的字符串比较不区分大小写？
• 比较java中的字符串
• 字符串在 Android 上的 Java 中似乎并不相等，即使它们打印相同
• 比较 java 字符串与 ==
• 如何将 String 值与 Java 中的 String 类型的 ArrayList 进行比较？
• 如何比较 java 中的两个字符串并按字母顺序定义它们中的哪个比另一个小？

回答by MasterCassim

You need to use:

您需要使用：

list.get(i).equals("x");

!= / == only checks the reference.

!= / == 只检查引用。

I don't knwo why you're using a ArrayList to count. You would probably something like that:

我不知道你为什么使用 ArrayList 来计数。你可能会这样：

int count = 0;
for (String s : table.getValue()) {
    if (s.equals("x")) {
        count++;
    }
}
System.out.println( count );

回答by Vaandu

For String you should use equals method.

对于 String，您应该使用 equals 方法。

int ct = 0;
for (String str : table.getValue()) {
    if ("x".equals(str)) { // "x".equals to avoid NullPoniterException
        count++;
    }
}
System.out.println(ct);

回答by John B

Since you are looking for both the elements as well as the size, I would recommend Guava's Iterables.filtermethod

由于您正在寻找元素和大小，我会推荐番石榴的Iterables.filter方法

List<String> filtered = Lists.newArrayList(
                     Iterables.filter(myList, 
                                      Predicates.equalTo("x")));
int count = filtered.size();

But as everyone else has pointed out, the reason your code is not working is the ==

但正如其他人所指出的，您的代码无法正常工作的原因是 ==