英文:
How to compute the frequency of occurrence of '+' and '++' in Java?
问题
我想计算输入文本文件中所有运算符出现的频率。文件中包含运算符+和++。如何区分它们各自的频率，因为我的程序将++视为两个不同的+运算符，而不是一个++？
以下是我的代码（input7.txt是一个测试文件）：
public static void main(String[] args) throws IOException {
String string = new String(Files.readAllBytes(Paths.get("input7.txt"))); //待计数的字符串
int frequencyArray[] = new int[string.length()];
int frequencyArray2[] = new int[string.length()];
char stringArray[] = string.toCharArray(); //字符数组
int i, j;
//统计字符出现频率
for (i = 0; i < string.length(); i++) {
frequencyArray[i] = 1;
//frequencyArray2[i] = 1;
for(j = i + 1; j < string.length(); j++)
{
if(stringArray[i] == stringArray[j])
{
frequencyArray[i]++;
stringArray[j] = '0'; //避免重复访问字符
}
}
}
//显示结果
System.out.println("字符及其对应的频率");
for (i = 0; i < frequencyArray.length; i++) {
if (stringArray[i] != ' ' && stringArray[i] != '0') {
System.out.println(stringArray[i] + "-" + frequencyArray[i]);
}
}
}
英文:
I want to calculate the frequency of the occurrence of all the operators from an input text file. The file contains the operators + and ++. How can I distinguish their respective frequency, as my program treats ++ as 2 distinct + operators rather than 1 ++?
Here is my code (input7.txt is a test file):
public static void main(String[] args) throws IOException {
String string = new String(Files.readAllBytes(Paths.get("input7.txt"))); //String to be counted
int frequencyArray[] = new int[string.length()];
int frequencyArray2[] = new int[string.length()];
char stringArray[] = string.toCharArray(); //Array of characters
int i, j;
//Count characters
for (i = 0; i < string.length(); i++) {
frequencyArray[i] = 1;
//frequencyArray2[i] = 1;
for(j = i + 1; j < string.length(); j++)
{
if(stringArray[i] == stringArray[j])
{
frequencyArray[i]++;
stringArray[j] = '0'; //To avoid revisiting a character
}
}
}
//Display results
System.out.println("Characters and their corresponding frequencies");
for (i = 0; i < frequencyArray.length; i++) {
if (stringArray[i] != ' ' && stringArray[i] != '0') {
System.out.println(stringArray[i] +"-" + frequencyArray[i]);
}
}
}
</details>
# 答案1
**得分**: 0
这对我有效：
```java
String s = "sdfasd++ sdfadsf+asdf sdf++sadfasdf++sadfsdf+asdfasdf++";
// 创建包含所有不同字符的集合
Set<Character> chars = new HashSet<Character>();
for (int i = 0; i < s.length(); i++) {
chars.add(s.charAt(i));
}
// 计算不同字符数量并将结果放入哈希映射中
Map<Character, Integer> counts = new HashMap<Character, Integer>();
for (Character c : chars) {
int count = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == c)
count++;
}
counts.put(c, count);
}
// 计算双字符运算符数量
int countPlusPlus = 0;
for (int i = 0; i < s.length() - 1; i++) {
if (s.substring(i, i + 2).equals("++"))
countPlusPlus++;
}
// 计算总数
int singleplusTotal = counts.get('+');
System.out.println("单个加号总数：" + singleplusTotal);
System.out.println("双加号总数：" + countPlusPlus);
System.out.println("仅单个加号：" + (singleplusTotal - countPlusPlus * 2));
英文:
This works for me:
String s = "sdfasd++ sdfadsf+asdf sdf++sadfasdf++sadfsdf+asdfasdf++";
// create Set with all distinct characters
Set<Character> chars = new HashSet<Character>();
for (int i = 0; i < s.length(); i++) {
chars.add(s.charAt(i));
}
// count distinct characters and put Results in HashMap
Map<Character, Integer> counts = new HashMap<Character, Integer>();
for (Character c : chars) {
int count = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == c)
count++;
}
counts.put(c, count);
}
// Count double-Character-Operators like this
int countPlusPlus = 0;
for (int i = 0; i < s.length() - 1; i++) {
if (s.substring(i, i + 2).equals("++"))
countPlusPlus++;
}
// Calculate totals like this
int singleplusTotal = counts.get('+');
System.out.println("Single Plus total" + singleplusTotal);
System.out.println("Double Plus total" + countPlusPlus);
System.out.println("Only single Plus" + (singleplusTotal - countPlusPlus * 2));
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