Java 如何计算字符串中字符的频率?
我需要写一些循环,可以计算字符串中每个字母的频率Java 如何计算字符串中字符的频率?,java,string,key-value,Java,String,Key Value,我需要写一些循环,可以计算字符串中每个字母的频率 例如:“aasjjikkk”将计算2'a',1's',2'j',1'i',3'k'。最终,我希望它们最终出现在一个以字符为键、以计数为值的映射中。有什么好办法吗 您可以使用(从)。它将为您提供每个对象的计数。例如: Multiset<Character> chars = HashMultiset.create(); for (int i = 0; i < string.length(); i++) { chars.add
例如:“aasjjikkk”将计算2'a',1's',2'j',1'i',3'k'。最终,我希望它们最终出现在一个以字符为键、以计数为值的映射中。有什么好办法吗 您可以使用(从)。它将为您提供每个对象的计数。例如:
Multiset<Character> chars = HashMultiset.create();
for (int i = 0; i < string.length(); i++) {
chars.add(string.charAt(i));
}
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
String s = "aasjjikkk";
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
Integer val = map.get(c);
if (val != null) {
map.put(c, new Integer(val + 1));
}
else {
map.put(c, 1);
}
}
Multiset chars=HashMultiset.create();
对于(int i=0;i
然后,对于每个字符,您可以调用
chars.count('a')
,它返回出现的次数如果不需要超快速,只需创建一个整数数组,每个字母一个整数(仅字母so 2*26整数?或任何可能的二进制数据?)。一次遍历字符串一个字符,获取负责整数的索引(例如,如果您只有字母字符,则可以将“a”置于索引0处,并通过将“a”减去“Z”中的任何“a”来获取该索引,这只是一个如何获得相当快的索引的示例)并增加该索引中的值
有各种各样的微优化可以加快速度(如有必要)。好吧,有两种方法可以考虑,这取决于您的偏好:
String text = "aasjjikkk";
char[] charArray = text.toCharArray();
Map<Character, Integer> freqList = new LinkedHashMap<Character, Integer>();
for(char key : charArray) {
if(freqList.containsKey(key)) {
freqList.put(key, freqList.get(key) + 1);
} else
freqList.put(key, 1);
}
for(int i=0;i
freq[str[i]-'a']++//假设所有字符都是小写的
因此,“a”的数量将存储在频率[0]处,“z”的数量将存储在频率[25]您可以使用java映射并将
char
映射到int
。然后可以迭代字符串中的字符,检查它们是否已添加到映射中,如果已添加,则可以增加其值
例如:
Multiset<Character> chars = HashMultiset.create();
for (int i = 0; i < string.length(); i++) {
chars.add(string.charAt(i));
}
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
String s = "aasjjikkk";
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
Integer val = map.get(c);
if (val != null) {
map.put(c, new Integer(val + 1));
}
else {
map.put(c, 1);
}
}
HashMap map=newhashmap();
字符串s=“aasjjikkk”;
对于(int i=0;i
最后,你将有一个你遇到的所有字符的计数,你可以从中提取它们的频率
或者,您可以使用Bozho的解决方案,即使用多集并计算总发生次数。您可以使用哈希表,每个字符作为键,总计数成为值
Hashtable<Character,Integer> table = new Hashtable<Character,Integer>();
String str = "aasjjikkk";
for( c in str ) {
if( table.get(c) == null )
table.put(c,1);
else
table.put(c,table.get(c) + 1);
}
for( elem in table ) {
println "elem:" + elem;
}
Hashtable=newhashtable();
String str=“aasjjikkk”;
对于(str中的c){
if(table.get(c)==null)
表1.put(c,1);
其他的
table.put(c,table.get(c)+1);
}
用于(表中的元素){
println“elem:+elem;
}
这里有一个解决方案:
定义您自己的对
:
public class Pair
{
private char letter;
private int count;
public Pair(char letter, int count)
{
this.letter = letter;
this.count= count;
}
public char getLetter(){return key;}
public int getCount(){return count;}
}
然后你可以做:
public static Pair countCharFreq(String s)
{
String temp = s;
java.util.List<Pair> list = new java.util.ArrayList<Pair>();
while(temp.length() != 0)
{
list.add(new Pair(temp.charAt(0), countOccurrences(temp, temp.charAt(0))));
temp.replaceAll("[" + temp.charAt(0) +"]","");
}
}
public static int countOccurrences(String s, char c)
{
int count = 0;
for(int i = 0; i < s.length(); i++)
{
if(s.charAt(i) == c) count++;
}
return count;
}
公共静态对countCharFreq(字符串s)
{
字符串温度=s;
java.util.List List=new java.util.ArrayList();
while(临时长度()!=0)
{
添加(新对(临时字符(0),计数出现次数(临时,临时字符(0)));
临时替换全部(“[”+临时字符(0)+“]”,”);
}
}
公共静态整数计数(字符串s,字符c)
{
整数计数=0;
对于(int i=0;i
这是另一种解决方案,尽管可能很狡猾
public char getNumChar(String s) {
char[] c = s.toCharArray();
String alphabet = "abcdefghijklmnopqrstuvwxyz";
int[] countArray = new int[26];
for (char x : c) {
for (int i = 0; i < alphabet.length(); i++) {
if (alphabet.charAt(i) == x) {
countArray[i]++;
}
}
}
java.util.HashMap<Integer, Character> countList = new java.util.HashMap<Integer, Character>();
for (int i = 0; i < 26; i++) {
countList.put(countArray[i], alphabet.charAt(i));
}
java.util.Arrays.sort(countArray);
int max = countArray[25];
return countList.get(max);
}
public char getNumChar(字符串s){
char[]c=s.toCharArray();
字符串字母表=“abcdefghijklmnopqrstuvxyz”;
int[]countArray=新int[26];
用于(字符x:c){
对于(int i=0;i
这与xunil154的答案类似,不同之处在于字符串是一个字符数组,链接的hashmap用于维护字符的插入顺序
String text = "aasjjikkk";
char[] charArray = text.toCharArray();
Map<Character, Integer> freqList = new LinkedHashMap<Character, Integer>();
for(char key : charArray) {
if(freqList.containsKey(key)) {
freqList.put(key, freqList.get(key) + 1);
} else
freqList.put(key, 1);
}
String text=“aasjjikkk”;
char[]charArray=text.toCharArray();
Map freqList=新建LinkedHashMap();
for(字符键:charArray){
if(频率列表.容器(键)){
freqList.put(key,freqList.get(key)+1);
}否则
频率列表。输入(键,1);
}
导入java.io.FileInputStream;
导入java.util.HashSet;
导入java.util.Iterator;
公共类计数字符的频率{
公共静态void main(字符串args[])引发异常
{
HashSet hs=新的HashSet();
String str=“你好吗?”;
char arr[]=新字符[str.length()];
对于(inti=0;iimport java.util.HashMap;
导入java.util.Iterator;
导入java.util.Map;
导入java.util.Map.Entry;
导入java.util.Scanner;
公共类频率YOFCharacters{
公共静态void main(字符串[]args){
System.out.println(“请输入字符串以计算每个字符的频率:”);
扫描仪sc=新的扫描仪(System.in);
字符串s=sc.nextLine();
String input=s.replaceAll(“\\s”,”);//删除空格。
频率计数(输入);
}
专用静态无效频率计数(字符串输入){
Map hashCount=new HashMap();
字符c;
对于(inti=0;iimportjava.util.*;
类Charfrequency
{
公共静态void main(字符串a[]{
扫描仪sc=新的扫描仪(System.in);
System.out.println(“输入字符串:”);
字符串s1=sc.nextLine();
整数计数,j=1;
char var='a';
char ch[]=s1.toCharArray();
而(jpackage com.rishi.zava;
导入java.util.HashMa
package com.rishi.zava;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
public class ZipString {
public static void main(String arg[]) {
String input = "aaaajjjgggtttssvvkkllaaiiikk";
int len = input.length();
Map<Character, Integer> zip = new HashMap<Character, Integer>();
for (int j = 0; len > j; j++) {
int count = 0;
for (int i = 0; len > i; i++) {
if (input.charAt(j) == input.charAt(i)) {
count++;
}
}
zip.put(input.charAt(j), count);
}
StringBuffer myValue = new StringBuffer();
String myMapKeyValue = "";
for (Entry<Character, Integer> entry : zip.entrySet()) {
myMapKeyValue = Character.toString(entry.getKey()).concat(
Integer.toString(entry.getValue()));
myValue.append(myMapKeyValue);
}
System.out.println(myValue);
}
}
private static Map<Character, Integer> findCharacterFrequency(String str) {
Map<Character, Integer> map = new HashMap<>();
for (char ch : str.toCharArray()) {
/* Using getOrDefault(), since Java1.8 */
map.put(ch, map.getOrDefault(ch, 0) + 1);
}
return map;
}
Map<Character,Integer> frequencies = new HashMap<>();
for (char ch : input.toCharArray())
frequencies.put(ch, frequencies.getOrDefault(ch, 0) + 1);
CharBag bag = Strings.asChars("aasjjikkk").toBag();
Assert.assertEquals(2, bag.occurrencesOf('a'));
Assert.assertEquals(1, bag.occurrencesOf('s'));
Assert.assertEquals(2, bag.occurrencesOf('j'));
Assert.assertEquals(1, bag.occurrencesOf('i'));
Assert.assertEquals(3, bag.occurrencesOf('k'));
String string = "aasjjikkk";
Map<Character, Long> characterFrequency = string.chars() // creates an IntStream
.mapToObj(c -> (char) c) // converts the IntStream to Stream<Character>
.collect(Collectors.groupingBy(c -> c, Collectors.counting())); // creates a
// Map<Character, Long>
// where the Long is
// the frequency
import java.util.Scanner;
class String55 {
public static int frequency(String s1,String s2)
{
int count=0;
char ch[]=s1.toCharArray();
char ch1[]=s2.toCharArray();
for (int i=0;i<ch.length-1; i++)
{
int k=i;
int j1=i+1;
int j=0;
int j11=j;
int j2=j+1;
{
while(k<ch.length && j11<ch1.length && ch[k]==ch1[j11])
{
k++;
j11++;
}
int l=k+j1;
int m=j11+j2;
if( l== m)
{
count=1;
count++;
}
}
}
return count;
}
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
System.out.println("enter the pattern");
String s1=sc.next();
System.out.println("enter the String");
String s2=sc.next();
int res=frequency(s1, s2);
System.out.println("FREQUENCY==" +res);
}
}
Map<Character, Long> frequency =
str.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
Map<Character, Integer> frequency =
str.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(c -> 1)));
Map<Character, Integer> frequency =
str.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.toMap(Function.identity(), c -> 1, Math::addExact));
package com.dipu.string;
import java.util.HashMap;
import java.util.Map;
public class RepetativeCharInString {
public static void main(String[] args) {
String data = "aaabbbcccdddffffrss";
char[] charArray = data.toCharArray();
Map<Character, Integer> map = new HashMap<>();
for (char c : charArray) {
if (map.containsKey(c)) {
map.put(c, map.get(c) + 1);
} else {
map.put(c, 1);
}
}
System.out.println(map);
}
}
*import java.util.ArrayList;
import java.util.Collections;
public class Freq {
public static void main(String[] args) {
// TODO Auto-generated method stub
String temp="zsaaqaaaaaaaabbbbbcc";
List<String> temp1= new ArrayList<String> ();
ArrayList<Integer>freq=new ArrayList<Integer>();
for(int i=0;i<temp.length()-1;i++)
{
temp1.add(Character.toString(temp.charAt(i)));
}
Set<String> uniqset=new HashSet<String>(temp1);
for(String s:uniqset)
{
freq.add(Collections.frequency(temp1, s));
System.out.println(s+" -->>"+Collections.frequency(temp1, s));
}
}
}
------Output-------
a -->>10
b -->>5
c -->>1
q -->>1
s -->>1
z -->>1
String s = "aaaabbbbcccddddd";
Map<Character, Integer> map = new HashMap<>();
s.chars().forEach(e->map.put((char)e, map.getOrDefault((char)e, 0) + 1));
char inputChar = '|';
int freq = "|fd|fdfd|f dfd|fd".replaceAll("[^" + inputChar +"]", "").length();
System.out.println("freq " + freq);
#include<stdio.h>`
#include <string.h>`
int main()
{
char s[1000];
int i,j,k,count=0,n;
printf("Enter the string : ");
gets(s);
for(j=0;s[j];j++);
n=j;
printf(" frequency count character in string:\n");
for(i=0;i<n;i++)
{
count=1;
if(s[i])
{
for(j=i+1;j<n;j++)
{
if(s[i]==s[j])
{
count++;
s[j]='\0';
}
}
printf(" '%c' = %d \n",s[i],count);
}
}
return 0;
}
void usingCollections(){
String input = "cuttack";
String [] stringArray = input.split("");
Set<String> s = new HashSet(Arrays.asList(stringArray));
for(String abc : s){
System.out.println (abc + ":"+Collections.frequency(Arrays.asList(stringArray),abc));
}
}
public static void main(String[] args) {
String s = "aaabbbcca";
Map<Character, Integer> freqMap = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
Character c = s.charAt(i);
freqMap.merge(c, 1, (a, b) -> a + b);
}
freqMap.forEach((k, v) -> System.out.println(k + " and " + v));
}
for (Character c : s.toCharArray()) {
freqMapSecond.merge(c, 1, Integer::sum);
}
public class demo {
public static void main(String[] args) {
String s = "babdcwertyuiuygf";
Map<Character, Integer> map = new TreeMap<>();
s.chars().forEach(e->map.put((char)e, map.getOrDefault((char)e, 0) + 1));
StringBuffer myValue = new StringBuffer();
String myMapKeyValue = "";
for (Map.Entry<Character, Integer> entry : map.entrySet()) {
myMapKeyValue = Character.toString(entry.getKey()).concat(
Integer.toString(entry.getValue()));
myValue.append(myMapKeyValue);
}
System.out.println(myValue);
}
}
Map<Character, Integer> map = new HashMap<>();
String s = "aasjjikkk";
for (int i = 0; i < s.length(); i++) {
map.merge(s.charAt(i), 1, (l, r) -> l + r);