问题
I need to write some kind of loop that can count the frequency of each letter in a string.
For example: \"aasjjikkk\" would count 2 \'a\', 1 \'s\', 2 \'j\', 1 \'i\', 3 \'k\'. Ultimately id like these to end up in a map with the character as the key and the count as the value. Any good idea how to do this?
回答1:
You can use a java Map and map a char to an int. You can then iterate over the characters in the string and check if they have been added to the map, if they have, you can then increment its value.
For example:
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);
}
}
At the end you will have a count of all the characters you encountered and you can extract their frequencies from that.
Alternatively, you can use Bozho's solution of using a Multiset and counting the total occurences.
回答2:
You can use a Multiset (from guava). It will give you the count for each object. For example:
Multiset<Character> chars = HashMultiset.create();
for (int i = 0; i < string.length(); i++) {
chars.add(string.charAt(i));
}
Then for each character you can call chars.count('a') and it returns the number of occurrences
回答3:
A concise way to do this is:
Map<Character,Integer> frequencies = new HashMap<>();
for (char ch : input.toCharArray())
frequencies.put(ch, frequencies.getOrDefault(ch, 0) + 1);
We use a for-each to loop through every character. The frequencies.getOrDefault() gets value if key is present or returns(as default) its second argument.
回答4:
Using the stream API as of JDK-8:
Map<Character, Long> frequency =
str.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
or if you want the keys as Integers:
Map<Character, Integer> frequency =
str.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.groupingBy(Function.identity(), Collectors.summingInt(c -> 1)));
Another variant:
Map<Character, Integer> frequency =
str.chars()
.mapToObj(c -> (char)c)
.collect(Collectors.toMap(Function.identity(), c -> 1, Math::addExact));
回答5:
Here is another solution, dodgy as it may be.
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);
}
回答6:
Well, two ways come to mind and it depends on your preference:
Sort the array by characters. Then, counting each character becomes trivial. But you will have to make a copy of the array first.
Create another integer array of size 26 (say freq) and str is the array of characters.
for(int i = 0; i < str.length; i ++)freq[str[i] - 'a'] ++; //Assuming all characters are in lower case
So the number of 'a' 's will be stored at freq[0] and the number of 'z' 's will be at freq[25]
回答7:
Here is a solution:
Define your own Pair:
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;}
}
Then you could do:
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;
}
回答8:
Since there was no Java 8 solution, thought of posting one. Also, this solution is much neater, readable and concise than some of the other solutions mentioned here.
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
回答9:
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);
}
}
Input = aaaajjjgggtttssvvkkllaaiiikk
Output = a6s2t3v2g3i3j3k4l2
回答10:
String s = "aaaabbbbcccddddd";
Map<Character, Integer> map = new HashMap<>();
Using one line in Java8
s.chars().forEach(e->map.put((char)e, map.getOrDefault((char)e, 0) + 1));
回答11:
You can use a CharAdapter and a CharBag from Eclipse Collections and avoid boxing to Character and Integer.
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'));
Note: I am a committer for Eclipse Collections.
回答12:
If this does not need to be super-fast just create an array of integers, one integer for each letter (only alphabetic so 2*26 integers? or any binary data possible?). go through the string one char at a time, get the index of the responsible integer (e.g. if you only have alphabetic chars you can have 'A' be at index 0 and get that index by subtracting any 'A' to 'Z' by 'A' just as an example of how you can get reasonably fast indices) and increment the value in that index.
There are various micro-optimizations to make this faster (if necessary).
回答13:
You can use a Hashtable with each character as the key and the total count becomes the value.
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;
}
回答14:
This is similar to xunil154's answer, except that a string is made a char array and a linked hashmap is used to maintain the insertion order of the characters.
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);
}
回答15:
NOTE, this will also count the frequencies of empty spaces
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
public class FrequenceyOfCharacters {
public static void main(String[] args) {
System.out.println("Please enter the string to count each character frequencey: ");
Scanner sc=new Scanner(System.in);
String input=sc.nextLine();
frequenceyCount(input);
}
private static void frequenceyCount(String input) {
Map<Character,Integer> hashCount=new HashMap<>();
Character c;
for(int i=0; i<input.length();i++)
{
c =input.charAt(i);
if(hashCount.get(c)!=null){
hashCount.put(c, hashCount.get(c)+1);
}else{
hashCount.put(c, 1);
}
}
Iterator it = hashCount.entrySet().iterator();
System.out.println("char : frequency");
while (it.hasNext()) {
Map.Entry pairs = (Map.Entry)it.next();
System.out.println(pairs.getKey() + " : " + pairs.getValue());
it.remove();
}
}
}
回答16:
import java.util.*;
class Charfrequency
{
public static void main(String a[]){
Scanner sc=new Scanner(System.in);
System.out.println("Enter Your String :");
String s1=sc.nextLine();
int count,j=1;
char var='a';
char ch[]=s1.toCharArray();
while(j<=26)
{
count=0;
for(int i=0; i<s1.length(); i++)
{
if(ch[i]==var || ch[i]==var-32)
{
count++;
}
}
if(count>0){
System.out.println("Frequency of "+var+" is "+count);
}
var++;
j++;
}
}
}
回答17:
The shorted possible code using a HashMap. (With no forceful line saves)
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;
}
回答18:
Please try the given code below, hope it will helpful to you,
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);
}
}
SAMPLE OUTPUT: enter the pattern man enter the String dhimanman FREQUENCY==2
Thank-you.Happy coding.
回答19:
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);
}
}
回答20:
*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
Use collections frequency method to count frequency of char*
回答21:
Uffh. Don't you think this is the simplest solution?
char inputChar = '|';
int freq = "|fd|fdfd|f dfd|fd".replaceAll("[^" + inputChar +"]", "").length();
System.out.println("freq " + freq);
回答22:
#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;
}
回答23:
We can use frequency method of Collections class for this. Split the string into string array. Use HashSet to remove duplicates and check frequency of each object in HashSet using frequency method of Collections
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));
}
}
回答24:
import java.io.FileInputStream;
import java.util.HashSet;
import java.util.Iterator;
public class CountFrequencyOfCharater {
public static void main(String args[]) throws Exception
{
HashSet hs=new HashSet();
String str="hey how are you?";
char arr[]=new char[str.length()];
for(int i=0;i<str.length();i++)
{
arr[i]=str.charAt(i);
}
for(int j=0;j<str.length();j++)
{
int c=0;
for(int k=0;k<str.length();k++)
{
if(arr[j]==arr[k])
c++;
}
hs.add(arr[j]+"="+c+",");
}
Iterator it=hs.iterator();
while(it.hasNext())
{
System.out.print(it.next());
}
}
}
来源：https://stackoverflow.com/questions/6712587/how-to-count-frequency-of-characters-in-a-string