Count the number of occurrences of each letter in

问题:

How can I count the number of occurrences in c of each letter (ignoring case) in the string? So that it would print out letter: # number of occurences, I have code to count the occurences of one letter, but how can I count the occurence of each letter in the string?

{
    char
    int count = 0;
    int i;

    //int length = strlen(string);

    for (i = 0; i < 20; i++)
    {
        if (string[i] == ch)
        {
            count++;
        }
    }

    return count;
}

output:

a : 1
b : 0
c : 2
etc...

回答1:

Let's assume you have a system where char is eight bit and all the characters you're trying to count are encoded using a non-negative number. In this case, you can write:

const char *str = "The quick brown fox jumped over the lazy dog.";

int counts[256] = { 0 };

int i;
size_t len = strlen(str);

for (i = 0; i < len; i++) {
    counts[(int)(str[i])]++;
}

for (i = 0; i < 256; i++) {
    printf("The %d. character has %d occurrences.\n", i, counts[i]);
}

Note that this will count all the characters in the string. If you are 100% absolutely positively sure that your string will have only letters (no numbers, no whitespace, no punctuation) inside, then 1. asking for "case insensitiveness" starts to make sense, 2. you can reduce the number of entries to the number of characters in the English alphabet (namely 26) and you can write something like this:

#include <ctype.h>
#include <string.h>
#include <stdlib.h>

const char *str = "TheQuickBrownFoxJumpedOverTheLazyDog";

int counts[26] = { 0 };

int i;
size_t len = strlen(str);

for (i = 0; i < len; i++) {
    // Just in order that we don't shout ourselves in the foot
    char c = str[i];
    if (!isalpha(c)) continue;

    counts[(int)(tolower(c) - 'a')]++;
}

for (i = 0; i < 26; i++) {
    printf("'%c' has %2d occurrences.\n", i + 'a', counts[i]);
}

回答2:

Like this:

int counts[26];
memset(counts, 0, sizeof(counts));
char *p = string;
while (*p) {
    counts[tolower(*p++) - 'a']++;
}

This code assumes that the string is null-terminated, and that it contains only characters a through z or A through Z, inclusive.

To understand how this works, recall that after conversion tolower each letter has a code between a and z, and that the codes are consecutive. As the result, tolower(*p) - 'a' evaluates to a number from 0 to 25, inclusive, representing the letter's sequential number in the alphabet.

This code combines ++ and *p to shorten the program.

回答3:

One simple possibility would be to make an array of 26 ints, each is a count for a letter a-z:

int alphacount[26] = {0}; //[0] = 'a', [1] = 'b', etc

Then loop through the string and increment the count for each letter:

for(int i = 0; i<strlen(mystring); i++)      //for the whole length of the string
    if(isalpha(mystring[i]))
        alphacount[tolower(mystring[i])-'a']++;  //make the letter lower case (if it's not)
                                                 //then use it as an offset into the array
                                                 //and increment

It's a simple idea that works for A-Z, a-z. If you want to separate by capitals you just need to make the count 52 instead and subtract the correct ASCII offset

回答4:

#include <stdio.h>
#include <string.h>
void main()
{
    printf("PLEASE ENTER A STRING\n");
    printf("GIVE ONLY ONE SPACE BETWEEN WORDS\n");
    printf("PRESS ENETR WHEN FINISHED\n");

    char str[100];
    int arr[26]={0};
    char ch;
    int i;

    gets(str);
    int n=strlen(str);

    for(i=0;i<n;i++)
    {
        ch=tolower(str[i]);
        if(ch>=97 && ch<=122)   
        {
            arr[ch-97]++;
        }
    }
    for(i=97;i<=122;i++)
        printf("%c OCCURS %d NUMBER OF TIMES\n",i,arr[i-97]);   
    return 0;
}

回答5:

After Accept Answer

A method that meets these specs: (IMO, the other answers do not meet all)

1. It is practical/efficient when char has a wide range. Example: CHAR_BIT is 16 or 32, so no use of bool Used[1 << CHAR_BIT];

2. Works for very long strings (use size_t rather than int).

3. Does not rely on ASCII. ( Use Upper[] )

4. Defined behavior when a char < 0. is...() functions are defined for EOF and unsigned char

   static const char Upper[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
   static const char Lower[] = "abcdefghijklmnopqrstuvwxyz";
   
   void LetterOccurrences(size_t *Count, const char *s) {
     memset(Count, 0, sizeof *Count * 26);
     while (*s) {
       unsigned char ch = *s;
       if (isalpha(ch)) {
         const char *caseset = Upper;
         char *p = strchr(caseset, ch);
         if (p == NULL) {
           caseset = Lower;
           p = strchr(caseset, ch);
         }
         if (p != NULL) {
           Count[p - caseset]++;
         }
       }
     }
   }
   
   // sample usage
   char *s = foo();
   size_t Count[26];
   LetterOccurrences(Count, s);
   for (int i=0; i<26; i++)
     printf("%c : %zu\n", Upper[i], Count[i]);
   }
   

回答6:

You can use the following code.

main()
{
    int i = 0,j=0,count[26]={0};
    char ch = 97;
    char string[100]="Hello how are you buddy ?";
    for (i = 0; i < 100; i++)
    {
        for(j=0;j<26;j++)
            {
            if (tolower(string[i]) == (ch+j))
                {
                    count[j]++;
                }
        }
    }
    for(j=0;j<26;j++)
        {

            printf("\n%c -> %d",97+j,count[j]);

    }

}

Hope this helps.

回答7:

#include<stdio.h>
#include<string.h>

#define filename "somefile.txt"

int main()
{
    FILE *fp;
    int count[26] = {0}, i, c;  
    char ch;
    char alpha[27] = "abcdefghijklmnopqrstuwxyz";
    fp = fopen(filename,"r");
    if(fp == NULL)
        printf("file not found\n");
    while( (ch = fgetc(fp)) != EOF) {
        c = 0;
        while(alpha[c] != '\0') {

            if(alpha[c] == ch) {
                count[c]++; 
            }
            c++;
        }
    }
    for(i = 0; i<26;i++) {
        printf("character %c occured %d number of times\n",alpha[i], count[i]);
    }
    return 0;
}

回答8:

for (int i=0;i<word.length();i++){
         int counter=0;
         for (int j=0;j<word.length();j++){
             if(word.charAt(i)==word.charAt(j))
             counter++;
             }// inner for
             JOptionPane.showMessageDialog( null,word.charAt(i)+" found "+ counter +" times");
         }// outer for

回答9:

#include<stdio.h>

void frequency_counter(char* str)
{
    int count[256] = {0};  //partial initialization
    int i;

    for(i=0;str[i];i++)
        count[str[i]]++;

    for(i=0;str[i];i++) {
        if(count[str[i]]) {
            printf("%c %d \n",str[i],count[str[i]]);
            count[str[i]]=0;
        }
    }
}

void main()
{
    char str[] = "The quick brown fox jumped over the lazy dog.";
    frequency_counter(str);
}

回答10:

Here is the C code with User Defined Function:

/* C Program to count the frequency of characters in a given String */

#include <stdio.h>
#include <string.h>

const char letters[] = "abcdefghijklmnopqrstuvwxzy";

void find_frequency(const char *string, int *count);

int main() {
    char string[100];
    int count[26] = { 0 };
    int i;

    printf("Input a string: ");
    if (!fgets(string, sizeof string, stdin))
        return 1;

    find_frequency(string, count);

    printf("Character Counts\n");

    for (i = 0; i < 26; i++) {
        printf("%c\t%d\n", letters[i], count[i]);
    }
    return 0;
}

void find_frequency(const char *string, int *count) {
    int i;
    for (i = 0; string[i] != '\0'; i++) {
        p = strchr(letters, string[i]);
        if (p != NULL) {
            count[p - letters]++;
        }
    }
}

回答11:

int charset[256] = {0};
int charcount[256] = {0};

for (i = 0; i < 20; i++)
{
    for(int c = 0; c < 256; c++)
    {
        if(string[i] == charset[c])
        {
           charcount[c]++;
        }
    }
}

charcount will store the occurence of any character in the string.

回答12:

//This is JavaScript Code.

function countWordOccurences()
{
    // You can use array of words or a sentence split with space.
    var sentence = "The quick brown fox jumped over the lazy dog.";
    //var sentenceArray = ['asdf', 'asdf', 'sfd', 'qwr', 'qwr'];
    var sentenceArray = sentence.split(' ', 1000); 
    var output;
    var temp;
    for(var i = 0; i < sentenceArray.length; i++) {
        var k = 1;
        for(var j = i + 1; j < sentenceArray.length; j++) {
            if(sentenceArray[i] == sentenceArray[j])
                    k = k + 1;
        }
        if(k > 1) {
            i = i + 1;
            output = output + ',' + k + ',' + k;
        }
        else
            output = output + ',' + k;
    }
    alert(sentenceArray + '\n' + output.slice(10).split(',', 500));
}

You can see it live --> http://jsfiddle.net/rammipr/ahq8nxpf/

回答13:

//c code for count the occurence of each character in a string.

void main()
   {
   int i,j; int c[26],count=0; char a[]="shahid";
   clrscr();
   for(i=0;i<26;i++)
    {
        count=0;
          for(j=0;j<strlen(a);j++)
                {
                 if(a[j]==97+i)
                    {
                     count++;
                         }
                           }
                  c[i]=count;
               }
              for(i=0;i<26;i++)
               {
              j=97+i;
          if(c[i]!=0) {  printf("%c of %d times\n",j,c[i]);
                 }
              }
           getch();
           }

回答14:

protected void btnSave_Click(object sender, EventArgs e)
    {           
        var FullName = "stackoverflow"

        char[] charArray = FullName.ToLower().ToCharArray();
        Dictionary<char, int> counter = new Dictionary<char, int>();
        int tempVar = 0;
        foreach (var item in charArray)
        {
            if (counter.TryGetValue(item, out tempVar))
            {
                counter[item] += 1;
            }
            else
            {
                counter.Add(item, 1);
            }
        }
        //var numberofchars = "";
        foreach (KeyValuePair<char, int> item in counter)
        {
            if (counter.Count > 0)
            {
                //Label1.Text=split(item.
            }
            Response.Write(item.Value + " " + item.Key + "<br />");
            // Label1.Text=item.Value + " " + item.Key + "<br />";
            spnDisplay.InnerText= item.Value + " " + item.Key + "<br />";
        }

    }