Count word in string

Question

I'm trying to write a function to count occurrences of a particular word in a string. For example: Given string -

"Stop, time to go home. Todo fix me."

Letters "to" appeared three times (twice in different words); however, the word "to" appears only once. What should I do to count only word "to" (if will appear in string more times then count every single one). Any advice? This is the code I was trying and playing around.

int word(char inputLine[]) {
    int word = 0, i = 0, j = 0;

    for (i = 0; inputLine[i] != '\0'; i++) {
            if (inputLine[i] == 't' || inputLine[i] == 'o' || inputLine[i] != ' ') {
                word++;
            }
    }
    return word;
}

Answer 1

Try this:

int word(char inputLine[]) {
    int word = 0, i = 0;

    // stop before the last char
    for (i = 0; inputLine[i] != '\0' && inputLine[i+1] != '\0'; i++) {

        // is (T or t) and (O or o)
        if ((inputLine[i] == 't' || inputLine[i] == 'T') && (inputLine[i+1] == 'o' || inputLine[i+1] == 'O')) {

            // after the 'to' is not a letter
            if ((inputLine[i+2] < 'a' || inputLine[i+2] > 'z') &&
                (inputLine[i+2] < 'A' || inputLine[i+2] > 'Z')) {

                // before is not a letter (or this is the start of the string)
                if (i == 0 ||
                    ((inputLine[i-1] < 'a' || inputLine[i-1] > 'z') &&
                     (inputLine[i-1] < 'A' || inputLine[i-1] > 'Z'))) {
                        word++;
                }
            }
        }
    }

    return word;
}

Answer 2

Let's posit these rules:

"to" can be a word only when there is no char before and after it except the space char

If you accept those rules as valid and correct you need to check 4 conditions:

if (str[i]=='t'&& str[i+1]=='o'&& str[i-1]!='a-z'&& str[i+2]!='a-z'){
        word++;
    }

Two more conditions can be included to check for the upper case letters.

Answer 3

The simplest way would be to use strtok. But, if you'd like to do it all by hand, the following will work. Although you only wanted the "to" this will work for any search string:

#include <stdio.h>

// word -- get number of string matches
int
word(char *input,char *str)
// input -- input buffer
// str -- string to search for within input
{
    int chr;
    int prev;
    int off;
    int stopflg;
    int wordcnt;

    off = -1;
    stopflg = 0;
    wordcnt = 0;
    prev = 0;

    for (chr = *input++;  ! stopflg;  prev = chr, chr = *input++) {
        // we've hit the end of the buffer
        stopflg = (chr == 0);

        // convert whitespace characters to EOS [similar to what strtok might
        // do]
        switch (chr) {
        case ' ':
        case '\t':
        case '\n':
        case '\r':
            chr = 0;
            break;
        }

        ++off;

        // reset on mismatch
        // NOTE: we _do_ compare EOS chars here
        if (str[off] != chr) {
            off = -1;
            continue;
        }

        // we just matched
        // if we're starting the word we must ensure we're not in the middle
        // of one
        if ((off == 0) && (prev != 0)) {
            off = -1;
            continue;
        }

        // at the end of a word -- got a match
        if (chr == 0) {
            ++wordcnt;
            off = -1;
            continue;
        }
    }

    return wordcnt;
}

void
tryout(int expcnt,char *buf)
{
    int actcnt;

    actcnt = word(buf,"to");
    printf("%d/%d -- '%s'\n",expcnt,actcnt,buf);
}

// main -- main program
int
main(int argc,char **argv)
{
    char *cp;

    --argc;
    ++argv;

    for (;  argc > 0;  --argc, ++argv) {
        cp = *argv;
        if (*cp != '-')
            break;

        switch (cp[1]) {
        default:
            break;
        }
    }

    tryout(1,"to");
    tryout(2,"to to");
    tryout(1," to ");
    tryout(1,"todo to");
    tryout(2,"todo to to");
    tryout(2,"doto to to");
    tryout(1,"doto to doto");
    tryout(0,"doto");

    return 0;
}

Answer 4

If you must use only "basic" C functions the above solutions seems ok, but in the case you want to build a more scalable application (and you want to solve the problem in a smarter way) you can use a library that manipulate regular expressions. You can check this answer: Regular expressions in C: examples?

Regexes has the advantage that you can make the regex case unsensible (That is one of your issues). I usually use pcre because it has the regex style of perl and java. Here it is a very useful example that uses pcre: http://www.mitchr.me/SS/exampleCode/AUPG/pcre_example.c.html

Answer 5

public class FindCountOfWordInString {

    public static void main(String[] args) {
        String str = "yhing ghingu jhhtring inghfg ajklingingd me";
        String find = "ing";

        int count = findCountOfWordInString(str, find);
        System.out.println(count);
    }

    private static int findCountOfWordInString(String str, String find) {

        String[] strArr = str.split(" ");
        int count = 0, k = 0;
        for (int i = 0; i < strArr.length; i++) {
            if (strArr[i].contains(find)) {
                String strCheck = strArr[i];
                char[] findCharArr = find.toCharArray();
                for (int j = 0; j < strCheck.length(); j++) {
                    if (strCheck.charAt(j) == findCharArr[k]) {
                        k++;
                        if (k == 3) {
                            count++;
                            k = 0;
                        }
                    } else {
                        k = 0;
                    }
                }
            }
        }
        return count;
    }
}