比较两个没有Apache StringUtils的字符串

Question

Hi I am working with a voice command project. So I want to receive user's voice at first then I want to check the matches and then I want to do something according to the command. For this, I found a way to match the strings using org.apache.commons.lang3.StringUtils but I find so many trouble with this. For ex:- I face problem when I go to import the apache's external library to my android studio. So my question is that:- is there any other way to compare the user's voice data and my specific command without using Apache's StringUtils method? Please help if you can

Answer 1

There are many string functions you can use to compare strings, for example

   if (result.equals("hello")) {
         doSomething();
   }

compares two strings

   result.startsWith("search for") {
         doSomething()
   }

checks the beginning of the result

   result.matches("yes|sure") {
         doSomething()
   }

checks result with regular expression.

You can find all that in a Java textbook. See for example

https://docs.oracle.com/javase/tutorial/java/data/comparestrings.html

If you want to use Levenshtein distance you can insert the following function in your code:

public int LevenshteinDistance (String s0, String s1) {                          
    int len0 = s0.length() + 1;                                                     
    int len1 = s1.length() + 1;                                                     

    // the array of distances                                                       
    int[] cost = new int[len0];                                                     
    int[] newcost = new int[len0];                                                  

    // initial cost of skipping prefix in String s0                                 
    for (int i = 0; i < len0; i++) cost[i] = i;                                     

    // dynamically computing the array of distances                                  

    // transformation cost for each letter in s1                                    
    for (int j = 1; j < len1; j++) {                                                
        // initial cost of skipping prefix in String s1                             
        newcost[0] = j;                                                             

        // transformation cost for each letter in s0                                
        for(int i = 1; i < len0; i++) {                                             
            // matching current letters in both strings                             
            int match = (s0.charAt(i - 1) == s1.charAt(j - 1)) ? 0 : 1;             

            // computing cost for each transformation                               
            int cost_replace = cost[i - 1] + match;                                 
            int cost_insert  = cost[i] + 1;                                         
            int cost_delete  = newcost[i - 1] + 1;                                  

            // keep minimum cost                                                    
            newcost[i] = Math.min(Math.min(cost_insert, cost_delete), cost_replace);
        }                                                                           

        // swap cost/newcost arrays                                                 
        int[] swap = cost; cost = newcost; newcost = swap;                          
    }                                                                               

    // the distance is the cost for transforming all letters in both strings        
    return cost[len0 - 1];                                                          
}

Answer 2

Take the source right from the library (Obviously follow the requirements of the Apache license)

https://commons.apache.org/proper/commons-lang/apidocs/src-html/org/apache/commons/lang3/StringUtils.html

Line 6865

/**
 * <p>Find the Levenshtein distance between two Strings.</p>
 *
 * <p>This is the number of changes needed to change one String into
 * another, where each change is a single character modification (deletion,
 * insertion or substitution).</p>
 *
 * <p>The previous implementation of the Levenshtein distance algorithm
 * was from <a href="http://www.merriampark.com/ld.htm">http://www.merriampark.com/ld.htm</a></p>
 *
 * <p>Chas Emerick has written an implementation in Java, which avoids an OutOfMemoryError
 * which can occur when my Java implementation is used with very large strings.<br>
 * This implementation of the Levenshtein distance algorithm
 * is from <a href="http://www.merriampark.com/ldjava.htm">http://www.merriampark.com/ldjava.htm</a></p>
 *
 * <pre>
 * StringUtils.getLevenshteinDistance(null, *)             = IllegalArgumentException
 * StringUtils.getLevenshteinDistance(*, null)             = IllegalArgumentException
 * StringUtils.getLevenshteinDistance("","")               = 0
 * StringUtils.getLevenshteinDistance("","a")              = 1
 * StringUtils.getLevenshteinDistance("aaapppp", "")       = 7
 * StringUtils.getLevenshteinDistance("frog", "fog")       = 1
 * StringUtils.getLevenshteinDistance("fly", "ant")        = 3
 * StringUtils.getLevenshteinDistance("elephant", "hippo") = 7
 * StringUtils.getLevenshteinDistance("hippo", "elephant") = 7
 * StringUtils.getLevenshteinDistance("hippo", "zzzzzzzz") = 8
 * StringUtils.getLevenshteinDistance("hello", "hallo")    = 1
 * </pre>
 *
 * @param s  the first String, must not be null
 * @param t  the second String, must not be null
 * @return result distance
 * @throws IllegalArgumentException if either String input {@code null}
 * @since 3.0 Changed signature from getLevenshteinDistance(String, String) to
 * getLevenshteinDistance(CharSequence, CharSequence)
 */
public static int getLevenshteinDistance(CharSequence s, CharSequence t) {
    if (s == null || t == null) {
        throw new IllegalArgumentException("Strings must not be null");
    }

    /*
       The difference between this impl. and the previous is that, rather
       than creating and retaining a matrix of size s.length() + 1 by t.length() + 1,
       we maintain two single-dimensional arrays of length s.length() + 1.  The first, d,
       is the 'current working' distance array that maintains the newest distance cost
       counts as we iterate through the characters of String s.  Each time we increment
       the index of String t we are comparing, d is copied to p, the second int[].  Doing so
       allows us to retain the previous cost counts as required by the algorithm (taking
       the minimum of the cost count to the left, up one, and diagonally up and to the left
       of the current cost count being calculated).  (Note that the arrays aren't really
       copied anymore, just switched...this is clearly much better than cloning an array
       or doing a System.arraycopy() each time  through the outer loop.)

       Effectively, the difference between the two implementations is this one does not
       cause an out of memory condition when calculating the LD over two very large strings.
     */

    int n = s.length(); // length of s
    int m = t.length(); // length of t

    if (n == 0) {
        return m;
    } else if (m == 0) {
        return n;
    }

    if (n > m) {
        // swap the input strings to consume less memory
        final CharSequence tmp = s;
        s = t;
        t = tmp;
        n = m;
        m = t.length();
    }

    int p[] = new int[n + 1]; //'previous' cost array, horizontally
    int d[] = new int[n + 1]; // cost array, horizontally
    int _d[]; //placeholder to assist in swapping p and d

    // indexes into strings s and t
    int i; // iterates through s
    int j; // iterates through t

    char t_j; // jth character of t

    int cost; // cost

    for (i = 0; i <= n; i++) {
        p[i] = i;
    }

    for (j = 1; j <= m; j++) {
        t_j = t.charAt(j - 1);
        d[0] = j;

        for (i = 1; i <= n; i++) {
            cost = s.charAt(i - 1) == t_j ? 0 : 1;
            // minimum of cell to the left+1, to the top+1, diagonally left and up +cost
            d[i] = Math.min(Math.min(d[i - 1] + 1, p[i] + 1), p[i - 1] + cost);
        }

        // copy current distance counts to 'previous row' distance counts
        _d = p;
        p = d;
        d = _d;
    }

    // our last action in the above loop was to switch d and p, so p now
    // actually has the most recent cost counts
    return p[n];
}