Physical distance between two places

问题

I need to measure the physical distance between two places whose names are provided as strings. Since sometimes the names are written slightly differently, I was looking for a library that could help me measure the difference and then combine it with a measure of the latitude and longitude to select the correct matches. Preferred languages: Java or PHP.

Any suggestions?

回答1:

Have a look at the Levenshtein distance. This is a way of measuring how different two strings are from one another.

Hopefully I understood your question correctly; using "distance" in the same sentence as "latitude and longitude" could be confusing!

回答2:

Although written in c (with python and tcl bindings), libdistance would be a tool for applying several distances metrics on strings/data.

Metrics included:

bloom
damerau
euclid
hamming
jaccard
levenshtein
manhattan
minkowski
needleman_wunsch

回答3:

You might get some decent results using a phonetic algorithm to find slightly misspelld names.

Also, if you use a more mechanical edit distance, you'll probably see better results using a weighted function that accounts for keyboard geometry (i.e. physically close keys are "cheaper" to replace than far off ones). That's a patented method btw, so be careful not to write something that becomes too popular ;)

回答4:

I took the liberty to translate a piece of C# code I've written to calculate the Levenshtein distance into Java code. It uses only two single-dimension arrays that alternate instead of a big jagged array:

public static int getDifference(String a, String b)
{
    // Minimize the amount of storage needed:
    if (a.length() > b.length())
    {
        // Swap:
        String x = a;
        a = b;
        b = x;
    }

    // Store only two rows of the matrix, instead of a big one
    int[] mat1 = new int[a.length() + 1];
    int[] mat2 = new int[a.length() + 1];

    int i;
    int j;

    for (i = 1; i <= a.length(); i++)
        mat1[i] = i;

    mat2[0] = 1;

    for (j = 1; j <= b.length(); j++)
    {
        for (i = 1; i <= a.length(); i++)
        {
            int c = (a.charAt(i - 1) == b.charAt(j - 1) ? 0 : 1);

            mat2[i] =
                Math.min(mat1[i - 1] + c,
                Math.min(mat1[i] + 1, mat2[i - 1] + 1));
        }

        // Swap:
        int[] x = mat1;
        mat1 = mat2;
        mat2 = x;

        mat2[0] = mat1[0] + 1;
    }

    // It's row #1 because we swap rows at the end of each outer loop,
    // as we are to return the last number on the lowest row
    return mat1[a.length()];
}

It is not rigorously tested, but it seems to be working okay. It was based on a Python implementation I made for a university exercise. Hope this helps!