CODEWITHSUNDEEP
google-bigquery
Andru
Andru

Reputation: 260

How to count letter differences of two strings in bigquery?

For example i have:

1: 6c71d997ba39
2: 6c71d997d269

I need to get 4.

Upvotes: 1

Views: 1300

Answers (3)

AISHWARYA CHANDRASEKHAR
AISHWARYA CHANDRASEKHAR

Reputation: 329

Source: https://stackoverflow.com/a/57499387/11059644

Ready to use shared UDFs - Levenshtein distance:

SELECT fhoffa.x.levenshtein('felipe', 'hoffa'), fhoffa.x.levenshtein('googgle', 'goggles'), fhoffa.x.levenshtein('is this the', 'Is This The')

Upvotes: 0

Mikhail Berlyant
Mikhail Berlyant

Reputation: 173106

You can consider using Levenshtein distance for your use-case

the Levenshtein distance is a string metric for measuring the difference between two sequences. Informally, the Levenshtein distance between two words is the minimum number of single-character edits (insertions, deletions or substitutions) required to change one word into the other

Below example is for BigQuery Standard SQL 

#standardSQL
CREATE TEMPORARY FUNCTION EDIT_DISTANCE(string1 STRING, string2 STRING)
RETURNS INT64
LANGUAGE js AS """
  var _extend = function(dst) {
    var sources = Array.prototype.slice.call(arguments, 1);
    for (var i=0; i<sources.length; ++i) {
      var src = sources[i];
      for (var p in src) {
        if (src.hasOwnProperty(p)) dst[p] = src[p];
      }
    }
    return dst;
  };

  var Levenshtein = {
    /**
     * Calculate levenshtein distance of the two strings.
     *
     * @param str1 String the first string.
     * @param str2 String the second string.
     * @return Integer the levenshtein distance (0 and above).
     */
    get: function(str1, str2) {
      // base cases
      if (str1 === str2) return 0;
      if (str1.length === 0) return str2.length;
      if (str2.length === 0) return str1.length;

      // two rows
      var prevRow  = new Array(str2.length + 1),
          curCol, nextCol, i, j, tmp;

      // initialise previous row
      for (i=0; i<prevRow.length; ++i) {
        prevRow[i] = i;
      }

      // calculate current row distance from previous row
      for (i=0; i<str1.length; ++i) {
        nextCol = i + 1;

        for (j=0; j<str2.length; ++j) {
          curCol = nextCol;

          // substution
          nextCol = prevRow[j] + ( (str1.charAt(i) === str2.charAt(j)) ? 0 : 1 );
          // insertion
          tmp = curCol + 1;
          if (nextCol > tmp) {
            nextCol = tmp;
          }
          // deletion
          tmp = prevRow[j + 1] + 1;
          if (nextCol > tmp) {
            nextCol = tmp;
          }

          // copy current col value into previous (in preparation for next iteration)
          prevRow[j] = curCol;
        }

        // copy last col value into previous (in preparation for next iteration)
        prevRow[j] = nextCol;
      }

      return nextCol;
    }

  };

  var the_string1;

  try {
    the_string1 = decodeURI(string1).toLowerCase();
  } catch (ex) {
    the_string1 = string1.toLowerCase();
  }

  try {
    the_string2 = decodeURI(string2).toLowerCase();
  } catch (ex) {
    the_string2 = string2.toLowerCase();
  }

  return Levenshtein.get(the_string1, the_string2) 

""";   

WITH strings AS (
  SELECT '1: 6c71d997ba39' string1, '2: 6c71d997d269' string2
)
SELECT string1, string2, EDIT_DISTANCE(string1, string2) changes
FROM   strings

with result 

Row     string1             string2             changes  
1       1: 6c71d997ba39     2: 6c71d997d269     4

Upvotes: 2

Elliott Brossard
Elliott Brossard

Reputation: 33765

SELECT
  (SELECT COUNTIF(c != s2[OFFSET(off)])
   FROM UNNEST(SPLIT(s1, '')) AS c WITH OFFSET off) AS count
FROM dataset.table

Upvotes: 0

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