How to count letter differences of two strings in bigquery? - google-bigquery

Question

For example i have:
1: 6c71d997ba39
2: 6c71d997d269
I need to get 4.

Answer 1

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

Answer 2

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

Answer 3

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')