Ok, so the standard algorithms are: 1) Hamming distance Only good for strings of the same length, but very efficient. Basically it simply counts the number of distinct characters. Not useful for fuzzy searching of natural language text. 2) Levenstein distance. The Levenstein distance measures distance in terms of the number of “operations” required to … Read more
To find out if a string contains substring you can use the index function: if (index($str, $substr) != -1) { print “$str contains $substr\n”; } It will return the position of the first occurrence of $substr in $str, or -1 if the substring is not found.
Try this expression: “Error importing row no\. (\d+):” DEMO Here you need to understand the quantifiers and escaped sequences: . any character; as you want only numbers, use \d; if you meant the period character you must escape it with a backslash (\.) ? Zero or one character; this isn’t what do you want, as … Read more
Years from 1000 to 2999 ^[12][0-9]{3}$ For 1900-2099 ^(19|20)\d{2}$
How about using StringUtils.countMatches from Apache Commons Lang? String str = “helloslkhellodjladfjhello”; String findStr = “hello”; System.out.println(StringUtils.countMatches(str, findStr)); That outputs: 3
import re pattern = re.compile(“^([A-Z][0-9]+)+$”) pattern.match(string)
See this regex: /^(?=.*Tim)(?=.*stupid).+/ Regex explanation: ^ Asserts position at start of string. (?=.*Tim) Asserts that “Tim” is present in the string. (?=.*stupid) Asserts that “stupid” is present in the string. .+Now that our phrases are present, this string is valid. Go ahead and use .+ or – .++ to match the entire string. To … Read more
I tried using existing fuzzy libraries like fuse.js and also found them to be terrible, so I wrote one which behaves basically like sublime’s search. https://github.com/farzher/fuzzysort The only typo it allows is a transpose. It’s pretty solid (1k stars, 0 issues), very fast, and handles your case easily: fuzzysort.go(‘int’, [‘international’, ‘splint’, ‘tinder’]) // [{highlighted: ‘*int*ernational’, … Read more
In case you’re interested in a quick visual comparison of Levenshtein and Difflib similarity, I calculated both for ~2.3 million book titles: import codecs, difflib, Levenshtein, distance with codecs.open(“titles.tsv”,”r”,”utf-8″) as f: title_list = f.read().split(“\n”)[:-1] for row in title_list: sr = row.lower().split(“\t”) diffl = difflib.SequenceMatcher(None, sr[3], sr[4]).ratio() lev = Levenshtein.ratio(sr[3], sr[4]) sor = 1 – distance.sorensen(sr[3], … Read more