What do 'lazy' and 'greedy' mean in the context of regular expressions?

Question

What are these two terms in an understandable way?

Commemorate

See also #3075630 – Overplay

Does this answer your question? Greedy vs. Reluctant vs. Possessive Qualifiers – Zebrawood

Answer 1

Greedy will consume as much as possible. From http://www.regular-expressions.info/repeat.html we see the example of trying to match HTML tags with <.+>. Suppose you have the following:

<em>Hello World</em>

You may think that <.+> (. means any non newline character and + means one or more) would only match the <em> and the </em>, when in reality it will be very greedy, and go from the first < to the last >. This means it will match <em>Hello World</em> instead of what you wanted.

Making it lazy (<.+?>) will prevent this. By adding the ? after the +, we tell it to repeat as few times as possible, so the first > it comes across, is where we want to stop the matching.

I'd encourage you to download RegExr, a great tool that will help you explore Regular Expressions - I use it all the time.

Sondra

so if you use greedy will u have 3 (1 element + 2 tags) matches or just 1 match (1 element)? – Commemorate

It would match only 1 time, starting from the first < and ending with the last >. – Sondra

But making it lazy would match twice, giving us both the opening and closing tag, ignoring the text in between (since it doesn't fit the expression). – Sondra

Another great tool I always use: debuggex.com It also has a "Embed on StackOverflow" function. – Metalloid

Just to add that there is a greedy way to go about it, too: <[^>]+> regex101.com/r/lW0cY6/1 – Carpal

In case one wants to delve a bit deeper into how lazy and greedy quantifiers work with optional subpatterns in-between them, check Perl regex matching optional phrase in longer sentence. – Press

For the record, about using regex with HTML #1732848 – Guadalquivir

It's curious, no more, that the highest-voted and selected answer is one that gives an example of a regex being used to parse HTML, even though it is generally recommended that an HTML parser be used for that task. A criticism, however, is that no definitions of the two concepts are given, only the results for an example. The first sentence, in particular, is so vague as to be meaningless. – Apophyllite

Answer 2

'Greedy' means match longest possible string.

'Lazy' means match shortest possible string.

For example, the greedy h.+l matches 'hell' in 'hello' but the lazy h.+?l matches 'hel'.

Obeah

Brilliant, so lazy will stop as soon as the condition l is satisfied, but greedy means it will stop only once the condition l is not satisfied any more? – Praetor

For all people reading the post: greedy or lazy quantifiers by themselves won't match the longest/shortest possible substring. You would have to use either a tempered greedy token, or use non-regex approaches. – Press

@AndrewS Don't be confused by the double ll in the example. It's rather lazy will match the shortest possible substring while greedy will match the longest possible. Greedy h.+l matches 'helol' in 'helolo' but the lazy h.+?l matches 'hel'. – Eucalyptus

Doesn't the ? make the .+ optional in h.+?l. Isn't that what the ? is for? Also, how would you differentiate between the two functions of ? – Maremma

@FloatingRock: No. x? means x is optional but +? is a different syntax. It means stop looking after you find something that matches - lazy matching. – Obeah

@FloatingRock: As for how you differentiate the different syntax, simple: ? means optional and +? means lazy. Therefore \+? means + is optional. – Obeah

Answer 3

Greedy quantifier	Lazy quantifier	Description
*	*?	Star Quantifier: 0 or more
+	+?	Plus Quantifier: 1 or more
?	??	Optional Quantifier: 0 or 1
{n}	{n}?	Quantifier: exactly n
{n,}	{n,}?	Quantifier: n or more
{n,m}	{n,m}?	Quantifier: between n and m

Add a ? to a quantifier to make it ungreedy i.e lazy.

Example:
test string : stackoverflow
greedy reg expression : s.*o output: stackoverflow
lazy reg expression : s.*?o output: stackoverflow

Armilda

is not ?? equivalent to ? . Similarly , isn't {n}? equivalen to {n} – Audryaudrye

@BreakingBenjamin: no ?? is not equivalent to ?, when it has a choice to either return 0 or 1 occurrence, it will pick the 0 (lazy) alternative. To see the difference, compare re.match('(f)?(.*)', 'food').groups() to re.match('(f)??(.*)', 'food').groups(). In the latter, (f)?? will not match the leading 'f' even though it could. Hence the 'f' will get matched by the second '.*' capture group. I'm sure you can construct an example with '{n}?' too. Admittedly these two are very-rarely-used. – Pedrick

@Audryaudrye Yes, {n}? is equivalent to {n}. See https://mcmap.net/q/25763/-how-do-a-n-and-a-n-differ – Complain

Answer 4

Greedy means your expression will match as large a group as possible, lazy means it will match the smallest group possible. For this string:

abcdefghijklmc

and this expression:

a.*c

A greedy match will match the whole string, and a lazy match will match just the first abc.

Xerophthalmia

For anyone else wondering, the expression mentioned above, a.*c, is greedy, while a.*?c would be the lazy counterpart. – Snowmobile

Answer 5

As far as I know, most regex engine is greedy by default. Add a question mark at the end of quantifier will enable lazy match.

As @Andre S mentioned in comment.

• Greedy: Keep searching until condition is not satisfied.
• Lazy: Stop searching once condition is satisfied.

Refer to the example below for what is greedy and what is lazy.

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {
    public static void main(String args[]){
        String money = "100000000999";
        String greedyRegex = "100(0*)";
        Pattern pattern = Pattern.compile(greedyRegex);
        Matcher matcher = pattern.matcher(money);
        while(matcher.find()){
            System.out.println("I'm greedy and I want " + matcher.group() + " dollars. This is the most I can get.");
        }
        
        String lazyRegex = "100(0*?)";
        pattern = Pattern.compile(lazyRegex);
        matcher = pattern.matcher(money);
        while(matcher.find()){
            System.out.println("I'm too lazy to get so much money, only " + matcher.group() + " dollars is enough for me");
        }
    }
}

---

The result is:

I'm greedy and I want 100000000 dollars. This is the most I can get.

I'm too lazy to get so much money, only 100 dollars is enough for me

Exclusion

Answer 6

Taken From www.regular-expressions.info

Greediness: Greedy quantifiers first tries to repeat the token as many times as possible, and gradually gives up matches as the engine backtracks to find an overall match.

Laziness: Lazy quantifier first repeats the token as few times as required, and gradually expands the match as the engine backtracks through the regex to find an overall match.

Ninny

This seems to be the most correct definition of "Laziness" compared with the higher-voted answers. The other answers seem to omit the concept that under laziness the engine "gradually expands the match...to find an overall match". – Selfeducated

Answer 7

From Regular expression

The standard quantifiers in regular expressions are greedy, meaning they match as much as they can, only giving back as necessary to match the remainder of the regex.

By using a lazy quantifier, the expression tries the minimal match first.

Fuller

Answer 8

Greedy matching. The default behavior of regular expressions is to be greedy. That means it tries to extract as much as possible until it conforms to a pattern even when a smaller part would have been syntactically sufficient.

Example:

import re
text = "<body>Regex Greedy Matching Example </body>"
re.findall('<.*>', text)
#> ['<body>Regex Greedy Matching Example </body>']

Instead of matching till the first occurrence of ‘>’, it extracted the whole string. This is the default greedy or ‘take it all’ behavior of regex.

Lazy matching, on the other hand, ‘takes as little as possible’. This can be effected by adding a ? at the end of the pattern.

Example:

re.findall('<.*?>', text)
#> ['<body>', '</body>']

If you want only the first match to be retrieved, use the search method instead.

re.search('<.*?>', text).group()
#> '<body>'

Source: Python Regex Examples

Garter

Answer 9

Greedy Quantifiers are like the IRS

They’ll take as much as they can. e.g. matches with this regex: .*

$50,000

Bye-bye bank balance.

See here for an example: Greedy-example

Non-greedy quantifiers - they take as little as they can

Ask for a tax refund: the IRS sudden becomes non-greedy - and return as little as possible: i.e. they use this quantifier:

(.{2,5}?)([0-9]*) against this input: $50,000

The first group is non-needy and only matches $5 – so I get a $5 refund against the $50,000 input.

See here: Non-greedy-example.

Why do we need greedy vs non-greedy?

It becomes important if you are trying to match certain parts of an expression. Sometimes you don't want to match everything - as little as possible. Sometimes you want to match as much as possible. Nothing more to it.

You can play around with the examples in the links posted above.

(Analogy used to help you remember).

Teahan

Answer 10

Best shown by example. String. 192.168.1.1 and a greedy regex \b.+\b You might think this would give you the 1st octet but is actually matches against the whole string. Why? Because the.+ is greedy and a greedy match matches every character in 192.168.1.1 until it reaches the end of the string. This is the important bit! Now it starts to backtrack one character at a time until it finds a match for the 3rd token (\b).

If the string a 4GB text file and 192.168.1.1 was at the start you could easily see how this backtracking would cause an issue.

To make a regex non greedy (lazy) put a question mark after your greedy search e.g

*?
??
+?

What happens now is token 2 (+?) finds a match, regex moves along a character and then tries the next token (\b) rather than token 2 (+?). So it creeps along gingerly.

Swelling

Answer 11

Greedy means it will consume your pattern until there are none of them left and it can look no further.

Lazy will stop as soon as it will encounter the first pattern you requested.

One common example that I often encounter is \s*-\s*? of a regex ([0-9]{2}\s*-\s*?[0-9]{7})

The first \s* is classified as greedy because of * and will look as many white spaces as possible after the digits are encountered and then look for a dash character "-". Where as the second \s*? is lazy because of the present of *? which means that it will look the first white space character and stop right there.

Grain

I don't understand how that regex is any different from not using a lazy quantifier. \s can only match a white space and there's a requirement for 7 digits after it. – Coulter

Answer 12

To give extra clarification on Laziness, here is one example which is maybe not intuitive on first look but explains idea of "gradually expands the match" from Suganthan Madhavan Pillai answer.

input -> [email protected]@
regex -> ^.*?@$

Regex for this input will have a match. At first glance somebody could say LAZY match(".*?@") will stop at first @ after which it will check that input string ends("$"). Following this logic someone would conclude there is no match because input string doesn't end after first @.

But as you can see this is not the case, regex will go forward even though we are using non-greedy(lazy mode) search until it hits second @ and have a MINIMAL match.

Palais

Answer 13

try to understand the following behavior:

    var input = "0014.2";

Regex r1 = new Regex("\\d+.{0,1}\\d+");
Regex r2 = new Regex("\\d*.{0,1}\\d*");

Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // "0014.2"

input = " 0014.2";

Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // " 0014"

input = "  0014.2";

Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // ""

Garrote

What language is this? – Coulter

The language he used is C# – Eryneryngo