Reputation: 2269

Regular expression for floating point numbers

I have a task to match floating point numbers. I have written the following regular expression for it:

[-+]?[0-9]*\.?[0-9]*

But, it returns an error:

Invalid escape sequence (valid ones are  \b  \t  \n  \f  \r  \"  \'  \\ )

As per my knowledge, we need to use an escape character for the . also. Please correct me where I am wrong.

Upvotes: 215

Answers (22)

Igor Ashikhhmin

Reputation: 1

Based on other answers, I guess we need to set start and end for our expression too, cause we can have "match" on something like this: '3.2.3' where match will be '2.3'. So here is my solution:

^[+-]?[0-9]+(\.[0-9]+)?$

If you have problems like the author of this thread, use this:

^[+-]?[0-9]+([.][0-9]+)?$

And a reminder for better understanding:

'^' - Start of our expression
'[+-]' - '+' or '-' symbol
'?' - previous symbol (or sequence) is optional
'[0-9]' - any number
'()' - group of expressions
'+' - one or more symbols (or sequences) like previous
'$' - end of our expression

Hope it will help you)

Upvotes: 0

Elyor Tukhtamuratov

Reputation: 1

I would suggest this pattern [-+]?[0-9]+[.]?[0-9]*

Upvotes: 0

Gayathri Saranath

Reputation: 19

If we are only looking to identify the floating points and not the integers, then can use this:

'\d*\.\d+'

Upvotes: 0

user2138149

Reputation: 16843

Match strings which are considered valid representations of floating point values by C and C++ (and many other language) compilers, using the C++ `regex` library:

In C++ with #include <regex> you can do this:

std::regex r("[+-]?[0-9]+[.][0-9]*([e][+-]?[0-9]+)?");
return std::regex_match(value, r);

which is considerably more simple than most of the above C++ related answers.

It matches strings which are considered to be valid string representations of floating point numbers according to C++ compilers.

That means things like

1.
-1.

are considered valid representations of floating point numbers but that

.1
-.1

are not.

To explain the expression in more detail, it is essentially composed of two parts:

[+-]?[0-9]+[.][0-9]*([e][+-]?[0-9]+)?

[+-]?[0-9]+[.][0-9]*
and                 ([e][+-]?[0-9]+)?

The first part is easy to understand:

Optional (meaning 0 or 1 occurances of) '+' or '-' character
At least 1 digit, or more than one digit
A literal '.' character, which is mandatory (otherwise you have a representation of an integer not a floating point value)
If you want the '.' to be optional, change it to [.]?
Followed by zero or more digits

The second part is also quite easy once broken down.

Firstly note that the expression is contained in parenthesys, followed by a ?. This means the expression inside the parentesys must match 0 or 1 times. (Meaning it is optional.)
Inside we have a literal 'e' character which must match
Followed by an optional '+' or '-' character
Followed by 1 or more digits

The last part [+-]?[0-9]+ is a regex for matching an integer.

To match integer values as well use:

[+-]?[0-9]+[.]?[0-9]*([e][+-]?[0-9]+)?

Note the ? after the [.].

But be aware this will also match things like

+100e+100

which is perhaps an unusual representation of an integer. Although it is technically an integer, you probably wouldn't expect this to be a match.

Other answers provide a solution if you don't want this behaviour.

To ensure an entire string is a match rather than just a string containing a match use anchors:

"^[+-]?[0-9]+[.][0-9]*([e][+-]?[0-9]+)?$"

Examples

Without anchor characters

With anchor characters

With optional '.' character:

Note that this matches the string .-100 and .1e100 if you do not include the anchor characters, which may not be what you want.

When considering this problem:

My aim was to validate user input to ensure it matches a valid C++ string representation of a floating point number. Hence I am assuming you will use anchor characters and that you do not consider strings like

hello world 3.14 this contains a floating point number

to be a valid floating point number - because although the string contains a floating point number, the whole string is not a valid floating point number.

Other answers may suit your needs better if you just want to detect floating points within larger strings/text.

Upvotes: 6

Obaida Al-Bllani

Reputation: 9

if you are using flutter you can use [0-9]([.]([0-9])*)? This would parse 123.123

Upvotes: -2

Max McKinley

Reputation: 31

This is for javascript (idk if there's a large difference between languages)

`int: /0|[1-9][0-9]*/`

For floats:

`float:   /[0-9]+\.[0-9]+/`

Upvotes: -1

wei He

Reputation: 1

In C Language, the answer would go about like this:

[+-]?((\d+\.?\d*)|(\.\d+))(([eE][+-]?)?\d+)?[fFlL]?

Upvotes: 0

JDB

Reputation: 25835

TL;DR

Use [.] instead of \. and [0-9] instead of \d to avoid escaping issues in some languages (like Java).

^{Thanks to the nameless one for originally recognizing this.}

One relatively simple pattern for matching a floating point number in a larger string is:

[+-]?([0-9]*[.])?[0-9]+

This will match:

123
123.456
.456

^{See a working example}

If you also want to match 123. (a period with no decimal part), then you'll need a slightly longer expression:

[+-]?([0-9]+([.][0-9]*)?|[.][0-9]+)

^{See pkeller's answer for a fuller explanation of this pattern}

If you want to include a wider spectrum of numbers, including scientific notation and non-decimal numbers such as hex and octal, see my answer to How do I identify if a string is a number?.

If you want to validate that an input is a number (rather than finding a number within the input), then you should surround the pattern with ^ and $, like so:

^[+-]?([0-9]+([.][0-9]*)?|[.][0-9]+)$

Irregular Regular Expressions

"Regular expressions", as implemented in most modern languages, APIs, frameworks, libraries, etc., are based on a concept developed in formal language theory. However, software engineers have added many extensions that take these implementations far beyond the formal definition. So, while most regular expression engines resemble one another, there is actually no standard. For this reason, a lot depends on what language, API, framework or library you are using.

(Incidentally, to help reduce confusion, many have taken to using "regex" or "regexp" to describe these enhanced matching languages. See Is a Regex the Same as a Regular Expression? at RexEgg.com for more information.)

That said, most regex engines (actually, all of them, as far as I know) would accept \.. Most likely, there's an issue with escaping.

The Trouble with Escaping

Some languages have built-in support for regexes, such as JavaScript. For those languages that don't, escaping can be a problem.

This is because you are basically coding in a language within a language. Java, for example, uses \ as an escape character within it's strings, so if you want to place a literal backslash character within a string, you must escape it:

// creates a single character string: "\"
String x = "\\";

However, regexes also use the \ character for escaping, so if you want to match a literal \ character, you must escape it for the regex engine, and then escape it again for Java:

// Creates a two-character string: "\\"
// When used as a regex pattern, will match a single character: "\"
String regexPattern = "\\\\";

In your case, you have probably not escaped the backslash character in the language you are programming in:

// will most likely result in an "Illegal escape character" error
String wrongPattern = "\.";
// will result in the string "\."
String correctPattern = "\\.";

All this escaping can get very confusing. If the language you are working with supports raw strings, then you should use those to cut down on the number of backslashes, but not all languages do (most notably: Java). Fortunately, there's an alternative that will work some of the time:

String correctPattern = "[.]";

For a regex engine, \. and [.] mean exactly the same thing. Note that this doesn't work in every case, like newline (\\n), open square bracket (\\[) and backslash (\\\\ or [\\]).

A Note about Matching Numbers

(Hint: It's harder than you think)

Matching a number is one of those things you'd think is quite easy with regex, but it's actually pretty tricky. Let's take a look at your approach, piece by piece:

[-+]?

Match an optional - or +

[0-9]*

Match 0 or more sequential digits

\.?

Match an optional .

[0-9]*

Match 0 or more sequential digits

First, we can clean up this expression a bit by using a character class shorthand for the digits (note that this is also susceptible to the escaping issue mentioned above):

[0-9] = \d

I'm going to use \d below, but keep in mind that it means the same thing as [0-9]. (Well, actually, in some engines \d will match digits from all scripts, so it'll match more than [0-9] will, but that's probably not significant in your case.)

Now, if you look at this carefully, you'll realize that every single part of your pattern is optional. This pattern can match a 0-length string; a string composed only of + or -; or, a string composed only of a .. This is probably not what you've intended.

To fix this, it's helpful to start by "anchoring" your regex with the bare-minimum required string, probably a single digit:

\d+

Now we want to add the decimal part, but it doesn't go where you think it might:

\d+\.?\d* /* This isn't quite correct. */

This will still match values like 123.. Worse, it's got a tinge of evil about it. The period is optional, meaning that you've got two repeated classes side-by-side (\d+ and \d*). This can actually be dangerous if used in just the wrong way, opening your system up to DoS attacks.

To fix this, rather than treating the period as optional, we need to treat it as required (to separate the repeated character classes) and instead make the entire decimal portion optional:

\d+(\.\d+)? /* Better. But... */

This is looking better now. We require a period between the first sequence of digits and the second, but there's a fatal flaw: we can't match .123 because a leading digit is now required.

This is actually pretty easy to fix. Instead of making the "decimal" portion of the number optional, we need to look at it as a sequence of characters: 1 or more numbers that may be prefixed by a . that may be prefixed by 0 or more numbers:

(\d*\.)?\d+

Now we just add the sign:

[+-]?(\d*\.)?\d+

Of course, those slashes are pretty annoying in Java, so we can substitute in our long-form character classes:

[+-]?([0-9]*[.])?[0-9]+

Matching versus Validating

This has come up in the comments a couple times, so I'm adding an addendum on matching versus validating.

The goal of matching is to find some content within the input (the "needle in a haystack"). The goal of validating is to ensure that the input is in an expected format.

Regexes, by their nature, only match text. Given some input, they will either find some matching text or they will not. However, by "snapping" an expression to the beginning and ending of the input with anchor tags (^ and $), we can ensure that no match is found unless the entire input matches the expression, effectively using regexes to validate.

The regex described above ([+-]?([0-9]*[.])?[0-9]+) will match one or more numbers within a target string. So given the input:

apple 1.34 pear 7.98 version 1.2.3.4

The regex will match 1.34, 7.98, 1.2, .3 and .4.

To validate that a given input is a number and nothing but a number, "snap" the expression to the start and end of the input by wrapping it in anchor tags:

^[+-]?([0-9]*[.])?[0-9]+$

This will only find a match if the entire input is a floating point number, and will not find a match if the input contains additional characters. So, given the input 1.2, a match will be found, but given apple 1.2 pear no matches will be found.

Note that some regex engines have a validate, isMatch or similar function, which essentially does what I've described automatically, returning true if a match is found and false if no match is found. Also keep in mind that some engines allow you to set flags which change the definition of ^ and $, matching the beginning/end of a line rather than the beginning/end of the entire input. This is typically not the default, but be on the lookout for these flags.

Upvotes: 464

Sen

Reputation: 37

(\d*)(\.)*(\d+)

This would parse the below.

11.00
12
.0

There must be one number. The decimal point and the number before the decimal point is optional.

Upvotes: 1

Mihai Ciobanu

Reputation: 451

^[+-]?([0-9]{1,})[.,]([0-9]{1,})$

This will match:

1.2
12.3
123.4
1,2
12,3
123,4

Upvotes: 2

ProjectPhysX

Reputation: 5746

This captures floating-point numbers as recognized in C/C++ code:

[+-]?((((\d+\.?\d*)|(\.\d+))([eE][+-]?\d+[fF]?)?)|((\d+\.\d*)|(\.\d+))[fF]?)

+/- sign
either only digits, digits., .digits or digits.digits
optional exponent with e or E, +/- sign and digits
optional f or F at the end, but only if the number contains a . or an exponent

Upvotes: 2

sergredov

Reputation: 11

For those who searching a regex which would validate an entire input that should be a signed float point number on every single character typed by a user.

I.e. a sign goes first (should match and be valid), then all the digits (still match and valid) and its optional decimal part.

In JS, we use onkeydown/oninput event to do that + the following regex:

^[+-]?[0-9]*([\.][0-9]*)?$

Upvotes: 0

Yannis T

Reputation: 441

I want to match what most languages consider valid numbers (integer and floats):

'5' / '-5'
'1.0' / '1.' / '.1' / '-1.' / '-.1'
'0.45326e+04', '666999e-05', '0.2e-3', '-33.e-1'

Notes:

preceding sign of number ('-' or '+') is optional
'-1.' and '-.1' are valid but '.' and '-.' are invalid
'.1e3' is valid, but '.e3' and 'e3' are invalid

In order to support both '1.' and '.1' we need an OR operator ('|') in order to make sure we exclude '.' from matching.

[+-]? +/- sing is optional since ? means 0 or 1 matches

( since we have 2 sub expressions we need to put them in parenthesis

\d+([.]\d*)?(e[+-]?\d+)? This is for numbers starting with a digit

| separates sub expressions

[.]\d+(e[+-]?\d+)? this is for numbers starting with '.'

) end of expressions

For numbers starting with '.'

[.] first character is dot (inside brackets or else it is a wildcard character)

\d+ one or more digits

(e[+-]?\d+)? this is an optional (0 or 1 matches due to ending '?') scientific notation

For numbers starting with a digit

\d+ one or more digits

([.]\d*)? optionally we can have a dot character an zero or more digits after it

(e[+-]?\d+)? this is an optional scientific notation

Scientific notation

e literal that specifies exponent

[+-]? optional exponent sign

\d+ one or more digits

All of those combined:

[+-]?(\d+([.]\d*)?(e[+-]?\d+)?|[.]\d+(e[+-]?\d+)?)

To accept E as well:

[+-]?(\d+([.]\d*)?([eE][+-]?\d+)?|[.]\d+([eE][+-]?\d+)?)

(Test cases)

Upvotes: 34

Zoran Medojević

Reputation: 11

In c notation, float number can occur in following shapes:

123
123.
123.24
.24
2e-2 = 2 * 10 pow -2 = 2 * 0.1
4E+4 = 4 * 10 pow 4 = 4 * 10 000

For creating float regular expresion, I will first create "int regular expresion variable":

(([1-9][0-9]*)|0) will be int

Now, I will write small chunks of float regular expresion - solution is to concat those chunks with or simbol "|".

Chunks:

- (([+-]?{int}) satysfies case 1
- (([+-]?{int})"."[0-9]*)  satysfies cases 2 and 3
- ("."[0-9]*) satysfies case 4
- ([+-]?{int}[eE][+-]?{int}) satysfies cases 5 and 6

Final solution (concanating small chunks):

(([+-]?{int})|(([+-]?{int})"."[0-9]*)|("."[0-9]*)|([+-]?{int}[eE][+-]?{int})

Upvotes: 0

LuisDev99

Reputation: 1837

In C++ using the regex library

The answer would go about like this:

[0-9]?([0-9]*[.])?[0-9]+

Notice that I don't take the sign symbol, if you wanted it with the sign symbol it would go about this:

[+-]?([0-9]*[.])?[0-9]+

This also separates a regular number or a decimal number.

Upvotes: 0

mjwrazor

Reputation: 1964

for javascript

const test = new RegExp('^[+]?([0-9]{0,})*[.]?([0-9]{0,2})?$','g');

Which would work for 1.23 1234.22 0 0.12 12

You can change the parts in the {} to get different results in decimal length and front of the decimal as well. This is used in inputs for entering in number and checking every input as you type only allowing what passes.

Upvotes: 1

pkeller

Reputation: 741

I don't think that any of the answers on this page at the time of writing are correct (also many other suggestions elsewhere on SO are wrong too). The complication is that you have to match all of the following possibilities:

No decimal point (i.e. an integer value)
Digits both before and after the decimal point (e.g. 0.35 , 22.165)
Digits before the decimal point only (e.g. 0. , 1234.)
Digits after the decimal point only (e.g. .0 , .5678)

At the same time, you must ensure that there is at least one digit somewhere, i.e. the following are not allowed:

a decimal point on its own
a signed decimal point with no digits (i.e. +. or -.)
+ or - on their own
an empty string

This seems tricky at first, but one way of finding inspiration is to look at the OpenJDK source for the java.lang.Double.valueOf(String) method (start at http://hg.openjdk.java.net/jdk8/jdk8/jdk, click "browse", navigate down /src/share/classes/java/lang/ and find the Double class). The long regex that this class contains caters for various possibilities that the OP probably didn't have in mind, but ignoring for simplicity the parts of it that deal with NaN, infinity, Hexadecimal notation and exponents, and using \d rather than the POSIX notation for a single digit, I can reduce the important parts of the regex for a signed floating point number with no exponent to:

[+-]?((\d+\.?\d*)|(\.\d+))

I don't think that there is a way of avoiding the (...)|(...) construction without allowing something that contains no digits, or forbidding one of the possibilities that has no digits before the decimal point or no digits after it.

Obviously in practice you will need to cater for trailing or preceding whitespace, either in the regex itself or in the code that uses it.

Upvotes: 45