Reduce Float Precision Using Regexp in Swift

Reduce float precision using RegExp in swift

You may use

let str = "The example is [93829.38, 1415.45467897, 1.2, 134.34]"
let pattern = "(\\d+\\.\\d{2})\\d+"
let result = str.replacingOccurrences(of: pattern, with: "$1", options: [.regularExpression])
print(result) // => The example is [93829.38, 1415.45, 1.2, 134.34]

A pattern like (\d+\.\d{2})\d+ will match and capture into Group 1 one or more diigts, a dot and then two digits, and then will match one or more digits. The replacement is $1, the backreference to the value stored in Group 1, thus, truncating the digits matched with the last \d+.

See the regex demo here.

If there are any edge cases, they can usually be handled by means of word boundaries (\b) or lookarounds.

Regular expression for floating point numbers

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.

RegEx needed to match number to exactly two decimal places

^[0-9]*\.[0-9]{2}$ or ^[0-9]*\.[0-9][0-9]$

Regex to match 2 digits, optional decimal, two digits

EDIT: Changed to fit other feedback.

I understood you to mean that if there is no decimal point, then there shouldn't be two more digits. So this should be it:

\d{0,2}(\.\d{1,2})?

That should do the trick in most implementations. If not, you can use:

[0-9]?[0-9]?(\.[0-9][0-9]?)?

And that should work on every implementation I've seen.

Round a digit upto two decimal place in Swift

Use a format string to round up to two decimal places and convert the double to a String:

let currentRatio = Double (rxCurrentTextField.text!)! / Double (txCurrentTextField.text!)!
railRatioLabelField.text! = String(format: "%.2f", currentRatio)

Example:

let myDouble = 3.141
let doubleStr = Double(String(format: "%.2f", myDouble)) // 3.14

let myDouble = 3.141
let doubleStr = String(format: "%.2f", myDouble) // "3.14"

If you want to round up your last decimal place, you could do something like this :

let myDouble = 3.141
let doubleStr = Double(String(format: "%.2f", ceil(myDouble*100)/100)) // 3.15

let myDouble = 3.141
let doubleStr = String(format: "%.2f", ceil(myDouble*100)/100) // "3.15"

How can I limit the number of decimal points in a UITextField?

Implement the shouldChangeCharactersInRange method like this:

// Only allow one decimal point
// Example assumes ARC - Implement proper memory management if not using.
- (BOOL)textField:(UITextField *)textField shouldChangeCharactersInRange:(NSRange)range replacementString:(NSString *)string 
{
    NSString *newString = [textField.text stringByReplacingCharactersInRange:range withString:string];
    NSArray  *arrayOfString = [newString componentsSeparatedByString:@"."];

    if ([arrayOfString count] > 2 ) 
        return NO;

    return YES;
}

This creates an array of strings split by the decimal point, so if there is more than one decimal point we will have at least 3 elements in the array.

RegEx validation of decimal numbers with comma or dot

Try this regex:

/^(\d+(?:[\.\,]\d{2})?)$/

If $1 exactly matches your input string then assume that it is validated.

Limiting user input to a valid decimal number in Swift

Here is a simple example:

import UIKit

class ViewController: UIViewController, UITextFieldDelegate {

    @IBOutlet weak var textField: UITextField!

    override func viewDidLoad() {
        super.viewDidLoad()

        self.textField.delegate = self

    }

    //Textfield delegates
    func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool { // return NO to not change text

        switch string {
        case "0","1","2","3","4","5","6","7","8","9":
            return true
        case ".":
            let array = Array(textField.text)
            var decimalCount = 0
            for character in array {
                if character == "." {
                    decimalCount++
                }
            }

            if decimalCount == 1 {
                return false
            } else {
                return true
            }
        default:
            let array = Array(string)
            if array.count == 0 {
                return true
            }
            return false
        }
    }
}

Regular expression for matching latitude/longitude coordinates?

Whitespace is \s, not \w

^(-?\d+(\.\d+)?),\s*(-?\d+(\.\d+)?)$

See if this works

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