1. Introduction

Oftentimes while operating upon Strings, we need to figure out whether a String is a valid number or not.

In this tutorial, we’ll explore multiple ways to detect if the given String is numeric, first using plain Java, then regular expressions and finally by using external libraries.

Once we're done discussing various implementations, we'll use benchmarks to get an idea of which methods are optimal.

Further reading:

Java String Conversions

Quick and practical examples focused on converting String objects to different data types in Java.

A Guide To Java Regular Expressions API

A practical guide to Regular Expressions API in Java.

Understanding the NumberFormatException in Java

Learn the various causes of NumberFormatException in Java and some best practices for avoiding it.

2. Prerequisites

Let's start with some prerequisites before we head on to the main content.

In the latter part of this article, we'll be using Apache Commons external library for which we'll add its dependency in our pom.xml:

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.9</version>
</dependency>

The latest version of this library can be found on Maven Central.

3. Using Plain Java

Perhaps the easiest and the most reliable way to check whether a String is numeric or not is by parsing it using Java's built-in methods:

  1. Integer.parseInt(String)
  2. Float.parseFloat(String)
  3. Double.parseDouble(String)
  4. Long.parseLong(String)
  5. new BigInteger(String)

If these methods don't throw any NumberFormatException, then it means that the parsing was successful and the String is numeric:

public static boolean isNumeric(String strNum) {
    if (strNum == null) {
        return false;
    }
    try {
        double d = Double.parseDouble(strNum);
    } catch (NumberFormatException nfe) {
        return false;
    }
    return true;
}

Let's see this method in action:

assertThat(isNumeric("22")).isTrue();
assertThat(isNumeric("5.05")).isTrue();
assertThat(isNumeric("-200")).isTrue(); 
assertThat(isNumeric("10.0d")).isTrue();
assertThat(isNumeric("   22   ")).isTrue();
 
assertThat(isNumeric(null)).isFalse();
assertThat(isNumeric("")).isFalse();
assertThat(isNumeric("abc")).isFalse();

In our isNumeric() method, we're just checking for values that are of type Double, but this method can also be modified to check for Integer, Float, Long and large numbers by using any of the parse methods that we have enlisted earlier.

These methods are also discussed in the Java String Conversions article.

4. Using Regular Expressions

Now let's use regex -?\d+(\.\d+)? to match numeric Strings consisting of the positive or negative integer and floats.

But this goes without saying, that we can definitely modify this regex to identify and handle a wide range of rules. Here, we'll keep it simple.

Let’s break down this regex and see how it works:

  • -? – this part identifies if the given number is negative, the dash “” searches for dash literally and the question mark “?” marks its presence as an optional one
  • \d+ – this searches for one or more digits
  • (\.\d+)? – this part of regex is to identify float numbers. Here we're searching for one or more digits followed by a period. The question mark, in the end, signifies that this complete group is optional

Regular expressions are a very broad topic. To get a brief overview, check our tutorial on the Java regular expressions API.

For now, let's create a method using the above regular expression:

private Pattern pattern = Pattern.compile("-?\\d+(\\.\\d+)?");

public boolean isNumeric(String strNum) {
    if (strNum == null) {
        return false; 
    }
    return pattern.matcher(strNum).matches();
}

Let's now look at some assertions for the above method:

assertThat(isNumeric("22")).isTrue();
assertThat(isNumeric("5.05")).isTrue();
assertThat(isNumeric("-200")).isTrue();

assertThat(isNumeric(null)).isFalse();
assertThat(isNumeric("abc")).isFalse();

 5. Using Apache Commons

In this section, we'll discuss various methods available in the Apache Commons library.

5.1. NumberUtils.isCreatable(String) 

NumberUtils from Apache Commons provides a static method NumberUtils.isCreatable(String) which checks whether a String is a valid Java number or not.

This method accepts:

  1. Hexadecimal numbers starting with 0x or 0X
  2. Octal numbers starting with a leading 0
  3. Scientific notation (for example 1.05e-10)
  4. Numbers marked with a type qualifier (for example 1L or 2.2d)

If the supplied string is null or empty/blank, then it's not considered a number and the method will return false.

Let's run some tests using this method:

assertThat(NumberUtils.isCreatable("22")).isTrue();
assertThat(NumberUtils.isCreatable("5.05")).isTrue();
assertThat(NumberUtils.isCreatable("-200")).isTrue();
assertThat(NumberUtils.isCreatable("10.0d")).isTrue();
assertThat(NumberUtils.isCreatable("1000L")).isTrue();
assertThat(NumberUtils.isCreatable("0xFF")).isTrue();
assertThat(NumberUtils.isCreatable("07")).isTrue();
assertThat(NumberUtils.isCreatable("2.99e+8")).isTrue();
 
assertThat(NumberUtils.isCreatable(null)).isFalse();
assertThat(NumberUtils.isCreatable("")).isFalse();
assertThat(NumberUtils.isCreatable("abc")).isFalse();
assertThat(NumberUtils.isCreatable(" 22 ")).isFalse();
assertThat(NumberUtils.isCreatable("09")).isFalse();

Note how we're getting true assertions for hexadecimal numbers, octal numbers and scientific notations in lines 6, 7 and 8 respectively.

Also, on line 14, the string “09” returns false because the preceding “0” indicates that this is an octal number and “09” is not a valid octal number.

For every input that returns true with this method, we can use NumberUtils.createNumber(String) which will give us the valid number.

5.2. NumberUtils.isParsable(String) 

The NumberUtils.isParsable(String) method checks whether the given String is parsable or not.

Parsable numbers are those that are parsed successfully by any parse method like Integer.parseInt(String), Long.parseLong(String), Float.parseFloat(String) or Double.parseDouble(String).

Unlike NumberUtils.isCreatable(), this method won't accept hexadecimal numbers, scientific notations or strings ending with any type qualifier, that is, ‘f', ‘F', ‘d' ,'D' ,'l'or‘L'.

Let's look at some affirmations:

assertThat(NumberUtils.isParsable("22")).isTrue();
assertThat(NumberUtils.isParsable("-23")).isTrue();
assertThat(NumberUtils.isParsable("2.2")).isTrue();
assertThat(NumberUtils.isParsable("09")).isTrue();

assertThat(NumberUtils.isParsable(null)).isFalse();
assertThat(NumberUtils.isParsable("")).isFalse();
assertThat(NumberUtils.isParsable("6.2f")).isFalse();
assertThat(NumberUtils.isParsable("9.8d")).isFalse();
assertThat(NumberUtils.isParsable("22L")).isFalse();
assertThat(NumberUtils.isParsable("0xFF")).isFalse();
assertThat(NumberUtils.isParsable("2.99e+8")).isFalse();

On line 4, unlike NumberUtils.isCreatable(), the number starting with string “0” isn't considered as an octal number, but a normal decimal number and hence it returns true.

We can use this method as a replacement for what we did in section 3, where we’re trying to parse a number and checking for an error.

5.3. StringUtils.isNumeric(CharSequence) 

The method StringUtils.isNumeric(CharSequence) checks strictly for Unicode digits. This means:

  1. Any digits from any language that is a Unicode digit is acceptable
  2. Since a decimal point is not considered as a Unicode digit, it's not valid
  3. Leading signs (either positive or negative) are also not acceptable

Let's now see this method in action:

assertThat(StringUtils.isNumeric("123")).isTrue();
assertThat(StringUtils.isNumeric("١٢٣")).isTrue();
assertThat(StringUtils.isNumeric("१२३")).isTrue();
 
assertThat(StringUtils.isNumeric(null)).isFalse();
assertThat(StringUtils.isNumeric("")).isFalse();
assertThat(StringUtils.isNumeric("  ")).isFalse();
assertThat(StringUtils.isNumeric("12 3")).isFalse();
assertThat(StringUtils.isNumeric("ab2c")).isFalse();
assertThat(StringUtils.isNumeric("12.3")).isFalse();
assertThat(StringUtils.isNumeric("-123")).isFalse();

Note that the input parameters in lines 2 and 3 are representing numbers 123 in Arabic and Devanagari respectively. Since they're valid Unicode digits, this method returns true on them.

5.4. StringUtils.isNumericSpace(CharSequence)

The StringUtils.isNumericSpace(CharSequence) checks strictly for Unicode digits and/or space. This is same as StringUtils.isNumeric() with the only difference being that it accepts spaces as well, not only leading and trailing spaces but also if they're in between numbers:

assertThat(StringUtils.isNumericSpace("123")).isTrue();
assertThat(StringUtils.isNumericSpace("١٢٣")).isTrue();
assertThat(StringUtils.isNumericSpace("")).isTrue();
assertThat(StringUtils.isNumericSpace("  ")).isTrue();
assertThat(StringUtils.isNumericSpace("12 3")).isTrue();
 
assertThat(StringUtils.isNumericSpace(null)).isFalse();
assertThat(StringUtils.isNumericSpace("ab2c")).isFalse();
assertThat(StringUtils.isNumericSpace("12.3")).isFalse();
assertThat(StringUtils.isNumericSpace("-123")).isFalse();

6. Benchmarks

Before we conclude this article, let's go through some benchmark results to help us to analyze which of the above-mentioned methods are best for our use-case.

6.1. Simple Benchmark

First, we take a simple approach. We pick one string value – for our test we use Integer.MAX_VALUE. Then, that value will be tested against all our implementations:

Benchmark                                     Mode  Cnt    Score   Error  Units
Benchmarking.usingCoreJava                    avgt   20   57.241 ± 0.792  ns/op
Benchmarking.usingNumberUtils_isCreatable     avgt   20   26.711 ± 1.110  ns/op
Benchmarking.usingNumberUtils_isParsable      avgt   20   46.577 ± 1.973  ns/op
Benchmarking.usingRegularExpressions          avgt   20  101.580 ± 4.244  ns/op
Benchmarking.usingStringUtils_isNumeric       avgt   20   35.885 ± 1.691  ns/op
Benchmarking.usingStringUtils_isNumericSpace  avgt   20   31.979 ± 1.393  ns/op

As we see, the most costly operations are regular expressions. After that is our core Java-based solution.

Moreover, note that the operations using the Apache Commons library are by-and-large the same.

6.2. Enhanced Benchmark

Let's use a more diverse set of tests, for a more representative benchmark:

  • 95 values are numeric (0-94 and Integer.MAX_VALUE)
  • 3 contain numbers but are still malformatted — ‘x0‘, ‘0..005′, and ‘–11
  • 1 contains only text
  • 1 is a null

Upon executing the same tests, we'll see the results:

Benchmark                                     Mode  Cnt      Score     Error  Units
Benchmarking.usingCoreJava                    avgt   20  10162.872 ± 798.387  ns/op
Benchmarking.usingNumberUtils_isCreatable     avgt   20   1703.243 ± 108.244  ns/op
Benchmarking.usingNumberUtils_isParsable      avgt   20   1589.915 ± 203.052  ns/op
Benchmarking.usingRegularExpressions          avgt   20   7168.761 ± 344.597  ns/op
Benchmarking.usingStringUtils_isNumeric       avgt   20   1071.753 ±   8.657  ns/op
Benchmarking.usingStringUtils_isNumericSpace  avgt   20   1157.722 ±  24.139  ns/op

The most important difference is that two of our tests – the regular expressions solution and the core Java-based solution – have traded places.

From this result, we learn that throwing and handling of the NumberFormatException, which occurs in only 5% of the cases, has a relatively big impact on the overall performance. So, we conclude, that the optimal solution depends on our expected input.

Also, we can safely conclude that we should use the methods from the Commons library or a method implemented similarly for optimal performance.

7. Conclusion

In this article, we explored different ways to find if a String is numeric or not. We looked at both solutions – built-in methods and also external libraries.

As always, the implementation of all examples and code snippets given above including the code used to perform benchmarks can be found over on GitHub.

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3 Comments
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JoeHx
2 years ago

I was wondering how using try-catches would compare to using regular expressions. It’s interesting the rather large difference between those two and the rest.

John Douglass
John Douglass
2 years ago

You will probably get a good performance boost for regex by reusing a Pattern object, e.g.:

private static Pattern pNumeric = Pattern.compile(“-?\\d+(\\.\\d+)?”);
public static boolean isNumeric(String strNum) {
return pNumeric.matcher(strNum).matches();
}

It was about 3-4x faster for me, but still much slower than the alternatives.

Loredana Crusoveanu
2 years ago
Reply to  John Douglass

Interesting, thanks. We’ll update the article.

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