CODEWITHSUNDEEP
javastringsorting
Sanju
Sanju

Reputation: 903

How do I sort strings that contain numbers in Java

I want to sort a String that has nr. How do I do that?

Lets say my integers are

Class2
"3"
"4"
"1"

in main I do class2.Sort();

Thanks in Advance.

Upvotes: 5

Views: 59888

Answers (7)

Hec T
Hec T

Reputation: 9

static final Comparator<Object> COMPARADOR = new Comparator<Object>() {
    public int compare(Object o1, Object o2) {
        double numero1;
        double numero2;
        try {
            numero1 = Double.parseDouble(o1.toString());
            numero2 = Double.parseDouble(o2.toString());
            return Double.compare(numero1, numero2);
        } catch (Exception e) {
            return o1.toString().compareTo(o2.toString());
        }
    }
};

... ArrayList listaDeDatos; listaDeDatos.sort(COMPARADOR);

Upvotes: 0

lobsten
lobsten

Reputation: 41

In Java 8 we have nice solution

public static List<String> sortAsNumbers(Collection<String> collection) {
    return collection
            .stream()
            .map(Integer::valueOf)
            .sorted()
            .map(String::valueOf)
            .collect(Collectors.toList());
}

Upvotes: 4

Kevin Day
Kevin Day

Reputation: 16413

A general purpose solution is to use what's called a 'natural order comparator'.

Here's an example:

http://pierre-luc.paour.9online.fr/NaturalOrderComparator.java

Natural ordering is actually quite important in cases where a string might contain runs of numbers and you want things to sort alphabetically on the letters but numerically on the numbers. Modern versions of Windows Explorer uses this for ordering file names, for example. It's also very handy for picking out the latest version of a library based on version strings (i.e. "1.2.3" compared to to "1.20.1").

If your strings really just contain numbers (like you put in your description), then you are best off not using strings at all - create and work with Integer objects instead.

Note: The link above seems to be broken. The code is so useful that I'm going to post it here:

/*
 * <copyright>
 *
 *  Copyright 1997-2007 BBNT Solutions, LLC
 *  under sponsorship of the Defense Advanced Research Projects
 *  Agency (DARPA).
 *
 *  You can redistribute this software and/or modify it under the
 *  terms of the Cougaar Open Source License as published on the
 *  Cougaar Open Source Website (www.cougaar.org).
 *
 *  THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
 *  "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
 *  LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
 *  A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
 *  OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
 *  SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
 *  LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
 *  DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
 *  THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
 *  (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
 *  OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
 *
 * </copyright>
 */
/*
NaturalOrderComparator.java -- Perform 'natural order' comparisons of strings in Java.
Copyright (C) 2003 by Pierre-Luc Paour <[email protected]>

Based on the C version by Martin Pool, of which this is more or less a straight conversion.
Copyright (C) 2000 by Martin Pool <[email protected]>

This software is provided 'as-is', without any express or implied
warranty.  In no event will the authors be held liable for any damages
arising from the use of this software.

Permission is granted to anyone to use this software for any purpose,
including commercial applications, and to alter it and redistribute it
freely, subject to the following restrictions:

1. The origin of this software must not be misrepresented; you must not
claim that you wrote the original software. If you use this software
in a product, an acknowledgment in the product documentation would be
appreciated but is not required.
2. Altered source versions must be plainly marked as such, and must not be
misrepresented as being the original software.
3. This notice may not be removed or altered from any source distribution.
 */
package org.cougaar.util;

//CHANGES: KD - added case sensitive ordering capability
// Made comparison so it doesn't treat spaces as special characters

//CHANGES:
//   set package to "org.cougaar.util"
//   replaced "import java.util.*" with explicit imports,
//   added "main" file reader support

import java.util.Comparator;

/**
 * A sorting comparator to sort strings numerically,
 * ie [1, 2, 10], as opposed to [1, 10, 2].
 */
public final class NaturalOrderComparator<T> implements  Comparator<T> {

    public static final Comparator<String> NUMERICAL_ORDER = new NaturalOrderComparator<String>(false);
    public static final Comparator<String> CASEINSENSITIVE_NUMERICAL_ORDER = new NaturalOrderComparator<String>(true);

    private final boolean caseInsensitive;

    private NaturalOrderComparator(boolean caseInsensitive) {
        this.caseInsensitive = caseInsensitive;
    }

    int compareRight(String a, String b) {
        int bias = 0;
        int ia = 0;
        int ib = 0;

        // The longest run of digits wins.  That aside, the greatest
        // value wins, but we can't know that it will until we've scanned
        // both numbers to know that they have the same magnitude, so we
        // remember it in BIAS.
        for (;; ia++, ib++) {
            char ca = charAt(a, ia);
            char cb = charAt(b, ib);

            if (!Character.isDigit(ca) && !Character.isDigit(cb)) {
                return bias;
            } else if (!Character.isDigit(ca)) {
                return -1;
            } else if (!Character.isDigit(cb)) {
                return +1;
            } else if (ca < cb) {
                if (bias == 0) {
                    bias = -1;
                }
            } else if (ca > cb) {
                if (bias == 0)
                    bias = +1;
            } else if (ca == 0 && cb == 0) {
                return bias;
            }
        }
    }

    public int compare(T o1, T o2) {
        String a = o1.toString();
        String b = o2.toString();

        int ia = 0, ib = 0;
        int nza = 0, nzb = 0;
        char ca, cb;
        int result;

        while (true) {
            // only count the number of zeroes leading the last number compared
            nza = nzb = 0;

            ca = charAt(a, ia);
            cb = charAt(b, ib);

            // skip over leading zeros
            while (ca == '0') {
                if (ca == '0') {
                    nza++;
                } else {
                    // only count consecutive zeroes
                    nza = 0;
                }

                // if the next character isn't a digit, then we've had a run of only zeros
                // we still need to treat this as a 0 for comparison purposes
                if (!Character.isDigit(charAt(a, ia+1)))
                    break;

                ca = charAt(a, ++ia);
            }

            while (cb == '0') {
                if (cb == '0') {
                    nzb++;
                } else {
                    // only count consecutive zeroes
                    nzb = 0;
                }

                // if the next character isn't a digit, then we've had a run of only zeros
                // we still need to treat this as a 0 for comparison purposes
                if (!Character.isDigit(charAt(b, ib+1)))
                    break;

                cb = charAt(b, ++ib);
            }

            // process run of digits
            if (Character.isDigit(ca) && Character.isDigit(cb)) {
                if ((result = compareRight(a.substring(ia), b
                        .substring(ib))) != 0) {
                    return result;
                }
            }

            if (ca == 0 && cb == 0) {
                // The strings compare the same.  Perhaps the caller
                // will want to call strcmp to break the tie.
                return nza - nzb;
            }

            if (ca < cb) {
                return -1;
            } else if (ca > cb) {
                return +1;
            }

            ++ia;
            ++ib;
        }
    }

    private char charAt(String s, int i) {
        if (i >= s.length()) {
            return 0;
        } else {
            return caseInsensitive ? Character.toUpperCase(s.charAt(i)) : s.charAt(i);
        }
    }


}

Upvotes: 27

Thorbj&#248;rn Ravn Andersen
Thorbj&#248;rn Ravn Andersen

Reputation: 75426

The way to sort "things" according to some order is to create a Comparator which knows which of any two things are first according to the order, or to let the "thing" itself implement the Comparable interface so you do not need the Comparator.

If your job is to sort as integers, then consider converting to integers and then sort as the Integer class already implements Comparable.

Upvotes: 1

eljenso
eljenso

Reputation: 17037

  public static void main(String[] args)
  {
    String string = "3 42 \n   11   \t  7  dsfss  365          \r   1";
    String[] numbers = string.split("\\D+");
    Arrays.sort(numbers, new Comparator<String>()
    {
      public int compare(String s1, String s2)
      {
        return Integer.valueOf(s1).compareTo(Integer.valueOf(s2));
      }
    });
    System.out.println(Arrays.toString(numbers));
  }

Upvotes: 16

polygenelubricants
polygenelubricants

Reputation: 383986

Your question is rather ill-formed, but here are a few things that you should know:

  • There is an Arrays.sort(Object[]) that you can use to sort arbitrary objects.
  • The natural ordering for String is lexicographic.
  • Java arrays are covariant: a String[] is an Object[].

So, given String[] sarr, if you want to sort it lexicographically (i.e. "1" < "10" < "2"), simply Arrays.sort(sarr); works. It doesn't matter if the strings contain numbers or not.

If you want to sort the strings as if they are numbers (i.e. "1" < "2" < "10"), then you need to convert the strings to numeric values. Depending on the range of these numbers, Integer.parseInt might do; you can always use BigInteger otherwise.

Let's assume that BigInteger is required.

You now have two options:

  • Convert String[] to BigInteger[], then since BigInteger implements Comparable<BigInteger>, you can use Arrays.sort using its natural ordering. You may then convert the sorted BigInteger[] back to String[].

  • Convert String to BigInteger "just-in-time" for comparison by a custom Comparator<String>. Since Arrays.sort uses the comparison based mergesort, you can expect O(N log N) comparisons, and therefore as many conversions.

Upvotes: 6

Midhat
Midhat

Reputation: 17840

If the numbers are all single digit, split the string into a char array and sort the array. otherwise there has to be a delimiter to seperate the numbers. call string.split using that delimiter and sort the resulting array. The function to sort is Arrays.sort() if memory serves me right

javadoc references

http://java.sun.com/j2se/1.4.2/docs/api/java/lang/String.html#split%28java.lang.String%29 http://java.sun.com/javase/6/docs/api/java/util/Arrays.html#sort%28double[]%29

Upvotes: 0

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