How to sort a String list by using part of its numerical contents?

Question

I am trying to sort an ArrayList<String>. The strings in the list look like this "abc 123 def 456" and I want to sort the list such that the string with the smallest last number comes first e.g. if the elements are

{"abc 123 def 456", "ghi 456 jkl 789", "mno 101 pqr 112"}

then I want the sorted list to have the third String first, then the first String, and then the second String.

{"mno 101 pqr 112", "abc 123 def 456","ghi 456 jkl 789"}

Just like this.

I tried to sort it using list.split(" "), Double.parseDouble(String)and Collections.sort(list), but I don't now how I could connect the remainder of my Strings correctly with the list of sorted numbers.

Edit:

String[] numbers = list.get(i).split(" ");
numberList.add(Integer.parseInt(numbers[3]));
Collections.sort(numberList);

Now I know how the numbers should be sorted, but I can't reconnect them with the rest of "their String".

Show us your best attempt. Don't use Double.parseDouble() to parse an integer. — JB Nizet
– JB Nizet, Commented Dec 12, 2015 at 16:53
Just sort by the last number, you are not force to separate the parts of the string so just don't do that. Why make your life harder? — Peter Lawrey
– Peter Lawrey, Commented Dec 12, 2015 at 16:54
@JBNizet Perhaps you could explain what is the downside of parsing an integer with parseDouble? — Peter Lawrey
– Peter Lawrey, Commented Dec 12, 2015 at 16:55
My actual String contains doubles not integers, that's why I wrote Double.parseDouble() here as well. @Peter Lawrey I want the Strings to be the same after sorting, just in an other order. Can I still sort by the last number? — DTsn
– DTsn, Commented Dec 12, 2015 at 16:58
Double.parseDouble() expects a decimal number, whereas you expect an integer. So it will happily return a value if what you're parsing happens to be a valid double, although it's supposed to accept only valid integers. For example, parseDouble() will happily parse "1.23e7", although that is clearly not what you expect, and you'd better have an exception, signalling the bug, if that's what you're actually parsing. If you expect an integer, parse an integer. — JB Nizet
– JB Nizet, Commented Dec 12, 2015 at 16:59

Edgar Rokjān · Accepted Answer · 2015-12-12 17:02:37Z

2

Simple solution with custom Comparator:

String[] arr = {"abc 123 def 456", "ghi 456 jkl 789", "mno 101 pqr 112"};

Arrays.sort(arr, new Comparator<String>() {
    @Override
    public int compare(String o1, String o2) {
        return getLastNum(o1) - getLastNum(o2);
    }

    private int getLastNum(String s) {
        String[] tokens = s.split(" ");
        return Double.parseDouble(tokens[tokens.length - 1]);
    }
});

Hope that this piece of code is self explanatory.

answered Dec 12, 2015 at 17:02

Edgar Rokjān

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Comments

Luiggi Mendoza · Accepted Answer · 2015-12-12 17:12:37Z

0

If all of the will have the same format, then parse the last string of the result of String#split and compare those values.

list.sort((s1, s2) -> {
    String[] split1 = s1.split("\\s+");
    String[] split2 = s2.split("\\s+");
    return Double.compare(
        Double.parseDouble(split1[split1.length - 1]),
        Double.parseDouble(split2[split2.length - 1]);
});

Or as suggested by @JBNizet:

list.sort(Comparator.comparingDouble(s -> Double.parseDouble(s.substring(s.lastIndexOf(' ') + 1))));

edited Dec 12, 2015 at 17:12

answered Dec 12, 2015 at 16:59

Luiggi Mendoza

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6 Comments

JB Nizet Over a year ago

Or, with Java 8: list.sort(Comparator.comparingDouble(s -> Double.parseDouble(s.substring(s.lastIndexOf(' ') + 1))));. Love those simple one-liners.

JB Nizet Over a year ago

Your mixed up split1 and split2 in the answer: split2[split1.length - 1]

JB Nizet Over a year ago

They're still in the wrong order. Your comparator sorts in reverse order. That's one of the benefits of Comparator.comparing(): you don't mess up with the order of the arguments, because you just have one.

Luiggi Mendoza Over a year ago

I've misread the question, I (don't know why I) though OP was looking for inverse order.

DTsn Over a year ago

I'm afraid I don't really understand your code. What is String#split ? And s1 and s2? Are those the Strings in the list?

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