从Java中的ArrayList中删除“ regex重复项”

Question

I want to "clean" an ArrayList in java, here is the explanation

Assuming we have this list :

a = ["a_12_b", "a_13_b", "a_13bis_b", "a_14_b", "a_14_new_b"]

In this list, "a_13bis_b" and "a_14_new_b" are considered as duplicates, Why ? because each entry have this regex : a_ "a string with a lenght =2" _b

The output should be :

a = ["a_12_b", "a_13_b", "a_14_b"]

I used this simple code, but it returns wrong output :

for (int j = 0; j < list.size(); j++) {
            //basically clean entry will remove the a_ and _b
            String value1= cleanEntry(list.get(j));
            for (int k = 0; k < list.size(); k++) {
                    String value2= cleanEntry(list.get(k));
                    if (k != j && value1.equalsIgnoreCase(value2)) {
                        duplicates.add(list.get(k))
                        list.remove(k);
                    }
            }
}

Any help ?

Answer 1

You could use the stream map method with a regular expression to "normalize" the strings to a common format and then create a set out of the normalized strings.

Something like this:

List<String> a = Arrays.asList("a_12_b", "a_13_b", "a_13bis_b", "a_14_b", "a_14_new_b");
Set<String> uniques = a.stream()
                .map(s -> s.replaceAll("^([a-z]_\\d{2})[^\\d].+(_[a-z])$", "$1$2"))
                .collect(Collectors.toSet());
System.out.println(uniques);

This prints:

[a_14_b, a_13_b, a_12_b]

Solution for Java 7, 6:

List<String> a = Arrays.asList("a_12_b", "a_13_b", "a_13bis_b", "a_14_b", "a_14_new_b");
Set<String> set = new LinkedHashSet<>();
for(String s : a) {
    set.add(s.replaceAll("^([a-z]_\\d{2})[^\\d].+(_[a-z])$", "$1$2"));
}
System.out.println(set);

Result:

[a_12_b, a_13_b, a_14_b]

If you need more than 2 numeric characters, you can change the regular expression. Here is an example with result:

List<String> a = Arrays.asList("a_12345678901234567890123456_b", "a_13345678901234567890123456_b",
                "a_13345678901234567890123456bis_b", "a_14345678901234567890123456_b", "a_14345678901234567890123456_new_b");
Set<String> set = new LinkedHashSet<>();
for(String s : a) {
    set.add(s.replaceAll("^([a-z]_\\d{26})[^\\d].+(_[a-z])$", "$1$2"));
}
System.out.println(set);

Result:

[a_12345678901234567890123456_b, a_13345678901234567890123456_b, a_14345678901234567890123456_b]

Answer 2

You can simply discard all the characters after 2nd character before comparison. Try this..

for (int j = 0; j < list.size(); j++) {
    //basically clean entry will remove the a_ and _b
    String value1= cleanEntry(list.get(j));
    for (int k = 0; k < list.size(); k++) {
        String value2= cleanEntry(list.get(k));
        if (k != j && value1.substring(0,2).equalsIgnoreCase(value2.substring(0,2))) {
            duplicates.add(list.get(k)) list.remove(k);
        }
    } 
}