如何在java中找到2D ArrayList列的唯一值？

Question

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.function.Function;
import java.util.stream.Collectors;

public class Delete
{
    public static void main(String[] args)
    {
         List<List<String>> list = new ArrayList<>();
         list.add(List.of("A","B","C","R"));
         list.add(List.of("E","F","G","F"));
         list.add(List.of("A","B","C","D"));
         System.out.println(list.stream().distinct().count());
         Map<String, Long> countMapOfColumn = list.stream()
                                                  .filter(innerList -> innerList.size() >= 3)
                                                  .map(innerList -> innerList.get(3 - 1))
                         .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
        System.out.println(countMapOfColumn.keySet());
    }
}

I want to find out the unique element of column 3 which are "C","G" and there are 2 unique element in column 3.

This can be done using loop but I can't use loop. Using java stream there may be a solution.

In addition, how can I get count of rows having "A","B" in column 1,2 at a time and in general for N columns?

Answer 1

For your given datastructure;

Map<String, Long> countMapOfColumn = list.stream()
        .filter(innerList -> innerList.size() >= column)
        .map(innerList -> innerList.get(column - 1))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

where column = 3 , will get you a map with {"C", "G"} as keyset, and values will give you the count of each character in the given column. It will also work for non uniform matrices, since it will just skip the rows that has no nth column.

Unique count of any column will the the size of resulting map of collect method; countMapOfColumn.size()

To get individual char counts, use the resulting map with keys as input, for example to get count of C in column = 3 , use countMapOfColumn.get("C") , which will return 2

But I'd rather use a Guava's Table , you can directly select any given row or column without any hassle, then it is only, getting nth column, and filter our duplicates.

update

To get the count of rows that start with consecutive list list = {"A", "B"} ;

List<String> startingList = Arrays.asList("A", "B");
Long count = list.stream()
        .filter(innerList -> innerList.size() >= startingList.size() 
                && innerList.subList(0, startingList.size()).equals(startingList))
        .count();

Answer 2

Your question doesn't compile. Assuming you have defined the list as follows, you can easily find the distinct value in column 3 by using the following. Here we are just mapping the 3rd column (by using index 2) and then applying .distinct() on the resulting stream.

List<List<String>> list = Arrays.asList(Arrays.asList("A", "B", "C"), 
                                        Arrays.asList("E", "F", "G"),
                                        Arrays.asList("A", "B", "C"));

List<String> distinct = list.stream().map(a -> a.get(2))
                            .distinct().collect(Collectors.toList());

When you print out the elements it will output :

C

G

You can use the same logic to find count of 'C' in column 3 using the same logic.

list.stream().map(a -> a.get(2)).filter(i -> i.contains("C")).count();