AssertEquals 2 Lists ignore order

只谈情不闲聊 提交于 2019-11-27 04:09:20

As you mention that you use Hamcrest, I would pick one of the collection Matchers

import static org.hamcrest.collection.IsIterableContainingInAnyOrder.containsInAnyOrder;
import static org.junit.Assert.assertThat;

public class CompareListTest {

    @Test
    public void compareList() {
        List<String> expected = Arrays.asList("String A", "String B");
        List<String> actual = Arrays.asList("String B", "String A");

        assertThat("List equality without order", 
            actual, containsInAnyOrder(expected.toArray()));
    }

}
robertoia

You can use List.containsAll with JUnit's assertTrue to check that the first list contains every element from the second one, and vice versa.

assertTrue(first.size() == second.size() && 
    first.containsAll(second) && second.containsAll(first));
Alex Worden

Here's a solution that avoids quadratic complexity (iterating over the lists multiple times). This uses the Apache Commons CollectionUtils class to create a Map of each item to a frequency count itself in the list. It then simply compares the two Maps.

Assert.assertEquals("Verify same metrics series",
    CollectionUtils.getCardinalityMap(expectedSeriesList),
    CollectionUtils.getCardinalityMap(actualSeriesList));

I also just spotted CollectionUtils.isEqualCollection that claims to do exactly what is being requested here...

https://commons.apache.org/proper/commons-collections/apidocs/index.html?org/apache/commons/collections4/CollectionUtils.html

Note that solution by Roberto Izquierdo has quadratic complexity in general. Solution on HashSets always has linear complexity:

assertTrue(first.size() == second.size() &&
        new HashSet(first).equals(new HashSet(second)));

Im late to the party but here's my solution using Junit only. Any thoughts are welcome.

List<String> actual = new ArrayList<>();
actual.add("A");
actual.add("A");
actual.add("B");

List<String> expected = new ArrayList<>();
actual.add("A");
actual.add("B");
actual.add("B");

//Step 1: assert for size
assertEquals(actual.size(), expected.size());

//Step 2: Iterate
for(String e: expected){
    assertTrue(actual.contains(e));
    actual.remove(e);
}

You can use ListAssert that comes in junit-addons jar.

ListAssert.assertEquals(yourList, Arrays.asList(3, 4, 5));
Kristjan Veskimäe

For a quick fix I would check both ways:

assertTrue(first.containsAll(second));
assertTrue(second.containsAll(first));

And trying with a situation where the number of the same elements is different (e.g. 1, 1, 2 and 1, 2, 2) I didn't get false positives.

    Collections.sort(excepted);
    Collections.sort(actual);
    assertEquals(excepted,actual);

With AssertJ, containsExactlyInAnyOrder() or containsExactlyInAnyOrderElementsOf() is what you need :

import org.assertj.core.api.Assertions;
import org.junit.jupiter.api.Test;

import java.util.Arrays;
import java.util.List;

public class CompareListTest {

    @Test
    public void compareListWithTwoVariables() {
        List<String> expected = Arrays.asList("String A", "String B");
        List<String> actual = Arrays.asList("String B", "String A");
        Assertions.assertThat(actual)
                  .containsExactlyInAnyOrderElementsOf(expected);
    }

    @Test
    public void compareListWithInlineExpectedValues() {
        List<String> actual = Arrays.asList("String B", "String A");
        Assertions.assertThat(actual)
                  .containsExactlyInAnyOrder("String A", "String B");
    }    
}
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!