Array assertion in Spock

Question

I have a list of some objects - let's assume companies. Now I want to check if this list contains companies with some names but not taking order into consideration. Currently I'm using such construction:

companyList.name.sort() == ["First", "Second"]

Is there any operator in Spock or Groovy that allows me to compare arrays without order?

I've never really thought of using .sort() as much of a trouble. It's fairly terse. — th3morg, Feb 11 '15 at 19:01

score 17 · Answer 1 · answered Oct 03 '15 at 18:47

You can make use of Hamcrest support in Spock and use the Matcher designed explicitly for this case - containsInAnyOrder. You need the following imports:

import static org.hamcrest.Matchers.containsInAnyOrder
import static spock.util.matcher.HamcrestSupport.that

Then you can write your test code as follows:

given:
    def companyList = [ "Second", "First"]
expect:
    that companyList, containsInAnyOrder("First", "Second")

This has the benefit over using .sort() in that duplicate elements in the List will be considered correctly. The following test will fail using Hamcrest but passes using .sort()

given:
    def companyList = [ "Second", "First", "Second"]
expect:
    that companyList, containsInAnyOrder("First", "Second")

Condition not satisfied:

that companyList, containsInAnyOrder("First", "Second")
|    |
|    [Second, First, Second]
false

Expected: iterable over ["First", "Second"] in any order
     but: Not matched: "Second"

If you're using then: instead of expect: you can use the expect instead of that import for readability.

then:
    expect companyList, containsInAnyOrder("First", "Second")

Opal · Accepted Answer · 2015-05-27T18:15:55.493

15

There's no such operator as far as I know. If the list doesn't contain any duplicates the following assertion may be used:

companyList.name as Set == ["First", "Second"] as Set

Or something like that:

companyList.name.size() == ["First", "Second"].size() && companyList.name.containsAll(["First", "Second"])

edited May 27 '15 at 18:15

answered Feb 11 '15 at 10:06

Opal

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I wanted something shorter than ```.sort()``` :) Answer like use ~== would be great :) – Jakub Kubrynski Feb 11 '15 at 10:23
Unfortunately there' no such operator. Will have a look in a minute. – Opal Feb 11 '15 at 10:24
5

IMHO `as Set` is the most logical, as it _states_, that order is not important – cfrick Feb 11 '15 at 10:52
Yes, I think exactly the same. – Opal Feb 11 '15 at 11:00
1

Yep - it's not as cool as operator but code readability is good – Jakub Kubrynski Feb 11 '15 at 20:53

score 2 · Answer 3 · answered Feb 11 '15 at 20:03

Upvoted for @Opal answer, it's probably the best in all terms. As a curiosity I would only add an example of usage of minus operator to compare two unsorted lists:

import spock.lang.Specification

class ComparingListsSpec extends Specification {

    def "should contain all companies in two unsorted lists"() {
        when:
        def companies = ["Lorem", "ipsum", "dolor", "sit", "amet"]
        def unsorted = ["ipsum", "sit", "Lorem", "dolor", "amet"]

        then:
        companies - unsorted == []

        and:
        unsorted - companies == []

        and:
        companies - unsorted in [[]]
    }
}

It also works if one of the lists contains redundant data.

i'd stay away from `-` for anything, that is not a trivial type. `assert [[x:0,y:0]]-[[x:1,y:1]] == []` — cfrick, Feb 11 '15 at 21:45

score 0 · Answer 4 · answered May 30 '23 at 13:21

Now there are dedicated matchers available in Spock:

You can compare list elements ignoring their ordering using the following operators:

==~ - Strict match (accounting for duplicated elements, etc.)
=~ - Lenient match (ignoring duplicates)

So [1,2,3] ==~ [2,3,1] is truthy.

Array assertion in Spock

4 Answers4