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POProlog: How do I count the number of passes and failures for a predicate?
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  1. POProlog: How do I count the number of passes and failures for a predicate?
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    copied!<p>I want to count the number of successes AND failures for a predicate that doesn't have any arguments. For simplicity, the predicate I want to test is test_arith/0. test_arith/0 has 3 tests for is/2 (again, for simplicity. I really want to test more complicated predicates that I'm writing, but those details would be a distraction for this question.) (P.S. I saw the other question about counting the number times a predicate is true. I want to count both the successes and the failures in a single pass. I don't want to have to run the real predicates more than once per test case because some of them take a long time to execute. aggregate/3 and aggregate_all seem to be single-minded as well, taking only the successes.)</p> <pre><code>test_arith(Passes, Failures) :- findall(P-F, (test_arith-&gt;(P=1,F=0);(P=0,F=1)), Scores), summarize_scores(Scores, 0, 0, Passes, Failures). test_arith :- 5 is 3 +2. % Test #1: Should pass test_arith :- 5 is 2 +2. % Test #2: Should fail test_arith :- 4 is 2 +2. % Test #3: Should pass summarize_scores([], Passes, Failures, Passes, Failures). summarize_scores([P-F|Scores], Passes_SF, Failures_SF, Passes, Failures) :- Next_Passes is P + Passes_SF, Next_Failures is F + Failures_SF, summarize_scores(Scores, Next_Passes, Next_Failures, Passes, Failures). </code></pre> <p>When I run</p> <pre><code>test_arith(P,F). </code></pre> <p>I get</p> <pre><code>P = 1, F = 0. </code></pre> <p>because test_arith seems to be called only once. I should get</p> <pre><code>P = 2, F = 1. </code></pre> <p>Thanks for any help you can give.</p> <p>I tried:</p> <pre><code>test_arith(Passes, Failures) :- bagof(P-F, A^(test_arith(A)-&gt;(P=1,F=0);(P=0,F=1)), Scores), summarize_scores(Scores, 0, 0, Passes, Failures). test_arith(_) :- 5 is 3 +2. test_arith(_) :- 5 is 2 +2. test_arith(_) :- 4 is 2 +2. test_arith2(Passes) :- aggregate(count, A^test_arith(A), Passes). </code></pre> <p>test_arith(P,F) yields: P = 1, F = 0. test_arith2(P) yields "2". (Which is good that it works, but only 1/4 of what I'm looking for. I need a count of failures and for each predicate to be run only once per test run, 3 calls in this case.)</p> <p>Then I tried adding a number for each test case:</p> <pre><code>test_arith(Passes, Failures) :- bagof(P-F, A^(test_arith(A)-&gt;(P=1,F=0);(P=0,F=1)), Scores), summarize_scores(Scores, 0, 0, Passes, Failures). test_arith(1) :- 5 is 3 +2. test_arith(2) :- 5 is 2 +2. test_arith(3) :- 4 is 2 +2. </code></pre> <p>and got:</p> <pre><code>test_arith(P,F). P = 1, F = 0. </code></pre>
 

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