Erlang列表理解 [英] Erlang list comprehension

问题描述

我正在测试一个表达式，其中两个不等式用于列表推导的条件。有没有办法在这里做任务，而不是重复那个表达？
（以下代码不起作用，但我希望这样）

I'm testing an expression with two inequalities for the condition of a list comprehension. Is there a way to have assignments here and not duplicate that expression? (The following code doesn't work, but I wish it would)

诊断（专长，患者症状） - >

diagnose(Expertise,PatientSymptoms)->

{[CertainDisease||
    {CertainDisease,KnownSymptoms}<-Expertise,
    C=length(PatientSymptoms)-length(PatientSymptoms--KnownSymptoms),
    C>=2,
    C<=5      
 ]}.

推荐答案

直接写一个 fun 将使用以布尔测试结尾的 begin ... end 块： / p>

A way of writing it directly without a fun would be to use a begin ... end block ending with a boolean test:

[ CertainDisease || {CertainDisease,KnownSymptoms} <- Expertise,
                    begin
                        C = length(PatientSymptoms) - length(PatientSymptoms -- KnownSymptoms),
                        C >= 2 andalso C <= 5
                    end ]

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