将函数应用于列表中的每个元素到另一个列表中的每个元素

Question

我想要一种更优雅的方式来编写以下两个函数，理想情况下是一个：

applyOperatorBetweenVariableLists:: [Variable] -> String -> [Variable] -> [Variable]
applyOperatorBetweenVariableLists firstList operator secondList = concat $ map (test firstList operator) secondList

test:: [Variable] -> String -> Variable -> [Variable]
test firstVariables operator secondVariable = concat $ map (applyOperatorBetweenVariables secondVariable operator) firstVariables

applyOperatorBetweenVariables的声明是：

applyOperatorBetweenVariables:: Variable -> String -> Variable -> [Variable]

我很确定必须有一个Prelude函数可以做到这一点，或者是一种非常优雅的编写方式。

Answer 1

可以使用do块来简洁地完成此操作：

applyOperatorBetweenVariableLists firstList operator secondList = do
  secondVariable <- secondList
  firstVariable <- firstList
  applyOperatorBetweenVariables secondVariable operator firstVariable

如果您想更加简洁，可以将参数applyOperatorBetweenVariableLists和applyOperatorBetweenVariables重新排序，然后使用liftJoin2或bind2来实现（就像我下面的最后一句话一样，但用liftA2代替了。

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我最初的回答是错误的，因为它留下了一层嵌套（即应该做额外的concat或join）：

这几乎是{em> liftA2，但您的论据却很奇怪。在此方面，您将如何实现所写内容：

import Control.Applicative (liftA2)
applyOperatorBetweenVariableLists firstList operator secondList = liftA2 (flip applyOperatorBetweenVariables operator) secondList firstList

根据该定义，应该清楚如何通过重新排序applyOperatorBetweenVariableLists = liftA2 . applyOperatorBetweenVariables和applyOperatorBetweenVariables的参数来将其更改和简化为const schema = new Schema({ associations: [{ type: mongoose.SchemaTypes.Mixed, ref: doc => doc.modelType }] }); const Product = mongoose.model('Product', schema); const products = await Product.find().populate('associations'); 。