Question

我正在用Lua切割牙齿并试图实现一些列表处理逻辑。我不确定我是否真正掌握了协同程序和匿名函数的功能，但我正在努力。我知道这些东西在Ruby / Groovy / Javascript等其他脚本语言中是如何工作的，我想在Lua中做一些同样聪明的事情。以下是我提出的例子：

model = { { player = "Cliff", age = 35, gender = "male" }, { player = "Ally", age = 36, gender = "female" }, { player = "Jasmine", age = 13, gender = "female" }, { player = "Lauren", age = 6.5, gender = "female" } }

function allplayers()
    return coroutine.create(function()
        for idx, each in ipairs(model) do
            coroutine.yield(idx, each)
        end
    end)
end

function handlePlayers(source)
    local status, idx, each = coroutine.resume(source)
    while each do
        print(idx, each.player)
        status, idx, each = coroutine.resume(source)
    end
end

function having(source, predicate)
    return coroutine.create(function()
        local status, idx, each = coroutine.resume(source)
        while each do
            if predicate(each)  then
                coroutine.yield(idx, each)
            end
            status, idx, each = coroutine.resume(source)
        end
    end)
end

handlePlayers(having(allplayers(), function(each) return each.age < 30 end))

理想情况下，我希望能够编写如下代码：

allplayers(having(function(each) return each.age < 30 end))

甚至更好：

allplayers(having({each.age < 30 }))

产生相同的输出，但我无法理解如何或甚至可以做到这一点。对于所有迭代和循环以及所有迭代和循环，我上面所看到的东西似乎过于冗余。有没有更古老的方法来做到这一点？（男孩，我想念Groovy中的编码，因为总有一种更加时髦的方式来做某事......）

Answer 1

如果您不需要重用过滤结果，请考虑使用迭代器：

model = {
  { player = "Cliff", age = 35, gender = "male" },
  { player = "Ally", age = 36, gender = "female" },
  { player = "Jasmine", age = 13, gender = "female" },
  { player = "Lauren", age = 6.5, gender = "female" }
}

function model:having(predicate)
    local index = 0 
    return function()
        while true do
            index = index + 1
            if index > #self then break end
            if predicate(self[index]) then return self[index] end
        end
    end
end

for item in model:having(function(m) return m.age < 30 end) do
    print(item.player)
end

Answer 2

您可能希望在Underscore.lua JavaScript库之后模拟Underscore.js。

您的示例如下所示：

_ = require 'underscore'
model = {
  { player = "Cliff", age = 35, gender = "male" },
  { player = "Ally", age = 36, gender = "female" },
  { player = "Jasmine", age = 13, gender = "female" },
  { player = "Lauren", age = 6.5, gender = "female" }
}
result1 = _.select(model, function(p) return p.age < 30 end) -- traditional way
result2 = _(model):select(function(p) return p.age < 30 end) -- object-oriented style, calls can be chained
assert(_.is_equal(result1, result2))

请注意，Lua不会为您提供任何其他语法糖（除了:运算符）。如果你真的想要更短的语法，你应该看看Metalua，它允许你改变语言，并提供开箱即用的以下短函数语法：

result = _.select(model, |p| p.age < 30)

Answer 3

这令人难以置信的过于复杂和过度设计。您想通过谓词过滤一个列表。那样做;这是一个简单的循环。写下来吧。

即使你绝对必须以功能方式执行此操作（Lua不是一种功能语言。它可以这样工作，但它不起作用），协同程序不合适。

观察：

model = {
  { player = "Cliff", age = 35, gender = "male" },
  { player = "Ally", age = 36, gender = "female" },
  { player = "Jasmine", age = 13, gender = "female" },
  { player = "Lauren", age = 6.5, gender = "female" }
}

function for_each_array(list, operation)
  for key, value in ipairs(list) do
    operation(value)
  end
end

function filter_if(list, predicate)
  return function(value)
    if(predicate(value)) then
      list[#list + 1] = value
    end
  end
end

local list = {}
for_each_array(model, filter_if(list, function(each) return each.age < 30 end))

for_each_array(list, function(each) print(each.player) end)

请参阅？根本不需要协同程序。

Lua DSL还是聪明的闭包？

3 个答案: