Lua DSL还是巧妙的闭包？

这太过于复杂和设计过度了。你想通过一个谓词来过滤一个列表。那就这样做；这是一个简单的循环。只需要写出来。

即使你确实必须用函数式风格来完成这个任务（Lua 不是一个函数式语言。它可以用这种方式工作，但它并不是函数式的），协同程序也不是合适的。

请看：

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)

看到了吗？完全不需要协同程序。

你可能想尝试基于 Underscore.js JavaScript 库的 Underscore.lua。

你的示例代码如下：

_ = 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) -- 传统方式
result2 = _(model):select(function(p) return p.age < 30 end) -- 面向对象风格，可以链接调用
assert(_.is_equal(result1, result2))

需要注意的是，Lua 没有其他语法糖（除了 : 运算符）。如果你真的想要更短的语法，可以看看 Metalua，它允许你改变语言，并提供以下默认的短函数语法：

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

如果不需要重复使用过滤结果，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