无法操作 Lua 表

Question

假设我有两个 tables:

veggie_multiples = {
{veggie = "carrot", quantity = 1},
{veggie = "tomato", quantity = 2},
{veggie = "celery", quantity = 3}}

veggie_singles = {
{veggie = "celery"},
{veggie = "carrot"},
{veggie = "potato"},
{veggie = "carrot"},
{veggie = "potato"}}

我想以一个 table 结束，它代表：

veggie_multiples = {
{veggie = "carrot", quantity = 3},
{veggie = "tomato", quantity = 2},
{veggie = "celery", quantity = 4},
{veggie = "potato", quantity = 2}}

我试过类似的方法：

veggie_multiples = {
{veggie = "carrot", quantity = 1},
{veggie = "tomato", quantity = 2},
{veggie = "celery", quantity = 3}}

veggie_singles = {
{veggie = "celery"},
{veggie = "carrot"},
{veggie = "potato"},
{veggie = "carrot"},
{veggie = "potato"}}

for i, n in ipairs(veggie_singles) do
    for ii, nn in ipairs(veggie_multiples) do
        if veggie_singles[i].veggie == veggie_multiples[ii].veggie then
            veggie_multiples[ii].quantity = veggie_multiples[ii].quantity + 1
            table.remove(veggie_singles, i)
        else
            table.insert(veggie_multiples, {veggie = veggie_singles[i], quantity = 1})
            table.remove(veggie_singles, i)
        end
    end
end

for i, n in ipairs(veggie_multiples) do
    print(veggie_multiples[i].veggie, " ", veggie_multiples[i].quantity)
end

无论我怎样尝试，它都无法正常工作。请帮忙！谢谢。

Answer 1

这个有效：

for i, n in ipairs(veggie_singles) do
    local existed = false
    for ii, nn in ipairs(veggie_multiples) do
        if veggie_singles[i].veggie == veggie_multiples[ii].veggie then
            veggie_multiples[ii].quantity = veggie_multiples[ii].quantity + 1
            existed = true
            break
        end
    end
    if not existed then
        table.insert(veggie_multiples, {veggie = veggie_singles[i].veggie, quantity = 1})
    end
end

仅当 veggie_singles[i].veggie 中的项目不等于 veggie_multiples 中的 all 项时，才会插入新项目。这就是标志 existed 的作用。

Answer 2

使用 ipairs 的 for 循环遍历索引和值 for i, n in ipairs(veggie_singles) 将在第一次迭代中给出 i=1 and n={veggie="celery"}，依此类推。代码不需要使用 i 所以在 Lua 中你使用 _ 作为一个扔掉。然后在 veggie multiples 中搜索与 veggie single 同名的条目。找不到就添加，找到就增加数量。

for _, vs in pairs(veggie_singles) do
    local found = false
    for _, vm in pairs(veggie_multiples) do
       if vm.veggie == vs.veggie then
          vm.quantity = vm.quantity + 1
          found = true
          break
       end
    end
    if not found then
        table.insert(veggie_multiples, {veggie=vs.veggie, quantity=1}) 
    end
end


for i, n in ipairs(veggie_multiples) do
    print(veggie_multiples[i].veggie, " ", veggie_multiples[i].quantity)
end

由于索引对解决问题不是很有用，而名称是您要查找内容的依据，因此您可以使用名称作为表中的键来简化代码。

quantity = {carrot=1, tomato=2, celery=3}
add = {"celery", "carrot", "potato", "carrot", "potato"}

for _, v in pairs(add) do
   quantity[v] = (quantity[v] or 0) + 1
end

for veggie, qty in pairs(quantity) do
   print(veggie, qty)
end

输出：

potato  2
carrot  3
celery  4
tomato  2

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