Question

如何在不更改原始表的情况下将包含少量子表的Lua表拆分为两个表。

e.g。将tbl = {{tbl1}, {tbl2}, {tbl3}, {tbl4}}分成subtbl1 = {{tbl1}, {tbl2}}，subtbl2 = {{tbl3}, {tbl4}}，同时保持tbl不变。

字符串有string.sub，但不知道表是否有类似的东西。我不认为unpack适用于我的案例，table.remove也会更改原始tbl。

为我的实际案例添加更多信息：

tbl在运行时填充了子表，子表的数量也发生了变化。我想保留前两个子表，并将其余的子表（在一个表中）传递给函数。

Answer 1

您可以使用建议的lhf方法保留前两个子表。然后，您可以unpack剩下的子表。

local unpack = table.unpack or unpack

local t = { {1}, {2}, {3}, {4}, {5}, {6} }

local t1 = { t[1], t[2] }    -- keep the first two subtables
local t2 = { unpack(t, 3) }  -- unpack the rest into a new table

-- check that t has been split into two sub-tables leaving the original unchanged
assert(#t == 6, 't has been modified')

-- t1 contains the first two sub-tables of t
assert(#t1 == 2, 'invalid table1 length')
assert(t[1] == t1[1], 'table1 mismatch at index 1')
assert(t[2] == t1[2], 'table1 mismatch at index 2')

-- t2 contains the remaining sub-tables in t
assert(#t2 == 4, 'invalid table2 length')
assert(t[3] == t2[1], 'table2 mismatch at index 1')
assert(t[4] == t2[2], 'table2 mismatch at index 2')
assert(t[5] == t2[3], 'table2 mismatch at index 3')
assert(t[6] == t2[4], 'table2 mismatch at index 4')

Answer 2

试试这个：

subtbl1 = { tbl[1], tbl[2] }
subtbl2 = { tbl[3], tbl[4] }

如何拆分包含子表的Lua表

2 个答案: