Algorithm 我如何在结构上深入比较可能包含循环引用的2个lua表,其中表的键本身可能是表?
此问题类似于先前发布的问题 问题是,那里的解决方案对于简单的深度比较非常有效。但是,它不能正确处理循环引用。更具体地说,以下是:Algorithm 我如何在结构上深入比较可能包含循环引用的2个lua表,其中表的键本身可能是表?,algorithm,lua,compare,structural-equality,Algorithm,Lua,Compare,Structural Equality,此问题类似于先前发布的问题 问题是,那里的解决方案对于简单的深度比较非常有效。但是,它不能正确处理循环引用。更具体地说,以下是: function table_eq(table1, table2) local avoid_loops = {} local function recurse(t1, t2) -- compare value types if type(t1) ~= type(t2) then return false end -- B
function table_eq(table1, table2)
local avoid_loops = {}
local function recurse(t1, t2)
-- compare value types
if type(t1) ~= type(t2) then return false end
-- Base case: compare simple values
if type(t1) ~= "table" then return t1 == t2 end
-- Now, on to tables.
-- First, let's avoid looping forever.
if avoid_loops[t1] then return avoid_loops[t1] == t2 end
avoid_loops[t1] = t2
-- Copy keys from t2
local t2keys = {}
local t2tablekeys = {}
for k, _ in pairs(t2) do
if type(k) == "table" then table.insert(t2tablekeys, k) end
t2keys[k] = true
end
-- Let's iterate keys from t1
for k1, v1 in pairs(t1) do
local v2 = t2[k1]
if type(k1) == "table" then
-- if key is a table, we need to find an equivalent one.
local ok = false
for i, tk in ipairs(t2tablekeys) do
if table_eq(k1, tk) and recurse(v1, t2[tk]) then
table.remove(t2tablekeys, i)
t2keys[tk] = nil
ok = true
break
end
end
if not ok then return false end
else
-- t1 has a key which t2 doesn't have, fail.
if v2 == nil then return false end
t2keys[k1] = nil
if not recurse(v1, v2) then return false end
end
end
-- if t2 has a key which t1 doesn't have, fail.
if next(t2keys) then return false end
return true
end
return recurse(table1, table2)
end
local t1 = {}
t1[t1]=t1
t1.x = {[t1] = {1, 2, 3}}
local t2 = {}
local t3 = {}
t2[t3]=t2
t3[t2]=t3
t2.x = {[t3] = {1, 2, 3}}
t3.x = {[t2] = {1, 2, 3}}
print(table_eq(t1, t2))
--[[>
lua: deeptest.lua:15: stack overflow
stack traceback:
deeptest.lua:15: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
...
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:26: in function <deeptest.lua:3>
(...tail calls...)
deeptest.lua:62: in main chunk
[C]: in ?
--]]
结构上与下表不同:
local t = {}
t[{}] = 1
t["1"] = {}
local t = {}
local t2 = {}
t[t2] = 1
t["1"] = t2
而在“内容平等”中,他们是平等的
测试用例:
local t1 = {}
t1[t1]=t1
t1.x = {[t1] = {1, 2, 3}}
local t2 = {}
local t3 = {}
t2[t3]=t2
t3[t2]=t3
t2.x = {[t3] = {1, 2, 3}}
t3.x = {[t2] = {1, 2, 3}}
assert(table_eq(t1, t2) == false)
assert(table_eq(t2, t3) == true)
local t4 = {}
t4[{}] = 1
t4["1"] = {}
local t5 = {}
local t6 = {}
t5[t6] = 1
t5["1"] = t6
assert(table_eq(t4, t5) == false)
在本例中,通过更改以下内容,可以轻松修复堆栈溢出:
-- if key is a table, we need to find an equivalent one.
local ok = false
for i, tk in ipairs(t2tablekeys) do
if table_eq(k1, tk) and recurse(v1, t2[tk]) then
致:
前两个测试用例返回true,但第三个测试用例失败,因为表作为键比较内容与内容。如果要测试键是否与实际表相同,则需要将其更改为:
-- if key is a table, we need to find an equivalent one.
local ok = false
for i, tk in ipairs(t2tablekeys) do
if k1 == tk and recurse(v1, t2[tk]) then
但请记住,第二个测试用例将失败,因为您希望表作为键来比较内容与内容,而不是实际的表
你的测试案例互相矛盾,你认为“平等”,所以没有真正的答案。
附言
此函数忽略元数据,这也可以与_eq元方法进行比较。因此,无论如何,这仍然不是一个完整的比较。你能给出你对平等的定义吗?从你的例子中,我不明白什么应该是表eq(t2,t3)的正确答案。也许可以选择将所有对象推到一个表中,然后断言它们都是浅相等的(以消除误报),然后进行深度比较?该解决方案可能仍然存在误报,但很容易固定到当前算法上。@EgorSkriptunoff
table_eq(t2,t3)
应该是。。。未定义。t2和t3指向同一对象的不同部分(即它们相互参照)。只有table_eq(t1,t2)
(即table_eq
与完全隔离的表)才是决定性的。仍在尝试理解预期的比较…一个浅显的比较就足以确定两个表具有相同的键值对。似乎你想要某种结构上的比较;包括,也许所有空表都是等效的?@TomBlodget No。空表只有在系统中是相同的引用时才等效于空表。也就是说,给定两个空表t1和t2,table_eq(t1,t2)
应该是真的,但是table_eq({a=t1,b=t2},{a=t2,b=t2})
应该是假的,它们并不矛盾。只是你只看“内容”和“实例”。但我要找的实际上不是结构,而是“结构”。Lua表没有任何结构,实际上就是这样。所以这是不可能的。你可以用一些东西来“排序”这些表,并尝试用这种方式进行结构比较。你说Lua表没有结构是什么意思?它们是一组键值对。它们被称为数据结构是有原因的。如果你不想得到帮助,那没关系,别那么聪明。您只能以两种方式比较表,内容或地址。如果您想比较用于创建表的变量名,您应该了解它们是什么。因为你给他们不同的名字,但他们代表的是完全相同的东西。你也可以比较结构,它是内容和地址以一种非常具体的方式混合在一起。在同一个表中,比较地址,但跨表比较内容。正如我所说,请看这里:它显示了单元测试中一些表的结构。
-- if key is a table, we need to find an equivalent one.
local ok = false
for i, tk in ipairs(t2tablekeys) do
if k1 == tk and recurse(v1, t2[tk]) then