How to merge two tables overwriting the elements which are in both?

Question

I need to merge two tables, with the contents of the second overwriting contents in the first if a given item is in both. I looked but the standard libraries don't seem to offer this. Where can I get such a function?

Answer 1

for k,v in pairs(second_table) do first_table[k] = v end

Answer 2

Here's what i came up with based on Doug Currie's answer:

function tableMerge(t1, t2)
    for k,v in pairs(t2) do
        if type(v) == "table" then
            if type(t1[k] or false) == "table" then
                tableMerge(t1[k] or {}, t2[k] or {})
            else
                t1[k] = v
            end
        else
            t1[k] = v
        end
    end
    return t1
end

Answer 3

Wouldn't this work properly?

function merge(t1, t2)
    for k, v in pairs(t2) do
        if (type(v) == "table") and (type(t1[k] or false) == "table") then
            merge(t1[k], t2[k])
        else
            t1[k] = v
        end
    end
    return t1
end

Answer 4

For numeric-index table merging:

for k,v in pairs(secondTable) do table.insert(firstTable, v) end

Answer 5

Doug Currie's answer is the simplest for most cases. If you need more robust merging of tables, consider using the merge() method from the Penlight library.

require 'pl'
pretty.dump(tablex.merge({a=1,b=2}, {c=3,d=4}, true))

-- {
--   a = 1,
--   d = 4,
--   c = 3,
--   b = 2
-- }

Answer 6

Here's iterative version for deep merge because I don't like potential stack overflows of recursive.

local merge_task = {}
function merge_to_left_o(orig, new)
   merge_task[orig] = new

   local left = orig
   while left ~= nil do
      local right = merge_task[left]
      for new_key, new_val in pairs(right) do
         local old_val = left[new_key]
         if old_val == nil then
            left[new_key] = new_val
         else
            local old_type = type(old_val)
            local new_type = type(new_val)
            if (old_type == "table" and new_type == "table") then
               merge_task[old_val] = new_val
            else
               left[new_key] = new_val
            end
         end
      end
      merge_task[left] = nil
      left = next(merge_task)
   end
end

Answer 7

I preferred James version for its simplicity and use it in my utils.lua - i did add a check for table type for error handling.

function merge(a, b)
    if type(a) == 'table' and type(b) == 'table' then
        for k,v in pairs(b) do if type(v)=='table' and type(a[k] or false)=='table' then merge(a[k],v) else a[k]=v end end
    end
    return a
end

Thanks for this nice function which should be part of the table class so you could call a:merge(b) but doing table.merge = function(a, b) ... did not work for me. Could even be compressed to a one liner for the real nerds :)

Answer 8

Like Doug Currie said, you can use his function, but there is a problem with his method. If first_table has things in it's k index, the function will over write it.

I'm assuming you're trying to merge these tables, not overwrite index's and value's. So this would be my method, it's very similar but is used for merging tables.

for _, v in pairs(second_table) do table.insert(first_table, v) end

The only problem with this solution is that the index is set as numbers, not as strings. This will work with tables with numbers as the index, and for tables with strings as their index, use Doug Currie's method.

Doug Currie's method:

for k,v in pairs(second_table) do first_table[k] = v end

Answer 9

Extending this great answer, https://stackoverflow.com/a/1283399/1570165, I would like to go with a (pure) functional approach like this one below:

-- example values
local t1 = { a = 0, b = 2 }
local t2 = { a = 1, c = 3 }

-- merge function that takes functional approach

local merge = function(a, b)
    local c = {}
    for k,v in pairs(a) do c[k] = v end
    for k,v in pairs(b) do c[k] = v end
    return c
end

-- t1 and t2 value still same after merge
print(merge(t1, t2)) -- { a = 1, b = 2, c = 3 }
print(t2) -- { a = 1, c = 3 }
print(t1) -- { a = 0, b = 2 }

Answer 10

for k,v in pairs(t2) do t1[k] = v end

key for string solution