How can I shift the list as a loop? The input:
local connections = {1, 1, 0, 1}
Needed result:
local variants = {
{1, 1, 0, 1}, -- as original table
{1, 1, 1, 0}, -- shifted once to the right
{0, 1, 1, 1}, -- shifted twice
{1, 0, 1, 1}, -- shifted three times
}
You want to perform a shift with "wraparound" / "circular" shift. [table.move], as pointed out by lhf, is almost what you need, but lacks the "circular part" and is unavailable on older versions of Lua (such as 5.1, which is still in widespread use). I thus propose implementing this in Lua as below, particularly if you want a new table:
local function cyclic_table_shift(tab, shift)
local len = #tab
local shifted = {}
for i = 1, len do
shifted[i] = tab[(i - 1 - shift) % len + 1] -- mod to make it wrap
end
return shifted
end
this yields the correct results for your example:
> connections = {1, 1, 0, 1}
> function cyclic_table_shift(tab, shift)
>> local len = #tab
>> local shifted = {}
>> for i = 1, len do
>> shifted[i] = tab[(i - 1 - shift) % len + 1] -- mod to make it wrap
>> end
>> return shifted
>> end
> table.unpack(cyclic_table_shift(connections, 0))
1 1 0 1
> table.unpack(cyclic_table_shift(connections, 1))
1 1 1 0
> table.unpack(cyclic_table_shift(connections, 2))
0 1 1 1
> table.unpack(cyclic_table_shift(connections, 3))
1 0 1 1
For Lua 5.3+, you can use table.move to do most of the work:
function shift(a)
local n=#a
local t=a[n]
table.move(a,1,n-1,2)
a[1]=t
end
Thanks all, there is the solution for Lua 5.1 (and 5.4)
if not table.move then
print ('used custom table.move')
-- thanks to index five
function table.move(a1,f,e,t,a2) -- a1, f, e, t [,a2]
-- Moves elements from the table a1 to the table a2,
-- performing the equivalent to the following multiple assignment:
-- a2[t],··· = a1[f],···,a1[e]. The default for a2 is a1.
-- The destination range can overlap with the source range.
-- The number of elements to be moved must fit in a Lua integer.
a2 = a2 or a1
if (a2 ~= a1) or (t < f) then -- use a2
for i = f, e do
a2[t+i-f] = a1[i]
end
elseif (t > f) then
for i = e, f, -1 do
a2[t+i-f] = a1[i]
end
end
return a2
end
end
table.unpack = table.unpack or unpack
function circularShift (tabl, shift)
local len = #tabl
local shifted = {}
table.move(tabl,len-shift,len,0,shifted)
table.move(tabl,1,len-shift,shift+1,shifted)
return shifted
end
local connections = {1, 1, 0, 1}
print (table.unpack(circularShift(connections, 0)))
print (table.unpack(circularShift(connections, 1)))
print (table.unpack(circularShift(connections, 2)))
print (table.unpack(circularShift(connections, 3)))
Result:
1 1 0 1
1 1 1 0
0 1 1 1
1 0 1 1
But the version of LMD is simpler and can accept left shifting too.
table.insert(connections, 1, table.remove(connections))and was not sure that it was a good solution