Popping the first element off an array in Lua

Question

I have an array x in Lua. I would like to set head = x[1] and rest = the rest of the array, so that rest[1] = x[2], rest[2] = x[3], etc.

How can I do this?

(note: I don't care if the original array gets mutated. In Javascript I would do head = x.shift() and x would contain the remaining elements.)

Otha

Answer 1

head = table.remove(x, 1)

"Pop" is a bit of a misnomer, as it implies a cheap operation, and removing the first element of an table requires relocating the rest of the contents--hence the name "shift" in JavaScript and some other languages.

Hurlyburly

Note, for an array of any reasonable size, this is a VERY slow operation; try and reconsider why you want to do this... – Ramberg

@Ramberg you mean "array of any unreasonable size" B-) – Tomkin

Instead of reconsidering the motivation for this, instead consider perhaps using a data structure other than an array. A linked list (easy to create in Lua) makes this a cheap operation, at the expense of more storage needed for each node. – Saintjust

Answer 2

You want table.remove:

local t = {1,2,3,4}
local head = table.remove(t,1)
print( head )
--> 1
print( #t )
--> 3
print( t[1] )
--> 2

---

As @daurnimator points out, this requires a lot of effort by the underlying implementation of arrays in the Lua runtime, shifting all the table elements. If you can instead represent your arrays backwards, calling the last item in the array head, then the call to table.remove() will be a cheap pop:

local t = {4,3,2,1}
local head = table.remove(t)
print(head)
--> 1
print( #t )
--> 3
print( t[#t] )
--> 2

Alternatively, you may choose to represent your sequence of elements as a linked list. In this case, popping an item off the head of the list is also a cheap operation (but pushing one onto the end is not, unless you keep track of the 'tail' in your list):

local setm,getm = setmetatable,getmetatable
local linkedlist=setm({__index={
  tail = function(l) while l.rest do l=l.rest end return l end, -- N.B. O(n)!
  push = function(l,v,t) t=l:tail() t.rest=setm({val=v},getm(l)) return t end,
  cram = function(l,v) return setm({val=v,rest=l},getm(l)) end,
  each = function(l,v)
    return function() if l then v,l=l.val,l.rest return v end end
  end
}},{ __call=function(lmeta,v,...)
  local head,tail=setm({val=v},lmeta) tail=head
  for i,v in ipairs{...} do tail=tail:push(v) end
  return head
end })

local numbers = linkedlist(1,2,3,4)
for n in numbers:each() do print(n) end
--> 1
--> 2
--> 3
--> 4

local head,rest = numbers.val, numbers.rest
print(head)
--> 1

for n in rest:each() do print(n) end
--> 2
--> 3
--> 4

local unrest = rest:cram('99')
for n in unrest:each() do print(n) end
--> 99
--> 2
--> 3
--> 4

Note in particular that

local head,rest = numbers.val, numbers.rest

does not modify any data structures but just gives you a rest handle on a particular link in the chain.

Saintjust

Answer 3

Normally in Lua the action of inserting an element x into a sequence...

Eg: S={a,b,c,d,e,f} to S={a,b,c,x,d,e,f}

...is very time-consuming because d has to be moved to index 5, e to index 6 etc.

Is there some other sequence of the form S where S[a]=b, S[b]=c, S[c]=d, S[d]=e and S[e]=f? That way, all you have to do is type:

S[c]=x S[x]=d

and boom, x is after c and before d in just two operations.

Bilodeau

Recently found out that what I was talking about is called a "linked list". – Bilodeau