Head and everything except Head?

Question

I have been working on picking expressions apart using Head and Part and encountered a little mystery. Consider the canonical example

a + b + c

which has FullForm

Plus[a, b, c]

We expect, then, (a+b+c)[[0]] === Plus === Head[a+b+c] to be True, and it is. But then, we would expect (a+b+c)[[1;;3]] would be {a, b, c}, wouldn't we? But it isn't. The following is true:

(a + b + c)[[1;;3]] === (a + b + c)

Somehow, [[1;;3]], which is supposed to pick off elements 1 through 3 of its argument and put them in a List, doesn't get rid of the Head, which is element 0!

The questions are, then (and I will be grateful for hints and answers!)

1. Why doesn't (a + b + c)[[1;;3]] get rid of (a + b + c)[[0]], the Head?

2. What is the right way to get rid of the Head?

3. (a + b + c)[[0;;3]] produces 0. I would expect it to produce {Plus, a, b, c}. Instead, it produces 0. This just deepens the mystery for me! Why?

Answer 1

Point #1

Part always wraps element sequences with the original head of the expression.

expr = Hold[1 + 1, 2 + 2, 3 + 3, 4 + 4, 5 + 5];

expr[[{2, 3}]]

Hold[2 + 2, 3 + 3]

This applies at every level of the extraction. For example, compare these outputs:

expr = g[h[1, 2], i[3, 4]];

expr[[  2 , 1  ]]
expr[[ {2}, 1  ]]
expr[[  2 ,{1} ]]
expr[[ {2},{1} ]]

3

g[3]

i[3]

g[i[3]]

I used this nontrivially for Elegant manipulation of the variables list.

Point #2

Consider instead using Extract which wraps sequences in List:

Extract[expr, {{2}, {3}}]

{4, 6}

The third argument of Extract can be used to specify a function to apply to individual elements before they are evaluated:

Extract[expr, {{2}, {3}}, HoldForm]

{2 + 2, 3 + 3}

If you want all parts you can also use Level:

Level[(a + b + c), {1}, Heads -> True]

{Plus, a, b, c}

Or Cases:

Cases[(a + b + c), _, Heads -> True]

{Plus, a, b, c}

Or Replace/ReplaceAll:

(a + b + c) /. head_[body___] :> {head, body}

{Plus, a, b, c}

Point #3

The last point is more tricky and I had to check it myself. There is a behavior that I also did not expect:

Range[5][[0 ;; 5]]

{}

What I expected was an error as seen here:

Range[5][[0 ;; 4]]

and here:

Range[5][[0 ;; 6]]

When Span is used in Part[x, 0 ;; n] where n is the length of x, Part returns the head of the expression. Therefore (a + b + c)[[0;;3]] returns Plus[] and Plus[] evaluates to 0.

---

I believe Span behaves this way because of how it handles non-positive values, and zero-length spans. Consider:

Range[10][[-2 ;; 10]]

{9, 10}

You can see that it wraps around. Now consider:

Range[10][[5 ;; 4]]

{}

An empty span returns the head of the expression with no arguments.

Using 0 ;; n where n is the last element in the list, or 0 ;; -1, is also an empty span wrapping around the open end of the list.

Range[10][[0 ;; -1]]

{}

Answer 2

Perhaps only the developers (or Leonid) can answer #1 and #3 and "why".

The quick answer to #2 is Apply, remembering that List is a Head like any other:

List @@ (a + b + c)

{a, b, c}

On #3, I can't explain why ;; syntax gives 0 when

(a + b + c)[[Range[0, 3]]]

gives

a + b + c + Plus

So you could try

List @@ ((a + b + c)[[Range[0, 3]]])

{a, b, c, Plus}