#[[1]]//FullForm = 1?

Question

I have a set of symbolic algebraic expressions that I'm trying to get some speed into. To illustrate the issue, I'll use a simple form like { k0 X, k1 Y, k1 Z}, where each term is large and complicated. By random walk, I've arrived at a promising form

{ r0 X, r1 Y,r1 Z}/.r0->k0 /.r1->k1

which is faster than the original, but with my particular set of expressions, it turns out that a function is faster:

{#[[1]] X, #[[2]] Y, #[[2]] Z}&@{k0, k1}

The expressions are large, making substitutions by hand time consuming and error prone, so I set out to replace the k s with a rule like

{k0->#[[1]],k1->#[[2]]}

I was surprised that this wasn't a straightforward operation, k0 being replaced by 1 instead of the intended #[[1]]. I thrashed around for a bit, eventually finding the odd (to me) result that

{#[[1]], #[[2]]}//FullForm

Gives

List[ 1, Part[Slot[1],2]]

Can anyone suggest a way of thinking in which this makes sense? Is there any place in the design for Part[Slot[1],1] ? I worked around it with a hack:

{#[[2]] X, #[[3]] Y, #[[3]] Z}&@{Null, k0, k1}

which works but is ugly to say nothing of embarrassing.

Thanks in advance.

Answer 1

This seems very unclear. Have you tried Apply? It's better (more readable) for working with elements of lists in pure functions.

{#1 X, #2 Y, #2 Z}&@@{k0, k1}

Note that the Slot[1][[1]]//FullForm == 1 peculiarity arises from the fact that the 1 is the first expression inside Slot. You'll notice that, e.g., f[1][[1]] returns 1, and that Slot[2][[1]] returns 2.

While I encourage you to use Apply (which is @@) for clarity, another solution does exist. By assigning a symbol to the pure function rather than a numbered slot (which evaluates to 1 with Part), you can get around the premature evaluation:

Function[a,{a[[1]] X, a[[2]] Y, a[[2]] Z}]@{k0, k1}

This is functionally equivalent to what your desired result is, but, again, less clear than Apply.