Add function arguments in iteration

Question

I have a list which has the number of n.

list={x1,x2,x3,x4,x5,...xn}

when I use these elements as arguments of a function f[x], I want to express function with iteration which is similar to NestList. According to the value of n, the arguments of the function are determined by the rule below.

Any ideas for this? Thanks.

Answer 1

ClearAll[fold1, folds2, fold3, fold4]

fold1  = List @@ Partition[# @@ #2, Length@#2, 2, -1, {}] &;

fold2[f_, l_] := f[## & @@ l[[;; #]]] & /@ Range[1, Length@l, 2]

fold3 = Extract[#2, List /@ Range[Range[1, Length@#2, 2]], Function[k, # @@ k]] &;

fold4 = # @@@ Extract[#2, List /@ Range[Range[1, Length @ #2, 2]]]&;

Examples:

fold1[f, {x1, x2, x3, x4, x5}]

{f[x1], f[x1, x2, x3], f[x1, x2, x3, x4, x5]}

fold1[f, {x1, x2, x3, x4, x5, x6, x7}]

{f[x1], f[x1, x2, x3], f[x1, x2, x3, x4, x5], f[x1, x2, x3, x4, x5, x6, x7]}

(fold1[f, #] ==  fold2[f, #] == fold3[f, #] == fold4[f, #])&[{x1, x2, x3, x4, x5, x6, x7}]

True

Answer 2

Another one:

list = {x1, x2, x3, x4, x5};
f @@@ FoldList[Flatten@*List, list[[{1}]], Rest@list][[;; ;; 2]]

And another:

Block[{f}
, SetAttributes[f, Flat]
; FoldList[f, f /@ list][[;; ;; 2]]
]

{f[x1], f[x1, x2, x3], f[x1, x2, x3, x4, x5]}

Answer 3

my original answer was unnecessarily complicated. A Table command is all what is needed

k = 4;(*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@Table[f[Sequence @@ list[[1 ;; (2 i - 1)]]], {i, Range[k]}]

Original

k = 4; (*must be <= 2*n-1 where n is length of list below*)
list = {x1, x2, x3, x4, x5, x6, x7};
Clear[f]
Flatten@MapIndexed[ {z = First[#2];
    f[Sequence @@ list[[1 ;; (2 z - 1)]]]} &, Range[k]]

Answer 4

Here we just need to specify n. The list of variables is built by Array, and then the mapping # generates the arguments of f. The List->Sequence removes extra brackets { } that would otherwise appear in the argument of f.

n = 4;
list = Array[x, 2 n - 1]; 
(f[list[[1 ;; 2 # - 1]]] /. List -> Sequence) & /@ Range[n]

{f[x[1]], f[x[1], x[2], x[3]], f[x[1], x[2], x[3], x[4], x[5]], 
     f[x[1], x[2], x[3], x[4], x[5], x[6], x[7]]}