Select certain items from nested list

Question

I have a list as follows:

lis1={{{a1,b1,c1,d1,e1},{a2,b2,c2,d2,e2}},
{{a3,b3,c3,d3,e3},{a4,b4,c4,d4,e4}}}

and would like to delete certain elements from it resp. pick certain elements to get a list as follows

lis2= {{{a1,c1,e1},{a2,c2,e2}},
    {{a3,c3,e3},{a4,c4,e4}}}

My problem is that the data I would like to select are not in an consecutive order (otherwise I could use the Part function). Does anyone have an hint?

Answer 1

You actually can still use the Part function!

If you know the positions explicitly you can try:

lis1[[All, All, {1, 3, 5}]]

Or, if they follow a regular pattern:

lis1[[All, All, 1 ;; 5 ;; 2]]

Answer 2

Just learned from this answer by ciao that Downsample is made for this kind of task:

Downsample[#, {1, 2}] & /@ lis1
(* {{{a1, c1, e1}, {a2, c2, e2}}, {{a3, c3, e3}, {a4, c4, e4}}} *)

You can also use Take, Drop and Extract to get the same result:

Take[#, All, {1, -1, 2}] & /@ lis1
Drop[#, {}, {2, -1, 2}] & /@ lis1
First@Rest@Extract[lis1, {{0}, {;; , ;; , ;; ;; 2}}]

You can use arbitrary lists for Part specifications with Part and Extract. For example

#[[All, ;; , {1, 4, 5}]] &@lis1  (* or *)
First@Rest@Extract[lis1, {{0}, {All, ;; , {1, 4, 5}}}]

both give all rows and columns 1, 4 and 5 in all matrices in lis1:

(* {{{a1, d1, e1}, {a2, d2, e2}}, {{a3, d3, e3}, {a4, d4, e4}}} *)

Answer 3

lis1 /. b1|d1|b2|d2|b3|d3|b4|d4->Sequence[]

Answer 4

lis1 = {{{a1, b1, c1, d1, e1}, {a2, b2, c2, d2, e2}}, {{a3, b3, c3, 
    d3, e3}, {a4, b4, c4, d4, e4}}};

MapAt[Nothing, {All, All, {2, 4}}] @ lis1

ReplaceAt[_ :> Nothing, {All, All, {2, 4}}] @ lis1

Both give the expected result:

{{{a1, c1, e1}, {a2, c2, e2}}, {{a3, c3, e3}, {a4, c4, e4}}}

ReplaceAt was introduced in V 13.1 and has the advantage that we can use the usual Condition syntax. Example:

ReplaceAt[lis1, x_ /; SymbolName[x] != "b1" :> Nothing, {All, All, {2, 4}}]

{{{a1, b1, c1, e1}, {a2, c2, e2}}, {{a3, c3, e3}, {a4, c4, e4}}}

Answer 5

Another way using ReplaceAll and Span:

(lis1 /. x_?VectorQ :> x[[1 ;; 5 ;; 2]]) === lis2

(*True*)

Or using ReplaceAll and Select:

(lis1 /. x_?VectorQ :> Select[x, OddQ[Det@Position[x, #]] &]) === lis2

 (*True*)

Thanks to @eldo for showing me the following shorter way:

  (lis1 /. x_?VectorQ :> x[[;; ;; 2]]) === lis2

  (*True*)