How can I manipulate a simple list of vectors?

Question

I'd like to remove the first element from each vector and adding it to the beginning of each subsequent element within the vector so that:

{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4, 15}, {9, 21}}

Becomes:

{{1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9, 21}}

I'd like to use >select< with the condition that the length is greater than 1 to get rid of the first elements after I have flattened the vector lists to the first level. However, how would I alter the vectors using the join command?

rogerRoger = {{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4,
     15}, {9, 21}}};

Select[Flatten[Join[], 1], Length@# > 1 &]

Answer 1

{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4, 15}, {9, 21}}

Becomes:

{{1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9, 21}}

Can you not use If ?

lis ={{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4, 15}, {9,21}}};
If[Length[#[[1]]] == 1, z = #[[1]]; Join[z, #] & /@ Rest[#], #] & /@ lis
Flatten[%, 1]

with the condition that length is greater than 1

I am not sure I understand why the above? Length of what?

You want to join the first list to the rest if the first list had length 1, right? That is what you example does.

So if first rest had length say 2, (greater than 1) it will not touch that list and will leave it as is. This is what I understood you wanted.

lis={{{1,3},{2,5},{5,10},{10,23}},{{2},{3,35},{4,15},{9,21}}};
If[Length[#[[1]]]==1,z=#[[1]];Join[z,#]&/@Rest[#],#]&/@lis
Flatten[%,1]

Answer 2

Cases can be used.

list = {{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4, 15}, {9, 21}}};

Flatten[Cases[list, {x_, y__} :> (Join[x, #] & /@ {y})], 1]

(* {1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9, 21}} *)

Answer 3

lst = {{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4, 15}, {9, 21}}};

Join @@ Map[Flatten /@ Thread[{#[[1, 1]], Rest @ #}] &] @ lst

{{1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9, 21}}

Also

Join @@ (Prepend[#[[1]]] /@ {##2} & @@@ lst)

Join @@ (Prepend @@@ Tuples[{{##2}, #}] & @@@ lst)

Flatten[Join @@@ Tuples[{{#}, {##2}}] & @@@ lst, 1]

Distribute[{{#}, {##2}}, List, List, Sequence, Join] & @@@ lst

all give

{{1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9, 21}}

And ... for fun:

☺☺☺ = ## & @@ (♯ |-> ## & @@@ {#, ♯}) /@ {##2} & @@@ # &;

☺☺☺ @ lst

{{1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9,    21}}

Answer 4

list = {{{1}, {2, 5}, {5, 10}, {10, 23}}, {{2}, {3, 35}, {4, 15}, {9, 21}}};

Define a helper function,

join[a_List, b_List] := Join[a, #] & /@ b

Then,

list2 = Flatten[join[#[[1]], Rest@#] & /@ list, 1]

(* {1, 2, 5}, {1, 5, 10}, {1, 10, 23}, {2, 3, 35}, {2, 4, 15}, {2, 9, 21}} *)