Partition of the long list

Question

I am trying to do the partition of the list variable called test so that answer will look like the variable called goal. The test list is actually a list of long data. In this example I have a short version example according to the code below which also includes my attempts called ans1 and ans2 using method called Partition and ArrayShape:

 Clear[ans];
 goal={{{a,b,c},{d,e,f},{g,h,i},j},{{k,l,m},{n,o,p},{q,r,s},t}};
 TableForm[goal]

And the output should be

 a d g
 b e h j
 c f i

 k n q
 l o r t
 m p s

My attempt is:

 test={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t};
 ans1=Partition[test,3,3,1,Nothing]
 {{a,b,c},{d,e,f},{g,h,i},{j,k,l},{m,n,o},{p,q,r},{s,t}}
 TableForm[ans1]

 ans2=ArrayReshape[test,{3,3,3},1]
 {{{a,b,c},{d,e,f},{g,h,i}},{{j,k,l},{m,n,o},{p,q,r}},{{s,t,1},{1,1,1},{1,1,1}}}
 TableForm[ans2]

Answer 1

Edit(reply to comment)

For the pattern {{#,#,#},{#,#,#},{#,#,#},{#,#,#},#},there are 3*4+1 elements, so we use

test = Range[39];
FlattenAt[#, {-1}] & /@ (Partition[#, UpTo[3]] & /@ 
   Partition[test, UpTo[3*4 + 1]])

Original

test={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t};
FlattenAt[#, {-1}] & /@ (Partition[#, UpTo[3]] & /@ 
   Partition[test, UpTo[10]])

FlattenAt[#, {-1}] & /@ (Partition[#, UpTo[3]] & /@ 
   Partition[test, 10])

FlattenAt[#, {-1}] & /@ (Partition[#, 3, 3, 1, Nothing] & /@ 
   Partition[test, 10])

Answer 2

Assuming that the number of items in test is not divisible by the total number of elements in each partition, e.g.,

test = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u,
    v, w, x, y};

a possible solution using TakeList can be:

FlattenAt[#, {4}] & /@ (TakeList[#, UpTo /@ {3, 3, 3, 1}] & /@ 
   Partition[test, UpTo[10]])

{{{a, b, c}, {d, e, f}, {g, h, i}, 
  j}, {{k, l, m}, {n, o, p}, {q, r, s}, t}, {{u, v, w}, {x, y}, {}}}