How can one nested list be used to find the positions of those values in another nested list?

Question

lista = {{0., 1., 0.1, 0.}, {0., 1., 0.2, 0.}, {0., 1., 0.3, 0.}};

listb = {
  {0., 1., 1., 0.}, {0., 0.85, 1., 0.}, {0., 1., 0.1, 0.}, 
  {0.,1., 0.2, 0.}, {0., 1., 0.3, 0.}, {0., 1., 0.4, 0.}, 
  {0., 1., 0.55, 0.}, {0., 1., 0.7, 0.}, {0., 1., 0.85, 0.}, 
  {0.4, 0.4, 0.2, 0.1}, {0.4, 0.4, 0.4, 0.1}, {0.4, 0.4, 0.7, 0.1}, 
  {0.4, 0.7, 0., 0.1}, {0.4, 0.7, 0.1, 0.1}, {0.4, 0.7, 0.2, 0.1}, 
  {0.4, 0.7, 0.4, 0.1}, {0.4, 0.7, 0.7, 0.1}, {0.5, 0.4, 0.4, 0.}, 
  {1., 0., 0., 0.1}, {0., 1., 0., 0.1}, {0., 0., 1., 0.1}, 
  {0.,1., 1., 0.1}, {1., 0., 1., 0.1}, {1., 1., 0., 0.1}};

How would one go about finding the first positions of lista sublists in listb?

Answer 1

How about mapping Position like so

Position[listb, #] & /@ lista
(*{{{3}}, {{4}}, {{5}}, {{6}}, {{7}}, {{8}}, {{9}}, {{1}}, {{2}}}*)

which gives the position of each sublist of lista in listb?

To get only the first instance, use

Position[listb, #, 1, 1] & /@ lista

(the last argument of Position specifies how many positions found to return; the penultimate argument specifies the level at which to search).

Answer 2

Position[listb, Alternatives@@ lista]