How could I generate the following list of points using some function or mapping without using the table function.
x = (1 - Cos[# Pi])/2 & /@ Range[0, 1, 0.1];
y = (1 - Cos[# Pi])/2 & /@ Range[0, 1, 0.1];
nx = Length[x];
ny = Length[y];
nodesElement =
Flatten[Table[{{x[[i]], y[[j]]}, {x[[i + 1]], y[[j]]}, {x[[i + 1]],
y[[j + 1]]}, {x[[i]], y[[j + 1]]}}, {j, 1, ny - 1}, {i, 1,
nx - 1}], 1];
I am interested in more pure functional programming
How about this one: Tuples /@ Tuples[Partition[x, 2, 1], 2]
Or in the case $x\neq y$: Tuples /@ Tuples[{Partition[x, 2, 1], Partition[y, 2, 1]}]
Note that this gives the same thing as
Flatten[Table[{{x[[i]], y[[j]]}, {x[[i]], y[[j+1]]}, {x[[i+1]],
y[[j]]}, {x[[i+1]], y[[j+1]]}}, {i, 1, ny - 1}, {j, 1, nx - 1}], 1]
Which is just a reordering of what you posted.
Using RotateLeft and MapThread:
xv = Indexed[x, #] & /@ Range[5]
yv = Indexed[y, #] & /@ Range[5]
MapThread[{{#1, #2}, {#3, #2}, {#3, #4}, {#1, #4}} &, {xv, yv,
RotateLeft[xv], RotateLeft[yv]}]
That wraps around, you can use Most for a finite set of quads.
Flatten[Outer[{y, x} |-> {{x[[1]], y[[1]]}, {x[[2]], y[[1]]}, {x[[2]], y[[2]]}, {x[[1]], y[[2]]}}, Partition[y, 2, 1], Partition[x, 2, 1], 1], 1]$\endgroup$Tuplesinstead of a custom function—or maybe there's a slicker way I'm not thinking of! $\endgroup$