• Zz (18/18) Dec 30 2015 Hi,
• Ilya Yaroshenko (17/36) Dec 30 2015 Hi,
• Zz (3/48) Dec 30 2015 Thanks

Zz <Zz nomail.com> writes:

Hi,

Just playing with ndslice and I couldn't figure how to get the 
following transformations.

given.

auto slicea = sliced(iota(6), 2, 3, 1);
foreach (item; slicea)
{
    writeln(item);
}

which gives.

[[0][1][2]]
[[3][4][5]]

what transformation should i do to get the following from slicea.

[[0][2][4]]
[[1][3][5]]

[[4][2][0]]
[[5][3][1]]

Zz

Ilya Yaroshenko <ilyayaroshenko gmail.com> writes:

On Wednesday, 30 December 2015 at 18:53:15 UTC, Zz wrote:
 Hi,

 Just playing with ndslice and I couldn't figure how to get the 
 following transformations.

 given.

 auto slicea = sliced(iota(6), 2, 3, 1);
 foreach (item; slicea)
 {
    writeln(item);
 }

 which gives.

 [[0][1][2]]
 [[3][4][5]]

 what transformation should i do to get the following from 
 slicea.

 [[0][2][4]]
 [[1][3][5]]

 [[4][2][0]]
 [[5][3][1]]

 Zz

Hi,

void main()
{
	auto slicea = sliced(iota(6), 2, 3, 1);
	auto sliceb = slicea.reshape(3, 2, 1).transposed!1;
	auto slicec = sliceb.reversed!1;

	writefln("%(%(%(%s%)\n%)\n\n%)", [slicea, sliceb, slicec]);
}

Output:
[0][1][2]
[3][4][5]

[0][2][4]
[1][3][5]

[4][2][0]
[5][3][1]

Ilya

Zz <Zz nomail.com> writes:

On Wednesday, 30 December 2015 at 20:43:21 UTC, Ilya Yaroshenko 
wrote:
 On Wednesday, 30 December 2015 at 18:53:15 UTC, Zz wrote:
 Hi,

 Just playing with ndslice and I couldn't figure how to get the 
 following transformations.

 given.

 auto slicea = sliced(iota(6), 2, 3, 1);
 foreach (item; slicea)
 {
    writeln(item);
 }

 which gives.

 [[0][1][2]]
 [[3][4][5]]

 what transformation should i do to get the following from 
 slicea.

 [[0][2][4]]
 [[1][3][5]]

 [[4][2][0]]
 [[5][3][1]]

 Zz

 Hi,

 void main()
 {
 	auto slicea = sliced(iota(6), 2, 3, 1);
 	auto sliceb = slicea.reshape(3, 2, 1).transposed!1;
 	auto slicec = sliceb.reversed!1;

 	writefln("%(%(%(%s%)\n%)\n\n%)", [slicea, sliceb, slicec]);
 }

 Output:
 [0][1][2]
 [3][4][5]

 [0][2][4]
 [1][3][5]

 [4][2][0]
 [5][3][1]

 Ilya

Thanks