4
$\begingroup$

I have many lists that I would like to RotateLeft, but i want the last element to be 0 (rather than the initial first element).

My starting list is:

listA1 = {1, 2, 3}
listA2 = {2, 3, 4}
listB1 = {4, 5, 6}
listB2 = {3, 5, 7}
listC1 = {1, 2, 6}
listC2 = {5, 4, 3}
nestedList = {{listA1, listA2}, {listB1, listB2}, {listC1, listC2}}

{{{1, 2, 3}, {2, 3, 4}}, {{4, 5, 6}, {3, 5, 7}}, {{1, 2, 6}, {5, 4, 3}}}

And my desired output is:

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

Using RotateLeft, i get close to the result:

currentOutput = RotateLeft[nestedList, {0, 0, 1}]

{{{2, 3, 1}, {3, 4, 2}}, {{5, 6, 4}, {5, 7, 3}}, {{2, 6, 1}, {4, 3, 5}}}

But the last element in each list is the first element.

Is there a very simple way to either extend my current code or alternatively do it all in one simple step?

$\endgroup$
1
  • 3
    $\begingroup$ You could use ReplacePart[]: ReplacePart[RotateLeft[{{{1, 2, 3}, {2, 3, 4}}, {{4, 5, 6}, {3, 5, 7}}, {{1, 2, 6}, {5, 4, 3}}}, {0, 0, 1}], {_, _, -1} -> 0] $\endgroup$ Nov 28, 2019 at 23:28

5 Answers 5

9
$\begingroup$

You can use ArrayPad to get the desired output in a single step:

ArrayPad[nestedList, {{0}, {0}, {-1, 1}}]

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

Alternatively, you can use PadRight:

PadRight[nestedList[[All, All, 2 ;;]], {Automatic, Automatic, 3}]

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

You can also Apply the function {##2, 0} & to nestedList at level 2:

Apply[{##2, 0} &, nestedList, {2}]

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

And for fun:

☺ = {##2, 0} & @@@ # & /@ # &;
☺ @ nestedList

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

$\endgroup$
4
$\begingroup$

I prefer @kglr solution but as an alternative you can do this as a second step:

currentOutput[[All, All, 3]] = 0
$\endgroup$
2
$\begingroup$
(list = {{{1, 2, 3}, {2, 3, 4}}, {{4, 5, 6}, {3, 5, 7}}, {{1, 2, 
      6}, {5, 4, 3}}}) // Map[MatrixForm]


(res = Map[Last@Partition[#, 3, 2, {2, -2}, 0] &, list, {-2}]) // 
 Map[MatrixForm]

enter image description here

$\endgroup$
2
$\begingroup$
list = {{{1, 2, 3}, {2, 3, 4}}, {{4, 5, 6}, {3, 5, 7}}, {{1, 2, 6}, {5, 4, 3}}};

Replace[list, {_, a_, b_} :> {a, b, 0}, {2}]

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

Map[Append[0], list[[;;, ;;, 2;;]], {2}]

{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}

$\endgroup$
1
$\begingroup$
list = {{{1, 2, 3}, {2, 3, 4}}, {{4, 5, 6}, {3, 5, 7}}, {{1, 2, 6}, {5, 4, 3}}};

Using Threaded:

Map[# + Threaded[{0, 0, -Last@#}] &@RotateLeft[#] &, list, {2}]

(*{{{2, 3, 0}, {3, 4, 0}}, {{5, 6, 0}, {5, 7, 0}}, {{2, 6, 0}, {4, 3, 0}}}*)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.