There is a list of lists list where lists are different length. Using one control I change a part of list to view, and with another one — what certain part of part to show at the end. Thought, Length could define here number of values in chosen part of list.

list = {{1, 2, 3, 4}, {1, 2, 3}}; (* for example *)
Manipulate[list[[j, k]], {j, 1, 2, 1}, {k, 1, Length[list[[j]]], 1}]


When control of j-th part switches to part of list with k-values > then it was before, it leads to a mistake, though when switching in opposite direction slider disposes itself correctly:

Mistake in an action


Do I desire that from Length in vain, and what is the solution of this problem? Assume, I can somehow set a value of k to 1, when j changes. How can I achieve this?

Thanks. I apologize for that the title not corresponding to an issue fully. Had no idea how to name it.


You can reset k whenever j changes if you customize the controller for it a little. Use the second argument to Dynamic to set k to 1 (or anything else, for that matter):

list = {{1, 2, 3, 4}, {1, 2, 3}}; (*for example*)
 list[[j, k]],
 {j, 1, 2, 1, Manipulator[Dynamic[j, (k = 1; j = #) &], ##2] &},
 {k, 1, Length[list[[j]]], 1}]


An alternative is to completely specify the control:

{j, 1, 2, 1, Manipulator[Dynamic[j, (k = 1; j = #) &], {1, 2, 1}] &}


{{j, 1}, Manipulator[Dynamic[j, (k = 1; j = #) &], {1, 2, 1}] &}

The difference is minimal (it affects Autorun). The first is exactly equivalent to original answer, except that to alter the domain 1, 2, 1, there are two places to edit.

  • $\begingroup$ Thanks! Your answer fits great. I vaguely understand meaning of # in Dynamic[j, (k = 1; j = #) &], but can't get sense of ##2 then. As I haven't completely grasped pure functions, can you explain that, please? $\endgroup$ – mikeonly Aug 6 '13 at 21:27
  • $\begingroup$ # stands for the first argument (in the Function in Dynamic) and ##2 stands for arguments 2 and above (in the Function Manipulator[..] &). See this answer for links to the funny symbols, and all the tutorials at the bottom of the Manipulate page for the rest, esp. this part $\endgroup$ – Michael E2 Aug 6 '13 at 21:40
  • $\begingroup$ @mikeonly I added a couple of alternatives. The first shows what ##2 becomes in this case, namely the single argument {1, 2, 1}. $\endgroup$ – Michael E2 Aug 7 '13 at 11:51

In this specific case, I prefer to keep both sliders' positions but instead of using directly their values, we can check if they do make sense and use a "reasonable guess" when they don't.

So, if k>Length[list[[j]]] instead of changing k into 1, we can use Length[list[[j]]]. I like the collateral effect of not losing the value of k in case we change j back into a value compatible with it. And I prefer to choose the last element when the index is greater than the length of the list than to choose the first.

This way, my 2 cents guess into your problem would be the following:

list = {{3, 5, 1, 7}, {8, 6}, {0}, {2, 9, 4}};
  Style[list[[j, Min[k, Length[list[[j]]]]]], 
      If[k > Length[list[[j]]], Red, "Output"]],
  {j, 1, Length[list], 1},
  {k, 1, Max[Length /@ list], 1}

As I am not using directly the k value chosen by the user, I prefer to change the color but this is only an idea.

  • $\begingroup$ So..! I found more one Brazilian guy using MMA. Nice! $\endgroup$ – Murta Aug 6 '13 at 21:30
  • $\begingroup$ @Murta Don't worry, you're not alone... :) $\endgroup$ – rm -rf Aug 6 '13 at 21:59

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