5
$\begingroup$

My goal is to replace the zeros of one list with the (i-1)th element of a second list. For example, if

list1 = {0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0}

and

list2 = {6, 1, 4, 7, 2, 9, 10, 8, 11, 3, 5, 0, 12}

the desired output is {0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}. Note that the first element of the output is defined as 0 still.

My attempt to create a code for this is to first find the zeros of list1then run a for-loop for $i \in$zeroslist1.

zeroslist1 = Flatten[Position[list1, 0]]
 DeleteCases[
 Flatten[Reap@
   Do[Sow[ReplacePart[vtest1, 
      i -> vtest2[[i - 1]] & /@ zeroslist1]], {i, zeroslist1}], 2], Null]

The results of the output are:

{{List,1,1,0,0,1,1,0,1,0,0,1,0}, {0,1,1,4,0,1,1,0,1,0,0,1,0}, {0,1,1,0,7,1,1,0,1,0,0,1,0},{0,1,1,0,0,1,1,10,1,0,0,1,0}, {0,1,1,0,0,1,1,0,1,11,0,1,0},{0,1,1,0,0,1,1,0,1,0,3,1,0}, {0,1,1,0,0,1,1,0,1,0,0,1,0}}.

Either a cleaner way to code the desire output or a method of merging the output of my current for-loop to get the desired output would be great.

$\endgroup$
6
$\begingroup$

You can use a multiplication instead of looping or conditionals:

list1 + (1 - list1) Prepend[Most[list2], 0]

{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}

The central point is that 0 and 1 in list1 aren't just symbols but numeric quantities.

$\endgroup$
  • $\begingroup$ What a simple way to do this. Thank you so much. $\endgroup$ – Will Feb 11 at 17:51
4
$\begingroup$
idx = Random`Private`PositionsOf[Rest[list1], 0];
result = list1;
result[[idx + 1]] = list2[[idx]];
result

{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}

$\endgroup$
3
$\begingroup$

Here's another possibility:

Module[{tmp=list1},
    With[{i = Pick[Range[Length[list1]-1], Rest @ list1, 0]},
        tmp[[i+1]]=list2[[i]]
    ];
    tmp
]

{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}

$\endgroup$
2
$\begingroup$

Also MapIndexed works fine

(# /. List -> 0) & /@ MapIndexed[Replace[#1, 0 -> list2[[#2[[1]] - 1]]] &, list1]

The first piece just replaces the special case when a zero element is in the first entry (in Mathematica the 0th element is the Head, which in this case is List). Then the function MapIndexed does its job.

$\endgroup$
0
$\begingroup$
MapThread[If[#1 == 0, #2, #1] &, {Join[list1, {0}], Join[{0}, list2]}] // Most

{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}

Your answer from above

{0, 1, 1, 4, 7, 1, 1, 10, 1, 11, 3, 1, 0}

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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