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POHow to replace each 0 with the preceding element in a list in an idiomatic way in Mathematica?
 

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  1. POHow to replace each 0 with the preceding element in a list in an idiomatic way in Mathematica?
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    <p>This is a fun little problem, and I wanted to check with the experts here if there is a better functional/Mathematica way to approach solving it than what I did. I am not too happy with my solution since I use big IF THEN ELSE in it, but could not find a Mathematica command to use easily to do it (such as <code>Select</code>, <code>Cases</code>, <code>Sow/Reap</code>, <code>Map</code>.. etc...)</p> <p>Here is the problem, given a list values (numbers or symbols), but for simplicity, lets assume a list of numbers for now. The list can contain zeros and the goal is replace the each zero with the element seen before it.</p> <p>At the end, <em>the list should contain no zeros in it</em>.</p> <p>Here is an example, given </p> <pre><code>a = {1, 0, 0, -1, 0, 0, 5, 0}; </code></pre> <p>the result should be</p> <pre><code>a = {1, 1, 1, -1, -1, -1, 5, 5} </code></pre> <p>It should ofcourse be done in the most efficient way.</p> <p>This is what I could come up with</p> <pre><code>Scan[(a[[#]] = If[a[[#]] == 0, a[[#-1]], a[[#]]]) &amp;, Range[2, Length[a]]]; </code></pre> <p>I wanted to see if I can use Sow/Reap on this, but did not know how.</p> <p><strong>question: can this be solved in a more functional/Mathematica way? The shorter the better ofcourse :)</strong></p> <p><strong>update 1</strong> Thanks everyone for the answer, all are very good to learn from. This is the result of speed test, on V 8.04, using windows 7, 4 GB Ram, intel 930 @2.8 Ghz:</p> <p>I've tested the methods given for <code>n</code> from <code>100,000</code> to <code>4 million</code>. The <code>ReplaceRepeated</code> method does not do well for large lists.</p> <p><strong>update 2</strong></p> <p>Removed earlier result that was shown above in update1 due to my error in copying one of the tests.</p> <p>The updated results are below. Leonid method is the fastest. Congratulation Leonid. A very fast method.</p> <p><img src="https://i.stack.imgur.com/YN2dq.png" alt="enter image description here"></p> <p>The test program is the following:</p> <pre><code>(*version 2.0 *) runTests[sizeOfList_?(IntegerQ[#] &amp;&amp; Positive[#] &amp;)] := Module[{tests, lst, result, nasser, daniel, heike, leonid, andrei, sjoerd, i, names}, nasser[lst_List] := Module[{a = lst}, Scan[(a[[#]] = If[a[[#]] == 0, a[[# - 1]], a[[#]]]) &amp;, Range[2, Length[a]]] ]; daniel[lst_List] := Module[{replaceWithPrior}, replaceWithPrior[ll_, n_: 0] := Module[{prev}, Map[If[# == 0, prev, prev = #] &amp;, ll] ]; replaceWithPrior[lst] ]; heike[lst_List] := Flatten[Accumulate /@ Split[lst, (#2 == 0) &amp;]]; andrei[lst_List] := Module[{x, y, z}, ReplaceRepeated[lst, {x___, y_, 0, z___} :&gt; {x, y, y, z}, MaxIterations -&gt; Infinity] ]; leonid[lst_List] := FoldList[If[#2 == 0, #1, #2] &amp;, First@#, Rest@#] &amp; @lst; sjoerd[lst_List] := FixedPoint[(1 - Unitize[#]) RotateRight[#] + # &amp;, lst]; lst = RandomChoice[Join[ConstantArray[0, 10], Range[-1, 5]], sizeOfList]; tests = {nasser, daniel, heike, leonid, sjoerd}; names = {"Nasser","Daniel", "Heike", "Leonid", "Sjoerd"}; result = Table[0, {Length[tests]}, {2}]; Do[ result[[i, 1]] = names[[i]]; Block[{j, r = Table[0, {5}]}, Do[ r[[j]] = First@Timing[tests[[i]][lst]], {j, 1, 5} ]; result[[i, 2]] = Mean[r] ], {i, 1, Length[tests]} ]; result ] </code></pre> <p>To run the tests for length 1000 the command is:</p> <pre><code>Grid[runTests[1000], Frame -&gt; All] </code></pre> <p>Thanks everyone for the answers.</p>
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    1. COJust a note that using `If` is *not* not functional. Conditionals are an essential part of functional programming, and do not require side effects. Think of `If` as a mathematical function mapping boolans (the set {True,False}) to something else. Otherwise I came up with the same solution as Andrei, which I think is the simplest, but definitely not the fastest (hence not the most practical if you process large data!)
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      1. POHow to replace each 0 with the preceding element in a list in an idiomatic way in Mathematica?
      1. USSzabolcs
    2. COreplaceWithPrior[ll_, n_: 0] := Module[{prev}, Map[If[# == 0, prev, prev = #] &, ll]] In[12]:= replaceWithPrior[a] Out[12]= {1, 1, 1, -1, -1, -1, 5, 5}
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      1. POHow to replace each 0 with the preceding element in a list in an idiomatic way in Mathematica?
      1. USDaniel Lichtblau
    3. COBTW what should happen if the first element is 0?
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      1. POHow to replace each 0 with the preceding element in a list in an idiomatic way in Mathematica?
      1. USSzabolcs
 

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