Cover up zeroes in a list

Question

Inspired by this SO question

As input you will be given a non-empty list of integers, where the first value is guaranteed to be non-zero. To construct the output, walk from the start of the list, outputting each non-zero value along the way. When you encounter a zero, instead repeat the value you most recently added to the output.

You may write a program or function, and have input/output take any convenient format which does not encode extra information, as long as is still an ordered sequence of integers. If outputting from a program, you may print a trailing newline. Except for this trailing newline, your output should be an acceptable input for your submission.

The shortest code in bytes wins.

Test Cases

[1, 0, 2, 0, 7, 7, 7, 0, 5, 0, 0, 0, 9] -> [1, 1, 2, 2, 7, 7, 7, 7, 5, 5, 5, 5, 9]
[1, 0, 0, 0, 0, 0] -> [1, 1, 1, 1, 1, 1]
[-1, 0, 5, 0, 0, -7] -> [-1, -1, 5, 5, 5, -7]
[23, 0, 0, -42, 0, 0, 0] -> [23, 23, 23, -42, -42, -42, -42]
[1, 2, 3, 4] -> [1, 2, 3, 4]
[-1234] -> [-1234]

Answer 1

Japt, 3 bytes

å!ª

Try it

å!ª     :Implicit input of array
å       :Cumulatively reduce
 !      :Flip the arguments
  ª     :Logical OR

Answer 2

05AB1E, 7 bytes

εĀiyU}X

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My first try, using 05AB1E

ε       for each y in input
  Āi    if > 0
    yU  save current value in X
  }
  X     push X
        implicitly close for-loop and print list

Answer 3

Perl 6, 16 bytes

*.map($ [R||]=*)

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Maps each element to the reverse logical OR with the previous calculated element. I wish I could do something like {[\R||] $_} but the R operator doesn't quite work that way.

Answer 4

Zsh, 21 bytes

for i;echo $[n=i?i:n]

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We use echo instead of <<< because we need to set n. Using <<<$[n=i?i:n] expands $[n=i?i:n] in a subshell, which will prevent n from ever being set.

Answer 5

k4, 15 bytes

{^\?[0=x;0N;x]}

---

   ?[        ]  / vector conditional ?[cond;true;false]; eg ?[010b;2 2 2;3 3 3] -> 3 2 3
 ^\             / fills nulls (0N) with preceding value

Answer 6

Prolog (SWI), 44 bytes

f(0,B,B).
f(A,_,A).
X*O:-scanl(f,X,0,[_|O]).

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Answer 7

Labyrinth, 21 bytes

The top half seems a bit wasteful, but I'm not sure what to do about it...

""?"
" #"
;  ;
,\!:
@

I/O via linefeed-delimited lists.

Try it online.

The idea is to keep the last printed thing on the stack and discard zeroes whenever we find them.

Answer 8

Arcyóu, 44 bytes

(F(L)(f x(_(_ L))(L((F(i)(?(L i)i($([ i))))x

This is an anonymous function taking one argument.

Explanation:

(F(L)               ; Anonymous function F(L)
  (f x (_ (_ L))    ; For x in range(len(L))
    (L              ; Implicit indexing
      (F(i)         ; Anonymous function F(i)
        (? (L i)    ; If L[i] is not 0:
          i         ; Return i; otherwise:
          ($ ([ i)) ; Recurse ($) with i decremented
        )           ; Then, we call this new function on x
      )             ; And use the return value as the index from line 3
      x             ; No final close-parens

Answer 9

C 45 Bytes

#define F(a,s) while(s--) *a=*a?*a++:*(a++-1)

Here's my test program

Go easy as it's my first try at code golf.

#include <stdio.h>

#define ARRAYSIZE(x) (sizeof(x)/sizeof(x[0]))
#define F(a,s) while(s--) *a=*a?*a++:*(a++-1)

void printSolutions(int * solutions, int size)
{
  do {
    printf("%d ", *solutions++);
  } while (--size > 0);
  printf("\n");
}

int main(int argc, char * argv[])
{
  static int testCase1[] = {1, 0, 2, 0, 7, 7, 7, 0, 5, 0, 0, 0, 9};
  static int testCase2[] = {1, 0, 0, 0, 0, 0}; 
  static int testCase3[] = {-1, 0, 5, 0, 0, -7};
  static int testCase4[] = {23, 0, 0, -42, 0, 0, 0};
  static int testCase5[] = {1, 2, 3, 4};
  static int testCase6[] = {-1234};
  int * p;
  int s;
  printSolutions(testCase1, ARRAYSIZE(testCase1));
  s = ARRAYSIZE(testCase1);
  p = testCase1;
  F(p, s);
  printSolutions(testCase1, ARRAYSIZE(testCase1));
  printSolutions(testCase2, ARRAYSIZE(testCase2));
  s = ARRAYSIZE(testCase2);
  p = testCase2;
  F(p, s);
  printSolutions(testCase2, ARRAYSIZE(testCase2));
  printSolutions(testCase3, ARRAYSIZE(testCase3));
  s = ARRAYSIZE(testCase3);
  p = testCase3;
  F(p, s);
  printSolutions(testCase3, ARRAYSIZE(testCase3));
  printSolutions(testCase4, ARRAYSIZE(testCase4));
  s = ARRAYSIZE(testCase4);
  p = testCase4;
  F(p, s);
  printSolutions(testCase4, ARRAYSIZE(testCase4));
  printSolutions(testCase5, ARRAYSIZE(testCase5));
  s = ARRAYSIZE(testCase5);
  p = testCase5;
  F(p, s);
  printSolutions(testCase5, ARRAYSIZE(testCase5));
  printSolutions(testCase6, ARRAYSIZE(testCase6));
  s = ARRAYSIZE(testCase6);
  p = testCase6;
  F(p, s);
  printSolutions(testCase6, ARRAYSIZE(testCase6));

  return 0;
}

Answer 10

Mathematica, 28 27 25 bytes

Saved 2 bytes thanks to Martin Büttner.

If[#2==0,##]&~FoldList~#&

Answer 11

Factor, 36 bytes

[ 0 [ dup 0 = -rot ? ] accumulate* ]

(Only works locally in Factor 0.98 or later. TIO is missing accumulate*.) ¯\_(ツ)_/¯

Explanation

• 0 Push a zero to the data stack for accumulate* to use later. This is a 'starter' element so that a binary operation can be performed on something with the first element of the input sequence.
• [ dup 0 = -rot ? ] Push a quotation to the data stack for accumulate* to use later. This is a function that says "Return the non-zero input if possible. Otherwise, return the second input." Shorter but harder to read than [ [ nip ] unless-zero ].
• accumulate* Like reduce, except the intermediate results are mapped into a sequence of the same size as the input sequence.

Answer 12

K (ngn/k), 9 bytes

{y+x*~y}\

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Take the cumulative sums of the input, resetting to 0 whenever a new non-zero value is encountered in the list.

If the 0's in the input are replaced with nulls (0N), ^\ (fill-scan) suffices for two bytes.

Answer 13

Groovy, 34 25 bytes

f={it.collect{it?n=it:n}}

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A more readable solution:

f = { // define new closure
  it.collect { // taking advantage of implicit "it". collect applies the closure
               // on every element in order and returns the new, mapped array
    it ? // Groovy Boolean evaluation: 0 = false, 1+ = true
    n = it // assign value to holding variable (if not 0). assignments are expressions
    :
    n // in Groovy, if no return, last expression evaluated is returned
  } // same as above, returns the mapped array
} // end of closure

Groovy is supposed to be similar to Java (and it is), but the Java answer is more than two times longer. Just something to think about.

EDIT: reduced it by 9 bytes(!) by not declaring the n variable beforehand (saves 6 bytes), and using a ternary operator instead of an if (saves 3 bytes)

Answer 14

Vyxal, 2 bytes

ɖ∨

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Answer 15

rs, 14 bytes

+(\S+) 0/\1 \1

1 byte shorter than Retina!!

Live demo and test cases.

Answer 16

Seriously, 12 bytes

0╗,`╜@;I;╗`M

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Explanation:

0╗,    initialize first register to 0 and get input
`...`M map the function over the input:
  ╜@;I    push the result of (a if a else <value in register 0>)
  ;╗      dupe result and push to register 0

Answer 17

Gema, 21 characters

0=$p
<N>=@set{p;$0}$0

Sample run:

bash-4.3$ gema '0=$p;<N>=@set{p;$0}$0' <<< '23 0 0 -42 0 0 0'
23 23 23 -42 -42 -42 -42

Answer 18

JavaScript, 20 Bytes

a=>a.map(b=>c=b?b:c)

Explanation:

a=>           // Define anonymous function which takes argument a
  a.map(      // Loop through input
    b=>       // Looping function
      c=b     // Assign c to the loop value (returns b)
      ?b:c    // Ternary Operator: If b (previous expression) is truthy (Non-zero integers are) return b else return c (last value)          
)

Answer 19

Perl 5 + -pl, 9 bytes

$;=$_||=$

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Explanation

This reads each number in as $_ (-p), then sets $; to $_, after first setting $_ to $; if $_ is falsy (0) and is implicitly printed (-p). Requires -l as otherwise $_ would include \n which would mean 0 isn't falsy. The final ; can be omitted as -p (well, specifically the -n part of -p) adds while(<STDIN>){ ... ;} around your code, resulting in $; at the end.

Answer 20

QBasic, 54 bytes

INPUT n
FOR i=1TO n
INPUT x
IF x=0THEN x=p
?x
p=x
NEXT

Input is the length of the list, followed by the numbers in the list one at a time. (Compare this I/O default, which is the same basic idea even though it refers specifically to passing an array to a function.)

You can try it at Archive.org.

Ungolfed

INPUT arrayLength
FOR i = 1 TO arrayLength
  INPUT number
  IF number = 0 THEN number = previous
  PRINT number
  previous = number
NEXT i

Answer 21

TI-Basic, 37 bytes

Prompt A
For(I,2,dim(ʟA
ʟA(I)+ʟA(I-1)not(ʟA(I→ʟA(I
End

Output is stored in ʟA.

Answer 22

Python 3.8 (pre-release), 44 bytes

g=lambda l,s=0:l and[a:=l[0]or s]+g(l[1:],a)

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Answer 23

Batch 147 bytes

@Set A=%* &@Setlocal EnableDelayedExpansion
@Set "L=%A: ="&If .!C!==. Set "C=%"
@Set "A=%A: ="&2>nul Set/AL/L&&Set/AC=L&<nul Set/p=!C! &Set "L=%"

Approach:
Update Value to output only when value is non zero

How:

Line one: Arg asssignment and environment prep. Defines the array from arg to the batch file IE: 4 1 0 2 4 3

Line two: Uses a substring operation for each whitespace present in the string with an if condition to isolate the first substring preceding a space for assignment to Last and CCurrent variables.

Line three: Uses Substring modification to iterate through the array from each space in the substring; divides the Last value by itself and on command success (&&) transfers the Last value to Current. (a 0 value results in divide by zero error, redirected to nul). Current value is output after each step of the substitution.

---

Example Input / output:

4 3 2 0 3
4 3 2 2 3

Answer 24

05AB1E, 6 bytes

Å»‚0Kθ

Unfortunately 05AB1E lacks a logical-OR builtin, and a bitwise-OR isn't what we want here.

Try it online or verify all test cases.

Explanation:

Å»      # Cumulative left-reduce the (implicit) input-list by:
  ‚     #  Pair the current and previous integers together
   0K   #  Remove any potential 0
     θ  #  And pop and push the last item of this singleton-list/pair
        # (after which the result is output implicitly)

Answer 25

MathGolf, 7 bytes

êæαç┤o╘

Prints all values on a separated line to STDOUT.

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Explanation:

ê       # Push the input as integer-array
 æ      # Foreach over these integers, using four characters as inner code-block:
  α     #  Wrap the top two values on the stack into a list
        #  (the first iteration will pair with an implicit 0)
   ç    #  Remove all 0s (falsey filter)
    ┤   #  Push it's last item to the stack (unfortunately doesn't pop the list)
     o  #  Print it with trailing newline (without popping)
 ╘      # After the foreach: empty the stack (otherwise it would implicitly output the
        # entire stack; feel free to remove it in the TIO-link to see the difference)

Answer 26

Ly, 19 bytes

&nry[p[s!]plu9`oy]p

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This works by running through all the numbers on the list and remember the current number if it's non-zero. Then it unconditionally deleted the current character and what's printed is whatever number is in the backup cell.

&nr                  - read all the nunbers into the stack, reverse them
   y[p          y]   - loop while the stack length is >0
      [  ]           - if/then, true is current number !=0
       s             - save the number to the backup cell
        !            - flip to 0 to exit the loop
                  p  - delete loop boolean so it doesn't print
          p          - delete current number
           lu        - load backup cell and print
             9`o     - print a LF

Answer 27

Arturo, 32 bytes

$[a][0map a'x[(0=x)?[dup][x x]]]

Try it

Answer 28

Clojure, 49 bytes

#(reduce(fn[c i](conj c(if(= i 0)(last c)i)))[]%)

Well this is very boring. TIO

Answer 29

Desmos, 80 bytes

L=l[1...i-1]
A=L[LL>0]
f(l)=[\{l[i]=0:A[A.\length],l[i]\}\for i=[1...l.\length]]

Most likely golfable, though I don't see anything at the moment.

Try It On Desmos!

Try It On Desmos! - Prettified