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Challenges

Truthify an array

+7
−0

Jelly has an atom called untruth, which when given indices, creates an array with 1s at those places: [2,4] → [0,1,0,1].

You are required to perform the inverse of this. Given a 2D boolean array, find the indices of the true values in it.

Challenge

You will be given a single 2D boolean array $M$. You may take its dimensions as a separate argument if needed.

The output must consist of all the indices of the true values in $M$.

Output may be 0-indexed or 1-indexed.

It is guaranteed that the input will only consist of two different values.

Test Cases

All test cases are 0-indexed.

I: [[1,0,1,1,1,0]]
O: [[0,0],[0,2],[0,3],[0,4]]

I: [[1,0,1],[1,0,1],[0,1,0]]
O: [[0,0],[0,2],[1,0],[1,2],[2,1]]

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General comments (5 comments)

10 answers

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+4
−0

C (gcc), 100 bytes

i,j;f(x,y,p)int*p;{puts("[");for(;i<x;i++)for(j=0;j<y;j++)*p++&&printf("[%d,%d],",i,j);puts("\b]");}

Try it online!

Note that TIO console doesn't handle backspace properly.

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+7
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JavaScript (Node.js), 74 70 bytes

Works for any number of dimensions.

f=(a,z=[])=>a.map?.((b,c)=>z.push(...f(b).map(a=>[c,...a])))?z:a?[z]:z

Try it online! (TIO uses an old version of Node.js that does not support the ?. operator, but you can try the 74 bytes solution)

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+5
−0

Jelly, 2 bytes

ŒṪ

Try it online!

1-indexed, but the footer converts to 0-indexed.

Less built-in:

Jelly, 3 bytes

œẹ1

Try it online!

œẹ     All multidimensional indices of
  1    one.

Less built-in:

Jelly, 7 bytes

J,Ɱ"T€Ẏ

Try it online!

 ,Ɱ        Pair each
    T      truthy index
     €     from all rows
J  "       with the indices of the rows.
      Ẏ    Flatten rows.

The absolute least built-in:

Jelly, 8 bytes

Jx),€"JẎ

Try it online!

Outputs each index backwards.

Jx          Repeat each index by the corresponding element
  )         for each row.
   ,€"J     Pair each remaining index with its row index.
       Ẏ    Flatten rows.
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APL (Dyalog Extended), 1 byte

¯\(ツ)

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General comments (1 comment)
+4
−0

JavaScript (V8), 44 bytes

a=>a.map((a,x)=>a.map((n,y)=>n&&print(x,y)))

Try it online!

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+2
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JavaScript (Node.js), 68 bytes

f=(a)=>a.map((x)=>[...x.entries()].filter((e)=>e[1]).map((e)=>e[0]))

Run the function f on your input, e.g. f([[1,0,1],[1,0,1],[0,1,0]]).

The output looks like this:

[[0, 2], [0, 2], [1]]

with one subarray for each input subarray.


Explanation:

f=(a)=> // creates a function to run the input on
	a.map((x)=> // for each outer array:
		[...x.entries()]   // creates things like this: [index, value]
		.filter((e)=>e[1]) // keep ones where the value acts like true
		.map((e)=>e[0])    // return the indices for each one
	)
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+2
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Japt -m, 8 7 bytes

ð òÎmiV

Try it

ð òÎmiV     :Implicit map of each sub-array at 0-based index V
ð           :Indices of truthy (i.e., non-zero) elements
  ò         :Partition between elements where
   Î        :The sign of their difference is truthy
    m       :Map
     iV     :  Prepend V
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Scala 3, 68 bytes

_.zipWithIndex.flatMap(_.zipWithIndex.filter(_._1)map(_._2)map _.->)

Try it in Scastie!

Takes a List[List[Boolean]], returns a List[(Int, Int)]. This abuses underscores a bit (which is why Scala 3 is needed), but I'm proud of it.

Explanation:

_.zipWithIndex //Zip each row with its index
  .flatMap(    //Map every (row, index) tuple to a list of truthy indices,
               //then flatten
  _.zipWithIndex //Zip each element with its column
    .filter(_._1) //Keep the tuples where the element (_1) is true
    map(_._2)     //Keep the second part of the tuple (the column)
    map _.->)     //Make a 2-tuple with the row's index
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+1
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Python 3, 75 bytes

lambda a:[[i,k]for i in r(l(a))for k in r(l(a[0]))if a[i][k]]
r=range;l=len

Try it online!

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+0
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shortC, 78 bytes

i,j;f(Ix,Iy,I*p){R"[");O;i<x;i++)Oj=0;j<y;j++)*p++&&R"[%d,%d],",i,j);J"\b]");}

Try it online!

From @Lundin's C (gcc) answer.

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