Number of similar pixels on a loop arround some pixel

Question

I defined a command ŠtejTakePiksleOkol, that counts the number of pixels on a square loop arround some pixel, that have similar normalized RGB.
If the square loop would go over the edge of the picture it only takes the loop bounded by the edge of the image.
ŠtejTakePiksleOkol[slikapiksli_, baarva_, [Epsilon]_, sred_, 
  polm_] := {
   resx = Length[slikapiksli[[1]]  ];
   resy = Length[slikapiksli];
   [CapitalDelta]naštetih = 0;
   
   (*edges of the loop*)
   levo = If[polm >= sred[[2]], 1, sred[[2]] - polm  ];
   desno = If[sred[[2]] + polm > resx, resx, sred[[2]] + polm ];
   gor = If[polm >= sred[[1]], 1, sred[[1]] - polm];
   dol = If[sred[[1]] + polm > resy, resy, sred[[1]] + polm];
   (*pixels we have to check*)
   kpnk = If[polm == 0,
     {sred},
     Flatten[
      {
       Table[
        {dol, x},
        {x, levo, desno - 1}],
       Table[
        {y, desno},
        {y, gor + 1, dol}],
       Table[
        {gor, x},
        {x, levo + 1, desno}],
       Table[
        {y, levo},
        {y, gor, dol - 1}]
       },
      1]
     ];
   (
      {i1, i2} = #;
      If[
        Normalize[baarva].Normalize[slikapiksli[[i1, i2]]  ] > 
         1 - [Epsilon],
        [CapitalDelta]naštetih++;
        ]
      ) & /@ kpnk;
   [CapitalDelta]naštetih
   }[[1]]

Where slikapiksli is the image data, baarva is the wanted RGB, $epsilon$ is sth like the allowed difference, sred is the middle pixel and polm is the max 'radius' of the loop.
Now let's make a simple image
slikakrogca = Image[
  Table[
   If[i1^2 + i2^2 < 100^2, {0, 1, 1}, {1, 1, 1}],
   {i1, -300, 500},{i2, -700, 500}]
  ]

Now apply the function for radius polm=15
AbsoluteTiming[
 ŠtejTakePiksleOkol[ImageData[slikakrogca], {0, 1, 1}, .001, {3, 40}, 15]
]

{0.0134907, 0}

The same for polm=16
AbsoluteTiming[
 ŠtejTakePiksleOkol[ImageData[slikakrogca], {0, 1, 1}, .001, {3, 40}, 16]
]

{0.536426, 0}

And for polm=700
AbsoluteTiming[
 ŠtejTakePiksleOkol[ImageData[slikakrogca], {0, 1, 1}, .001, {3, 40}, 700]
]

{0.583814, 185}

Basically it slows down horribly from 15 to 16. Why is that and how to fix it?

C. E. · Accepted Answer

I made a version of your code mostly to understand it. But the timing issue seems to have disappeared as well, so I will post it. If I have time later I might try to optimize it further.
getPixels[imageData_, center_, 0] := {center}
getPixels[imageData_, center_, radius_] := Module[{
   resx = Length[imageData[[1]]],
   resy = Length[imageData],
   columnStart,
   columnEnd,
   rowStart,
   rowEnd
   },
  columnStart = If[
    radius >= center[[2]],
    1,
    center[[2]] - radius
    ];
  columnEnd = If[
    center[[2]] + radius > resx,
    resx,
    center[[2]] + radius
    ];
  rowStart = If[
    radius >= center[[1]],
    1,
    center[[1]] - radius
    ];
  rowEnd = If[
    center[[1]] + radius > resy,
    resy,
    center[[1]] + radius
    ];
  Flatten[{
    Table[{rowEnd, x}, {x, columnStart, columnEnd - 1}],
    Table[{y, columnEnd}, {y, rowStart + 1, rowEnd}],
    Table[{rowStart, x}, {x, columnStart + 1, columnEnd}],
    Table[{y, columnStart}, {y, rowStart, rowEnd - 1}]
    }, 1]
  ]

countSimilarPixels[imageData_, rgbValue_, allowedDifference_, center_,
    radius_] := Total@Boole@MapThread[
     Normalize[rgbValue].Normalize[imageData[[#, #2]]] > 1 - allowedDifference &,
     Transpose@getPixels[imageData, center, radius]
     ];

testImage = Image[Table[If[i1^2 + i2^2 < 100^2, {0, 1, 1}, {1, 1, 1}], {i1, -300, 500}, {i2, -700, 500}]];

AbsoluteTiming[countSimilarPixels[ImageData[testImage], {0, 1, 1}, .001, {3, 40}, 50]]

(* Out: {0.012622, 0} *)

Second look
I had another look at the function getPixels and made it a bit more readable:
getPixels[imageData_, center_, 0] := {center}
getPixels[imageData_, {row_, col_}, radius_] := Module[{
   nrOfColumns,
   nrOfRows,
   columnStart,
   columnEnd,
   rowStart,
   rowEnd
   },
  {nrOfRows, nrOfColumns} = Dimensions[imageData]~Take~2;
  columnStart = Max[1, col - radius];
  columnEnd = Min[nrOfColumns, col + radius];
  rowStart = Max[1, row - radius];
  rowEnd = Min[nrOfRows, row + radius];
  Join[
   Table[{rowEnd, c}, {c, columnStart, columnEnd - 1}],
   Table[{r, columnEnd}, {r, rowStart + 1, rowEnd}],
   Table[{rowStart, c}, {c, columnStart + 1, columnEnd}],
   Table[{r, columnStart}, {r, rowStart, rowEnd - 1}]
   ]
  ]

And alternative way of writing countSimilarPixels is
countSimilarPixels[imageData_, referenceColor_, allowedDifference_, center_, radius_] := Module[
  {pixels, colors, similarity},
  pixels = getPixels[imageData, center, radius];
  colors = Normalize /@ Extract[imageData, pixels];
  similarity = colors.Normalize[referenceColor];
  Total@UnitStep[similarity - (1 - allowedDifference)]
  ]

GalZoidberg · Answer

Okay it turns out, that if i replace last (...)&/@kpnk with Do[...,{i,Length[kpnk]}] it works normally
Do[
  If[Normalize[baarva].Normalize[
      slikapiksli[[kpnk[[i, 1]], kpnk[[i, 2]]   ]]  ] > 1 - [Epsilon],
   [CapitalDelta]naštetih++;
   [CapitalDelta]vsotatakih += slikapiksli[[i1, i2]];
   [CapitalDelta]vsotakord += {i1, i2};
   ],
  {i, Length[kpnk]}];

{0.0143393, 0}

I don't know why Map fails, but now it works

Number of similar pixels on a loop arround some pixel

2 Answers

Second look

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