好吧,这个答案并不能完美地回答这个问题,但它应该给你一个好的开始!我知道我在代码中重复了自己,但我的目标只是让某些东西正常工作,以便您可以在其上进行构建,这不是生产代码!
前提条件
从大图开始:
我们需要尽可能地找到另一张图片的位置:
我决定将这个过程分成许多子步骤,您可以根据您希望代码执行的操作来改进或删除这些子步骤。
出于测试目的,我确实在不同的输入图像上测试了我的算法,因此您将看到一个定义要加载的文件的变量...
我们从:
function microtime_float()
{
list($usec, $sec) = explode(" ", microtime());
return ((float)$usec + (float)$sec);
}
$time_start = microtime_float();
$largeFilename = "large.jpg";
$small = imagecreatefromjpeg("small.jpg");
$large = imagecreatefromjpeg($largeFilename);
和
imagedestroy($small);
imagedestroy($large);
$time_end = microtime_float();
echo "in " . ($time_end - $time_start) . " seconds\n";
对我们的表演有一个好主意。幸运的是,大多数算法都非常快,所以我不必进行更多优化。
背景检测
我从检测背景颜色开始。我假设背景颜色将是图片中最常见的颜色。为此,我只计算了在大图片中可以找到的每种颜色的引用数量,使用递减值对其进行排序,并将第一个作为背景颜色(如果您更改了源图片,应该允许代码适应)
function FindBackgroundColor($image)
{
// assume that the color that's present the most is the background color
$colorRefcount = array();
$width = imagesx($image);
$height = imagesy($image);
for($x = 0; $x < $width; ++$x)
{
for($y = 0; $y < $height; ++$y)
{
$color = imagecolorat($image, $x, $y);
if(isset($colorRefcount[$color]))
$colorRefcount[$color] = $colorRefcount[$color] + 1;
else
$colorRefcount[$color] = 1;
}
}
arsort($colorRefcount);
reset($colorRefcount);
return key($colorRefcount);
}
$background = FindBackgroundColor($large); // Should be white
分区
我的第一步是尝试找到所有非背景像素所在的区域。通过一点填充,我能够将区域分组为更大的区域(这样一个段落将是一个区域而不是多个单独的字母)。我从 5 的填充开始并获得了足够好的结果,所以我坚持使用它。
这被分解为多个函数调用,所以我们开始:
function FindRegions($image, $backgroundColor, $padding)
{
// Find all regions within image where colors are != backgroundColor, including a padding so that adjacent regions are merged together
$width = imagesx($image);
$height = imagesy($image);
$regions = array();
for($x = 0; $x < $width; ++$x)
{
for($y = 0; $y < $height; ++$y)
{
$color = imagecolorat($image, $x, $y);
if($color == $backgroundColor)
{
continue;
}
if(IsInsideRegions($regions, $x, $y))
{
continue;
}
$region = ExpandRegionFrom($image, $x, $y, $backgroundColor, $padding);
array_push($regions, $region);
}
}
return $regions;
}
$regions = FindRegions($large, $background, 5);
在这里,我们迭代图片的每个像素,如果它的背景颜色,我们将其丢弃,否则,我们检查它的位置是否已经存在于我们找到的区域中,如果是这样,我们也跳过它。现在,如果我们没有跳过像素,这意味着它是一个彩色像素,应该是一个区域的一部分,所以我们开始ExpandRegionFrom
这个像素。
检查我们是否在区域内的代码非常简单:
function IsInsideRegions($regions, $x, $y)
{
foreach($regions as $region)
{
if(($region["left"] <= $x && $region["right"] >= $x) &&
($region["bottom"] <= $y && $region["top"] >= $y))
{
return true;
}
}
return false;
}
现在,扩展代码将尝试在每个方向上扩大区域,只要找到要添加到区域的新像素就会这样做:
function ExpandRegionFrom($image, $x, $y, $backgroundColor, $padding)
{
$width = imagesx($image);
$height = imagesy($image);
$left = $x;
$bottom = $y;
$right = $x + 1;
$top = $y + 1;
$expanded = false;
do
{
$expanded = false;
$newLeft = ShouldExpandLeft($image, $backgroundColor, $left, $bottom, $top, $padding);
if($newLeft != $left)
{
$left = $newLeft;
$expanded = true;
}
$newRight = ShouldExpandRight($image, $backgroundColor, $right, $bottom, $top, $width, $padding);
if($newRight != $right)
{
$right = $newRight;
$expanded = true;
}
$newTop = ShouldExpandTop($image, $backgroundColor, $top, $left, $right, $height, $padding);
if($newTop != $top)
{
$top = $newTop;
$expanded = true;
}
$newBottom = ShouldExpandBottom($image, $backgroundColor, $bottom, $left, $right, $padding);
if($newBottom != $bottom)
{
$bottom = $newBottom;
$expanded = true;
}
}
while($expanded == true);
$region = array();
$region["left"] = $left;
$region["bottom"] = $bottom;
$region["right"] = $right;
$region["top"] = $top;
return $region;
}
这些ShouldExpand
方法本可以以更简洁的方式编写,但我采用了一些快速原型的方法:
function ShouldExpandLeft($image, $background, $left, $bottom, $top, $padding)
{
// Find the farthest pixel that is not $background starting at $left - $padding closing in to $left
for($x = max(0, $left - $padding); $x < $left; ++$x)
{
for($y = $bottom; $y <= $top; ++$y)
{
$pixelColor = imagecolorat($image, $x, $y);
if($pixelColor != $background)
{
return $x;
}
}
}
return $left;
}
function ShouldExpandRight($image, $background, $right, $bottom, $top, $width, $padding)
{
// Find the farthest pixel that is not $background starting at $right + $padding closing in to $right
$from = min($width - 1, $right + $padding);
$to = $right;
for($x = $from; $x > $to; --$x)
{
for($y = $bottom; $y <= $top; ++$y)
{
$pixelColor = imagecolorat($image, $x, $y);
if($pixelColor != $background)
{
return $x;
}
}
}
return $right;
}
function ShouldExpandTop($image, $background, $top, $left, $right, $height, $padding)
{
// Find the farthest pixel that is not $background starting at $top + $padding closing in to $top
for($x = $left; $x <= $right; ++$x)
{
for($y = min($height - 1, $top + $padding); $y > $top; --$y)
{
$pixelColor = imagecolorat($image, $x, $y);
if($pixelColor != $background)
{
return $y;
}
}
}
return $top;
}
function ShouldExpandBottom($image, $background, $bottom, $left, $right, $padding)
{
// Find the farthest pixel that is not $background starting at $bottom - $padding closing in to $bottom
for($x = $left; $x <= $right; ++$x)
{
for($y = max(0, $bottom - $padding); $y < $bottom; ++$y)
{
$pixelColor = imagecolorat($image, $x, $y);
if($pixelColor != $background)
{
return $y;
}
}
}
return $bottom;
}
现在,为了查看算法是否成功,我添加了一些调试代码。
调试渲染
我创建了第二个图像来存储调试信息并将其存储在磁盘上,以便以后可以看到我的进度。
使用以下代码:
$large2 = imagecreatefromjpeg($largeFilename);
$red = imagecolorallocate($large2, 255, 0, 0);
$green = imagecolorallocate($large2, 0, 255, 0);
$blue = imagecolorallocate($large2, 0, 0, 255);
function DrawRegions($image, $regions, $color)
{
foreach($regions as $region)
{
imagerectangle($image, $region["left"], $region["bottom"], $region["right"], $region["top"], $color);
}
}
DrawRegions($large2, $regions, $red);
imagejpeg($large2, "regions.jpg");
我可以验证我的分区代码做得不错:
纵横比
我决定根据纵横比(宽高比)过滤掉一些区域。可以应用其他过滤,例如平均像素颜色或其他东西,但纵横比检查非常快,所以我使用了它。
我只是定义了一个“窗口”,如果区域的纵横比在最小值和最大值之间,则将在其中保留区域;
$smallAspectRatio = imagesx($small) / imagesy($small);
function PruneOutWrongAspectRatio($regions, $minAspectRatio, $maxAspectRatio)
{
$result = array();
foreach($regions as $region)
{
$aspectRatio = ($region["right"] - $region["left"]) / ($region["top"] - $region["bottom"]);
if($aspectRatio >= $minAspectRatio && $aspectRatio <= $maxAspectRatio)
{
array_push($result, $region);
}
}
return $result;
}
$filterOnAspectRatio = true;
if($filterOnAspectRatio == true)
{
$regions = PruneOutWrongAspectRatio($regions, $smallAspectRatio - 0.1 * $smallAspectRatio, $smallAspectRatio + 0.1 * $smallAspectRatio);
DrawRegions($large2, $regions, $blue);
}
imagejpeg($large2, "aspectratio.jpg");
通过添加DrawRegions
调用,我现在将仍在列表中的区域绘制为蓝色作为潜在位置:
如您所见,只剩下4个位置!
寻找角落
我们快完成了!现在,我正在做的是从小图片中查看四个角的颜色,并尝试在剩余区域的角中找到最佳匹配像素。此代码最有可能失败,因此如果您必须花时间改进解决方案,此代码将是一个不错的候选者。
function FindCorners($large, $small, $regions)
{
$result = array();
$bottomLeftColor = imagecolorat($small, 0, 0);
$blColors = GetColorComponents($bottomLeftColor);
$bottomRightColor = imagecolorat($small, imagesx($small) - 1, 0);
$brColors = GetColorComponents($bottomRightColor);
$topLeftColor = imagecolorat($small, 0, imagesy($small) - 1);
$tlColors = GetColorComponents($topLeftColor);
$topRightColor = imagecolorat($small, imagesx($small) - 1, imagesy($small) - 1);
$trColors = GetColorComponents($topRightColor);
foreach($regions as $region)
{
$bottomLeft = null;
$bottomRight = null;
$topLeft = null;
$topRight = null;
$regionWidth = $region["right"] - $region["left"];
$regionHeight = $region["top"] - $region["bottom"];
$maxRadius = min($regionWidth, $regionHeight);
$topLeft = RadialFindColor($large, $tlColors, $region["left"], $region["top"], 1, -1, $maxRadius);
$topRight = RadialFindColor($large, $trColors, $region["right"], $region["top"], -1, -1, $maxRadius);
$bottomLeft = RadialFindColor($large, $blColors, $region["left"], $region["bottom"], 1, 1, $maxRadius);
$bottomRight = RadialFindColor($large, $brColors, $region["right"], $region["bottom"], -1, 1, $maxRadius);
if($bottomLeft["found"] && $topRight["found"] && $topLeft["found"] && $bottomRight["found"])
{
$left = min($bottomLeft["x"], $topLeft["x"]);
$right = max($bottomRight["x"], $topRight["x"]);
$bottom = min($bottomLeft["y"], $bottomRight["y"]);
$top = max($topLeft["y"], $topRight["y"]);
array_push($result, array("left" => $left, "right" => $right, "bottom" => $bottom, "top" => $top));
}
}
return $result;
}
$closeOnCorners = true;
if($closeOnCorners == true)
{
$regions = FindCorners($large, $small, $regions);
DrawRegions($large2, $regions, $green);
}
我试图通过从角落增加“径向”(基本上是正方形)来找到匹配的颜色,直到找到匹配的像素(在公差范围内):
function GetColorComponents($color)
{
return array("red" => $color & 0xFF, "green" => ($color >> 8) & 0xFF, "blue" => ($color >> 16) & 0xFF);
}
function GetDistance($color, $r, $g, $b)
{
$colors = GetColorComponents($color);
return (abs($r - $colors["red"]) + abs($g - $colors["green"]) + abs($b - $colors["blue"]));
}
function RadialFindColor($large, $color, $startx, $starty, $xIncrement, $yIncrement, $maxRadius)
{
$result = array("x" => -1, "y" => -1, "found" => false);
$treshold = 40;
for($r = 1; $r <= $maxRadius; ++$r)
{
$closest = array("x" => -1, "y" => -1, "distance" => 1000);
for($i = 0; $i <= $r; ++$i)
{
$x = $startx + $i * $xIncrement;
$y = $starty + $r * $yIncrement;
$pixelColor = imagecolorat($large, $x, $y);
$distance = GetDistance($pixelColor, $color["red"], $color["green"], $color["blue"]);
if($distance < $treshold && $distance < $closest["distance"])
{
$closest["x"] = $x;
$closest["y"] = $y;
$closest["distance"] = $distance;
break;
}
}
for($i = 0; $i < $r; ++$i)
{
$x = $startx + $r * $xIncrement;
$y = $starty + $i * $yIncrement;
$pixelColor = imagecolorat($large, $x, $y);
$distance = GetDistance($pixelColor, $color["red"], $color["green"], $color["blue"]);
if($distance < $treshold && $distance < $closest["distance"])
{
$closest["x"] = $x;
$closest["y"] = $y;
$closest["distance"] = $distance;
break;
}
}
if($closest["distance"] != 1000)
{
$result["x"] = $closest["x"];
$result["y"] = $closest["y"];
$result["found"] = true;
return $result;
}
}
return $result;
}
如您所见,我不是 PHP 专家,我不知道有一个内置函数可以获取 rgb 通道,哎呀!
最后呼叫
所以现在算法运行了,让我们看看它使用以下代码发现了什么:
foreach($regions as $region)
{
echo "Potentially between " . $region["left"] . "," . $region["bottom"] . " and " . $region["right"] . "," . $region["top"] . "\n";
}
imagejpeg($large2, "final.jpg");
imagedestroy($large2);
输出(非常接近真正的解决方案):
Potentially between 108,380 and 867,827
in 7.9796848297119 seconds
给出这张图片(和之间的矩形108,380
以867,827
绿色绘制)
希望这可以帮助!