[英]PerlMagick: Use QueryColorname() from Histogram() output
I want to get histogram of unique color in an image with color names or their Hex code. 我想在具有颜色名称或其十六进制代码的图像中获得唯一颜色的直方图。
I am unable to convert histogram method output value into color name or Hex code using QueryColorname method; 我无法使用QueryColorname方法将直方图方法的输出值转换为颜色名称或十六进制代码; It always returns black and does not return Hex code.
它始终返回黑色,不返回十六进制代码。
It is possibly due to (0 ... 65535) result range from histogram() method which I am not able to convert into (0 .. 255), an acceptable range for Querycolorname() method. 可能是由于histogram()方法的(0 ... 65535)结果范围,我无法将其转换为Querycolorname()方法的可接受范围(0 .. 255)。
#!/usr/bin/perl
use Image::Magick;
$image=Image::Magick->new();
$image->ReadImage('Sun.jpeg');
my @histogram = $image->Histogram();
print "Red\tGreen\tBlue\tOpacity\tCount\tName\n";
for(my $i=0; $i<=29; $i++){ #Get 5 unique colors
print "$histogram[$i]\t";
if (($i+1)%5 == 0){ #Array elements of unique color
my $name = $image->QueryColorname('rgb16($histogram[$i-4],$histogram[$i-3],$histogram[$i- 2],$histogram[$i-1])');
print "$name\n";
}
}
Result looks like, 结果看起来像
Red Green Blue Opacity Count Name
红色绿色蓝色不透明度计数名称
0 0 0 0 16134 black0 0 0 0 16134黑色
257 257 257 0 27 black257257257257 0 27黑色
0 257 0 0 303 black0257 00303黑色
257 0 0 0 286 black257 0 0 0 286黑色
257 257 0 0 8 black257257 0 0 8黑色
71 0 0 0 82 black71 0 0 0 82黑色
Method descriptions at http://www.imagemagick.org/script/perl-magick.php http://www.imagemagick.org/script/perl-magick.php上的方法描述
First of all: as you are using single quotes around variables they don't get expanded. 首先:当您在变量周围使用单引号时,它们不会扩展。
QueryColorname
sees a string that it possibly converts to zero's. QueryColorname
看到一个可能会转换为零的字符串。 That's why all colors are "black". 这就是为什么所有颜色都是“黑色”的原因。
Second: I don't see rgb16
in the docu and I suppose it does not do what you want. 第二:我在文档中没有看到
rgb16
,我想它没有做您想要的。 Instead, you have to scale down to 8 Bit 相反,您必须缩小到8位
Putting both together I propose something like this for the inner if-Block: 将两者放在一起,我为内部的if-Block提出了这样的内容:
my $colVec = "rgb(";
$colVec .= int($histogram[$i-4]/65535*256) . ",";
$colVec .= int($histogram[$i-3]/65535*256) . ",";
$colVec .= int($histogram[$i-2]/65535*256) . ",";
$colVec .= $histogram[$i-1] . ")";
print $image->QueryColorname($colVec) . "\n";
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