I got this code to check if an image file contains blue pixels with Imagemagick and counting them - then saving the result.
我得到这个代码来检查图像文件是否包含Imagemagick的蓝色像素并计算它们 - 然后保存结果。
It works well, but it seems like many processes of Imagemagick hang forever on the server and are making it very slow.
它运行良好,但似乎Imagemagick的许多进程永远挂在服务器上,并使它非常慢。
Is there a way to improve this code and avoid this trouble?
有没有办法改善这些代码并避免这种麻烦?
module.exports = function (File) {
File.observe('after save', function countPixels(ctx, next) {
if (ctx.instance && !ctx.instance.blue_pixels) {
var exec = require('child_process').exec;
// Convert file to retrieve only blue pixels:
exec('convert ' + ctx.instance.path + ' -fx "u.b>(u.g+0.2)&&u.b>(u.r+0.2)&&saturation>0.6" -format "%[fx:mean*w*h]" info:',
function (error, stdout, stderr) {
if (error !== null) {
return next(error);
} else {
ctx.instance.blue_pixels = stdout;
File.upsert(ctx.instance);
}
});
}
next();
});
};
2 个解决方案
#1
1
The -fx
operator that you are using is notoriously slow - especially for large images. I had a try at casting the same formula using faster methods which may help you. So, I made a sample image:
您正在使用的-fx运算符非常慢 - 特别是对于大型图像。我尝试使用更快的方法来铸造相同的公式,这可能会对你有帮助。所以,我做了一个示例图像:
convert xc:red xc:lime -append \( xc:blue xc:cyan -append \) +append -resize 256x256! input.png
And then rewrote your expression like this:
然后重写你的表达式:
convert input.png \
\( -clone 0 -separate -delete 0 -evaluate-sequence subtract -threshold 20% -write BG.png \) \
\( -clone 0 -separate -delete 1 -evaluate-sequence subtract -threshold 20% -write BR.png \) \
\( -clone 0 -colorspace hsl -separate -delete 0,2 -threshold 60% -write S.png \) \
-delete 0 \
-evaluate-sequence min result.png
Note that the -write XYZ.png
are just debug statements that can be removed.
请注意,-write XYZ.png只是可以删除的调试语句。
Basically, I am building a mask of all pixels that meet your criteria and making them white, and making all the ones that don't match your criteria black and at the end, I run -evaluate-sequence min
to find the minimum of each pixel so that all three of your conditions must effectively be met:
基本上,我正在构建一个符合您标准的所有像素的蒙版,并将它们设置为白色,并使所有与您的标准不匹配的像素变黑,最后,我运行-evaluate-sequence min来查找每个像素的最小值像素,以便有效地满足您的所有三个条件:
- that blue exceeds green by 20%
- 蓝色超过绿色20%
- that blue exceeds red by 20%
- 蓝色超过红色20%
- that the saturation exceeds 60%
- 饱和度超过60%
The -separate -delete N
splits your image into RGB channels and then deletes one of the resulting channels, so if I -delete 1
(that is the Green channel) I am left with Red and Blue. Here are the intermediate, debug images. The first one is the condition Blue exceeds Red by 20%:
-separate -delete N将您的图像拆分为RGB通道,然后删除其中一个结果通道,因此如果I -delete 1(即绿色通道),我将留下红色和蓝色。这是中间调试图像。第一个是条件蓝色超过红色20%:
Then that Blue exceeds Green by 20%:
那么蓝色超过绿色20%:
And finally that the Saturation exceeds 60%:
最后,饱和度超过60%:
And then the result:
然后结果:
You'll need to put your -format "%[fx:mean*w*h]" info:
back on the end in place of the output image name to get the count of saturated blue pixels.
你需要将你的-format“%[fx:mean * w * h]”信息放回到最后代替输出图像名称以获得饱和蓝色像素的数量。
If I run your command:
如果我运行你的命令:
convert input.png -fx "u.b>(u.g+0.2)&&u.b>(u.r+0.2)&&saturation>0.6" result.png
My brain is not quite right today, so please run some checks - I may have something back-to-front somewhere!
我的大脑今天不太正确,所以请进行一些检查 - 我可能会在某个地方有一些东西!
As a benchmark, on a 10,000x10,000 pixel PNG, my code runs in 30 seconds, whereas the -fx
equivalent takes nearly 7 minutes.
作为基准,在10,000x10,000像素的PNG上,我的代码在30秒内运行,而-fx相当于将近7分钟。
#2
1
I don't know imagelagick part. But for node part I see that you call next non regarding to imagemgick opertion.
我不知道imagelagick的一部分。但是对于节点部分我看到你关于imagemgick opertion调用next non。
module.exports = function (File) {
File.observe('after save', function countPixels(ctx, next) {
if (ctx.instance && !ctx.instance.blue_pixels) {
var exec = require('child_process').exec;
// Convert file to retrieve only blue pixels:
exec('convert ' + ctx.instance.path + ' -fx "u.b>(u.g+0.2)&&u.b>(u.r+0.2)&&saturation>0.6" -format "%[fx:mean*w*h]" info:',
function (error, stdout, stderr) {
if (error !== null) {
return next(error);
} else {
ctx.instance.blue_pixels = stdout;
File.upsert(ctx.instance);
next();
}
});
}
else{next();}
//next(); //run next hook ASAP (before imagemagick returns back the result)
});
};
#1
1
The -fx
operator that you are using is notoriously slow - especially for large images. I had a try at casting the same formula using faster methods which may help you. So, I made a sample image:
您正在使用的-fx运算符非常慢 - 特别是对于大型图像。我尝试使用更快的方法来铸造相同的公式,这可能会对你有帮助。所以,我做了一个示例图像:
convert xc:red xc:lime -append \( xc:blue xc:cyan -append \) +append -resize 256x256! input.png
And then rewrote your expression like this:
然后重写你的表达式:
convert input.png \
\( -clone 0 -separate -delete 0 -evaluate-sequence subtract -threshold 20% -write BG.png \) \
\( -clone 0 -separate -delete 1 -evaluate-sequence subtract -threshold 20% -write BR.png \) \
\( -clone 0 -colorspace hsl -separate -delete 0,2 -threshold 60% -write S.png \) \
-delete 0 \
-evaluate-sequence min result.png
Note that the -write XYZ.png
are just debug statements that can be removed.
请注意,-write XYZ.png只是可以删除的调试语句。
Basically, I am building a mask of all pixels that meet your criteria and making them white, and making all the ones that don't match your criteria black and at the end, I run -evaluate-sequence min
to find the minimum of each pixel so that all three of your conditions must effectively be met:
基本上,我正在构建一个符合您标准的所有像素的蒙版,并将它们设置为白色,并使所有与您的标准不匹配的像素变黑,最后,我运行-evaluate-sequence min来查找每个像素的最小值像素,以便有效地满足您的所有三个条件:
- that blue exceeds green by 20%
- 蓝色超过绿色20%
- that blue exceeds red by 20%
- 蓝色超过红色20%
- that the saturation exceeds 60%
- 饱和度超过60%
The -separate -delete N
splits your image into RGB channels and then deletes one of the resulting channels, so if I -delete 1
(that is the Green channel) I am left with Red and Blue. Here are the intermediate, debug images. The first one is the condition Blue exceeds Red by 20%:
-separate -delete N将您的图像拆分为RGB通道,然后删除其中一个结果通道,因此如果I -delete 1(即绿色通道),我将留下红色和蓝色。这是中间调试图像。第一个是条件蓝色超过红色20%:
Then that Blue exceeds Green by 20%:
那么蓝色超过绿色20%:
And finally that the Saturation exceeds 60%:
最后,饱和度超过60%:
And then the result:
然后结果:
You'll need to put your -format "%[fx:mean*w*h]" info:
back on the end in place of the output image name to get the count of saturated blue pixels.
你需要将你的-format“%[fx:mean * w * h]”信息放回到最后代替输出图像名称以获得饱和蓝色像素的数量。
If I run your command:
如果我运行你的命令:
convert input.png -fx "u.b>(u.g+0.2)&&u.b>(u.r+0.2)&&saturation>0.6" result.png
My brain is not quite right today, so please run some checks - I may have something back-to-front somewhere!
我的大脑今天不太正确,所以请进行一些检查 - 我可能会在某个地方有一些东西!
As a benchmark, on a 10,000x10,000 pixel PNG, my code runs in 30 seconds, whereas the -fx
equivalent takes nearly 7 minutes.
作为基准,在10,000x10,000像素的PNG上,我的代码在30秒内运行,而-fx相当于将近7分钟。
#2
1
I don't know imagelagick part. But for node part I see that you call next non regarding to imagemgick opertion.
我不知道imagelagick的一部分。但是对于节点部分我看到你关于imagemgick opertion调用next non。
module.exports = function (File) {
File.observe('after save', function countPixels(ctx, next) {
if (ctx.instance && !ctx.instance.blue_pixels) {
var exec = require('child_process').exec;
// Convert file to retrieve only blue pixels:
exec('convert ' + ctx.instance.path + ' -fx "u.b>(u.g+0.2)&&u.b>(u.r+0.2)&&saturation>0.6" -format "%[fx:mean*w*h]" info:',
function (error, stdout, stderr) {
if (error !== null) {
return next(error);
} else {
ctx.instance.blue_pixels = stdout;
File.upsert(ctx.instance);
next();
}
});
}
else{next();}
//next(); //run next hook ASAP (before imagemagick returns back the result)
});
};