How to force celery tasks be executed in the same host?
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I'm working to implement an architecture based on Flask + Celery to process images. See image below:
The Celery client (producer) is a flask app that will receive images to be processed and push them into RabbitMQ to be consumed by celery workers.
The system can automatically start and stop new workers nodes based on the demand. Each celery worker node will be running with multiple threads (--concurrency parameter).
I'm thinking to have a single thread to consume from the default celery queue. Then any node is eligible to process the images.
Once a given node get/consume the image from the queue, the image will be placed locally in a in-memory database.
As I'm dealing with images, I want to avoid images flowing between different nodes through the network.
The image processing will be composed base many tasks and will be executed using a chain().
So, once I have a given image locally how can I force all the other sub-tasks (chain) be executed always in the same host ?
I know that I can create specific queues and assign those queues to workers in each node, but I don't know if this is the best/only solution, neither how to do it be dynamic.
Best Regards.
Kleyson Rios.
flask celery celery-task
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
add a comment |
I'm working to implement an architecture based on Flask + Celery to process images. See image below:
The Celery client (producer) is a flask app that will receive images to be processed and push them into RabbitMQ to be consumed by celery workers.
The system can automatically start and stop new workers nodes based on the demand. Each celery worker node will be running with multiple threads (--concurrency parameter).
I'm thinking to have a single thread to consume from the default celery queue. Then any node is eligible to process the images.
Once a given node get/consume the image from the queue, the image will be placed locally in a in-memory database.
As I'm dealing with images, I want to avoid images flowing between different nodes through the network.
The image processing will be composed base many tasks and will be executed using a chain().
So, once I have a given image locally how can I force all the other sub-tasks (chain) be executed always in the same host ?
I know that I can create specific queues and assign those queues to workers in each node, but I don't know if this is the best/only solution, neither how to do it be dynamic.
Best Regards.
Kleyson Rios.
flask celery celery-task
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
Possible duplicate of Route to worker depending on result in Celery?
– 2ps
Nov 18 '18 at 21:53
add a comment |
I'm working to implement an architecture based on Flask + Celery to process images. See image below:
The Celery client (producer) is a flask app that will receive images to be processed and push them into RabbitMQ to be consumed by celery workers.
The system can automatically start and stop new workers nodes based on the demand. Each celery worker node will be running with multiple threads (--concurrency parameter).
I'm thinking to have a single thread to consume from the default celery queue. Then any node is eligible to process the images.
Once a given node get/consume the image from the queue, the image will be placed locally in a in-memory database.
As I'm dealing with images, I want to avoid images flowing between different nodes through the network.
The image processing will be composed base many tasks and will be executed using a chain().
So, once I have a given image locally how can I force all the other sub-tasks (chain) be executed always in the same host ?
I know that I can create specific queues and assign those queues to workers in each node, but I don't know if this is the best/only solution, neither how to do it be dynamic.
Best Regards.
Kleyson Rios.
flask celery celery-task
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
I'm working to implement an architecture based on Flask + Celery to process images. See image below:
The Celery client (producer) is a flask app that will receive images to be processed and push them into RabbitMQ to be consumed by celery workers.
The system can automatically start and stop new workers nodes based on the demand. Each celery worker node will be running with multiple threads (--concurrency parameter).
I'm thinking to have a single thread to consume from the default celery queue. Then any node is eligible to process the images.
Once a given node get/consume the image from the queue, the image will be placed locally in a in-memory database.
As I'm dealing with images, I want to avoid images flowing between different nodes through the network.
The image processing will be composed base many tasks and will be executed using a chain().
So, once I have a given image locally how can I force all the other sub-tasks (chain) be executed always in the same host ?
I know that I can create specific queues and assign those queues to workers in each node, but I don't know if this is the best/only solution, neither how to do it be dynamic.
Best Regards.
Kleyson Rios.
flask celery celery-task
flask celery celery-task
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
asked Nov 16 '18 at 14:38
Kleyson RiosKleyson Rios
2112418
2112418
Possible duplicate of Route to worker depending on result in Celery?
– 2ps
Nov 18 '18 at 21:53
add a comment |
Possible duplicate of Route to worker depending on result in Celery?
– 2ps
Nov 18 '18 at 21:53
Possible duplicate of Route to worker depending on result in Celery?
– 2ps
Nov 18 '18 at 21:53
Possible duplicate of Route to worker depending on result in Celery?
– 2ps
Nov 18 '18 at 21:53
add a comment |
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Possible duplicate of Route to worker depending on result in Celery?
– 2ps
Nov 18 '18 at 21:53