Hi,
Currently I have around 1bio docs in my bucket, but total mutation is 7G during create new index
I see in doc metadata, there’s revision info.
I do POC, when I update 1 document mutiple times, data node will keep all of the revision and being included inside mutation.
I’ve tried doing compaction, but total mutation still have those revisions for 1 doc.
How to remove that revisions from mutation?
I assume if I can remove those revisions data, my mutation for 1 bio doc only 1G
CMIIW
Thanks,
@Han_Chris1,
Can you please be more clear on what do you mean by “1 bio docs”. Also, when you say 7G total mutations, where do you see this 7G. Are you using an array index?
Thanks,
Varun
Hi @varun.velamuri,
I mean total documents inside my bucket is 1.000.000.000 ,
I saw total mutation inside index statistics during index creation.( total mutations remaining)
7G means 7.000.000.000 data
@Han_Chris1,
Ok. Total mutations remaining 7G does not necessarily mean that 7G mutations are remaining. Let’s say a document is updated 100 times. The data service has processed all the updates and maintains the final version of the document, increments the version number of the document to 100.
The the total mutations remaining will be calculated based on this version number of the document. You may see total mutations remaining as 100 but the data service might be maintaining only 1 version of the document (with version number 100) and sending only that document. In such a case, you will see a sudden reduction in total mutations remaining from 100 to 0 just after processing one document. Bucket compaction may not help in this case because there might not be anything to compact.
Thanks,
Varun
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Thanks @varun.velamuri for your clear explanation.
But 1 thing that I still curious, this means total index mutations will keep growing until unlimited value, but the real total mutation should be the same with total doc in bucket?
Thanks,
Han
@Han_Chris1,
Yes. The real total mutation count will eventually become same as the total documents in the bucket.
Thanks,
Varun
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