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From Jack Krupansky <>
Subject Re: Data Modeling: Partition Size and Query Efficiency
Date Tue, 05 Jan 2016 16:52:41 GMT
Jim, I don't quite get why you think you would need to query 50 partitions
to return merely hundreds or thousands of rows. Please elaborate. I mean,
sure, for that extreme 100th percentile, yes, you would query a lot of
partitions, but for the 90th percentile it would be just one. Even the 99th
percentile would just be one or at most a few.

It would help if you could elaborate on the actual access pattern - how
rapidly is the data coming in and from where. You can do just a little more
work at the app level and and use Cassandra more effectively.

As always, we look to queries to determine what the Cassandra data model
should look like, so elaborate what your app needs to see. What exactly is
the app querying for - a single key, a slice, or... what?

And, as always, you commonly need to store the data in multiple query
tables so that the data model matches the desired query pattern.

Are the row sizes very dynamic, with some extremely large, or is it just
the number of rows that is making size an issue?

Maybe let the app keep a small cache of active partitions and their current
size so that the app can decide when to switch to a new bucket. Do a couple
of extra queries when a key is not in that cache to determine what the
partition size and count to initialize the cache entry for a key. If
necessary, keep a separate table that tracks the partition size or maybe
just the (rough) row count to use to determine when a new partition is

-- Jack Krupansky

On Tue, Jan 5, 2016 at 11:07 AM, Jim Ancona <> wrote:

> Thanks for responding!
> My natural partition key is a customer id. Our customers have widely
> varying amounts of data. Since the vast majority of them have data that's
> small enough to fit in a single partition, I'd like to avoid imposing
> unnecessary overhead on the 99% just to avoid issues with the largest 1%.
> The approach to querying across multiple partitions you describe is pretty
> much what I have in mind. The trick is to avoid having to query 50
> partitions to return a few hundred or thousand rows.
> I agree that sequentially filling partitions is something to avoid. That's
> why I'm hoping someone can suggest a good alternative.
> Jim
> On Mon, Jan 4, 2016 at 8:07 PM, Clint Martin <
>> wrote:
>> You should endeavor to use a repeatable method of segmenting your data.
>> Swapping partitions every time you "fill one" seems like an anti pattern to
>> me. but I suppose it really depends on what your primary key is. Can you
>> share some more information on this?
>> In the past I have utilized the consistent hash method you described (add
>> an artificial row key segment by modulo some part of the clustering key by
>> a fixed position count) combined with a lazy evaluation cursor.
>> The lazy evaluation cursor essentially is set up to query X number of
>> partitions simultaneously, but to execute those queries only add needed to
>> fill the page size. To perform paging you have to know the last primary key
>> that was returned so you can use that to limit the next iteration.
>> You can trade latency for additional work load by controlling the number
>> of concurrent executions you do as the iterating occurs. Or you can
>> minimize the work on your cluster by querying each partition one at a time.
>> Unfortunately due to the artificial partition key segment you cannot
>> iterate or page in any particular order...(at least across partitions)
>> Unless your hash function can also provide you some ordering guarantees.
>> It all just depends on your requirements.
>> Clint
>> On Jan 4, 2016 10:13 AM, "Jim Ancona" <> wrote:
>>> A problem that I have run into repeatedly when doing schema design is
>>> how to control partition size while still allowing for efficient multi-row
>>> queries.
>>> We want to limit partition size to some number between 10 and 100
>>> megabytes to avoid operational issues. The standard way to do that is to
>>> figure out the maximum number of rows that your "natural partition key"
>>> will ever need to support and then add an additional artificial partition
>>> key that segments the rows sufficiently to get keep the partition size
>>> under the maximum. In the case of time series data, this is often done by
>>> bucketing by time period, i.e. creating a new partition every minute, hour
>>> or day. For non-time series data by doing something like
>>> Hash(clustering-key) mod desired-number-of-partitions.
>>> In my case, multi-row queries to support a REST API typically return a
>>> page of results, where the page size might be anywhere from a few dozen up
>>> to thousands. For query efficiency I want the average number of rows per
>>> partition to be large enough that a query can be satisfied by reading a
>>> small number of partitions--ideally one.
>>> So I want to simultaneously limit the maximum number of rows per
>>> partition and yet maintain a large enough average number of rows per
>>> partition to make my queries efficient. But with my data the ratio between
>>> maximum and average can be very large (up to four orders of magnitude).
>>> Here is an example:
>>> Rows per Partition
>>> Partition Size
>>> Mode
>>> 1
>>> 1 KB
>>> Median
>>> 500
>>> 500 KB
>>> 90th percentile
>>> 5,000
>>> 5 MB
>>> 99th percentile
>>> 50,000
>>> 50 MB
>>> Maximum
>>> 2,500,000
>>> 2.5 GB
>>> In this case, 99% of my data could fit in a single 50 MB partition. But
>>> if I use the standard approach, I have to split my partitions into 50
>>> pieces to accommodate the largest data. That means that to query the 700
>>> rows for my median case, I have to read 50 partitions instead of one.
>>> If you try to deal with this by starting a new partition when an old one
>>> fills up, you have a nasty distributed consensus problem, along with
>>> read-before-write. Cassandra LWT wasn't available the last time I dealt
>>> with this, but might help with the consensus part today. But there are
>>> still some nasty corner cases.
>>> I have some thoughts on other ways to solve this, but they all have
>>> drawbacks. So I thought I'd ask here and hope that someone has a better
>>> approach.
>>> Thanks in advance,
>>> Jim

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