Do we lose performance or gain disk size significantly if I define lots of columns but use few but different set per row?
You will not lose performance. In fact, it's actually quite efficient (depending on your use case) since empty columns don't take up any space. Whether or not you'll "gain" significant disk size is subjective. It's more about the disk space you'll be saving in place of the alternative method you would use to avoid empty columns.
"having columns without values is virtually free in Cassandra"
http://www.datastax.com/dev/blog/cql3_collections
Related
If the answer is yes,
Does that mean unlike Mongo or RDMS, whether we retrieve every column or some column will have big performance impact in Cassandra?(I am not talking about transfer time over network as it will affect all of the above)
Does that mean during compaction, it cannot just stop when it finds the latest row for a primary key, it has to go through the full set in SSTables? (I understand there will be optimisations as previously compacted SSTable will have maximum one occurrence for row)
Please ask only one question per question.
That is entirely up to you. If you write one column value, it'll persist just that one. If you write them all, they will all persist, even if they are the same as the current value.
whether we retrieve every column or some column will have big performance impact
This is definitely the case. Queries for column values that are small or haven't been written to or deleted will be much faster than the opposite.
during compaction, it cannot just stop when it finds the latest row for a primary key, it has to go through the full set in SSTables?
Yes. And not just during compaction, but read queries will also check multiple SSTable files.
I had a question that is related to pyspark's repartitionBy() function which I originally posted in a comment on this question. I was asked to post it as a separate question, so here it is:
I understand that df.partitionBy(COL) will write all the rows with each value of COL to their own folder, and that each folder will (assuming the rows were previously distributed across all the partitions by some other key) have roughly the same number of files as were previously in the entire table. I find this behavior annoying. If I have a large table with 500 partitions, and I use partitionBy(COL) on some attribute columns, I now have for example 100 folders which each contain 500 (now very small) files.
What I would like is the partitionBy(COL) behavior, but with roughly the same file size and number of files as I had originally.
As demonstration, the previous question shares a toy example where you have a table with 10 partitions and do partitionBy(dayOfWeek) and now you have 70 files because there are 10 in each folder. I would want ~10 files, one for each day, and maybe 2 or 3 for days that have more data.
Can this be easily accomplished? Something like df.write().repartition(COL).partitionBy(COL) seems like it might work, but I worry that (in the case of a very large table which is about to be partitioned into many folders) having to first combine it to some small number of partitions before doing the partitionBy(COL) seems like a bad idea.
Any suggestions are greatly appreciated!
You've got several options. In my code below I'll assume you want to write in parquet, but of course you can change that.
(1) df.repartition(numPartitions, *cols).write.partitionBy(*cols).parquet(writePath)
This will first use hash-based partitioning to ensure that a limited number of values from COL make their way into each partition. Depending on the value you choose for numPartitions, some partitions may be empty while others may be crowded with values -- for anyone not sure why, read this. Then, when you call partitionBy on the DataFrameWriter, each unique value in each partition will be placed in its own individual file.
Warning: this approach can lead to lopsided partition sizes and lopsided task execution times. This happens when values in your column are associated with many rows (e.g., a city column -- the file for New York City might have lots of rows), whereas other values are less numerous (e.g., values for small towns).
(2) df.sort(sortCols).write.parquet(writePath)
This options works great when you want (1) the files you write to be of nearly equal sizes (2) exact control over the number of files written. This approach first globally sorts your data and then finds splits that break up the data into k evenly-sized partitions, where k is specified in the spark config spark.sql.shuffle.partitions. This means that all values with the same values of your sort key are adjacent to each other, but sometimes they'll span a split, and be in different files. This, if your use-case requires all rows with the same key to be in the same partition, then don't use this approach.
There are two extra bonuses: (1) by sorting your data its size on disk can often be reduced (e.g., sorting all events by user_id and then by time will lead to lots of repetition in column values, which aids compression) and (2) if you write to a file format the supports it (like Parquet) then subsequent readers can read data in optimally by using predicate push-down, because the parquet writer will write the MAX and MIN values of each column in the metadata, allowing the reader to skip rows if the query specifies values outside of the partition's (min, max) range.
Note that sorting in Spark is more expensive than just repartitioning and requires an extra stage. Behind the scenes Spark will first determine the splits in one stage, and then shuffle the data into those splits in another stage.
(3) df.rdd.partitionBy(customPartitioner).toDF().write.parquet(writePath)
If you're using spark on Scala, then you can write a customer partitioner, which can get over the annoying gotchas of the hash-based partitioner. Not an option in pySpark, unfortunately. If you really want to write a custom partitioner in pySpark, I've found this is possible, albeit a bit awkward, by using rdd.repartitionAndSortWithinPartitions:
df.rdd \
.keyBy(sort_key_function) \ # Convert to key-value pairs
.repartitionAndSortWithinPartitions(numPartitions=N_WRITE_PARTITIONS,
partitionFunc=part_func) \
.values() # get rid of keys \
.toDF().write.parquet(writePath)
Maybe someone else knows an easier way to use a custom partitioner on a dataframe in pyspark?
df.repartition(COL).write().partitionBy(COL)
will write out one file per partition. This will not work well if one of your partition contains a lot of data. e.g. if one partition contains 100GB of data, Spark will try to write out a 100GB file and your job will probably blow up.
df.repartition(2, COL).write().partitionBy(COL)
will write out a maximum of two files per partition, as described in this answer. This approach works well for datasets that are not very skewed (because the optimal number of files per partition is roughly the same for all partitions).
This answer explains how to write out more files for the partitions that have a lot of data and fewer files for the small partitions.
I am trying to implement moving average for a dataset containing a number of time series. Each column represents one parameter being measured, while one row contains all parameters measured in a second. So a row would look something like:
timestamp, parameter1, parameter2, ..., parameterN
I found a way to do something like that using window functions, but the following bugs me:
Partitioning Specification: controls which rows will be in the same partition with the given row. Also, the user might want to make sure all rows having the same value for the category column are collected to the same machine before ordering and calculating the frame. If no partitioning specification is given, then all data must be collected to a single machine.
The thing is, I don't have anything to partition by. So can I use this method to calculate moving average without the risk of collecting all the data on a single machine? If not, what is a better way to do it?
Every nontrivial Spark job demands partitioning. There is just no way around it if you want your jobs to finish before the apocalypse. The question is simple: When it comes time to do the inevitable aggregation (in your case, an average), how can you partition your data in such a way as to minimize shuffle by grouping as much related data as possible on the same machine?
My experience with moving averages is with stocks. In that case it's easy; the partition would be on the stock ticker symbol. After all, the calculation of the 50-Day Moving Average for Stock A has nothing to with that for Stock B, so those data don't need to be on the same machine. The obvious partition makes this simpler than your situation--not to mention that it only requires one data point (probably) per day (the closing price of the stock at the end of trading) while you have one per second.
So I can only say that you need to consider adding a feature to your data set whose sole purpose is to serve as a partition key even if it is irrelevant to what you're measuring. I would be surprised if there isn't one, but if not, then consider a time-based partition on days for example.
We are evaluating if we can migrate from SQL SERVER to cassandra for OLAP. As per the internal storage structure we can have wide rows. We almost need to access data by the date. We often need to access data within date range as we have financial data. If we use date as Partition key to support filter by date,we end up having less row with huge number of columns.
Will it hamper performance if we have millions of columns for a single row key in future as we process millions of transactions every day.
Do we need to have some changes in the access pattern to have more rows with less number of columns per row.
Need some performance insight to proceed in either direction
Using wide rows is typically fine with Cassandra, there are however a few things to consider:
Ensure that you don't reach the 2 billion column limit in any case
The whole wide row is stored on the same node: it needs to fit on the disk. Also, if you have some dates that are accessed more frequently then other dates (e.g. today) then you can create hotspots on the node that stores the data for that day.
Very wide rows can affect performance however: Aaron Morton from The Last Pickle has an interesting article about this: http://thelastpickle.com/blog/2011/07/04/Cassandra-Query-Plans.html
It is somewhat old, but I believe that the concepts are still valid.
For a good table design decision one needs to know all typical filter conditions. If you have any other fields you typically filter for as an exact match, you could add them to the partition key as well.
In my case I have a table structure like this:
table_1 {
entity_uuid text
,fk1_uuid text
,fk2_uuid text
,int_timestamp bigint
,cnt counter
,primary key (entity_uuid, fk1_uuid, fk2_uuid, int_timestamp)
}
The text columns are made up of random strings. However, only entity_uuid is truly random and evenly distributed. fk1_uuid and fk2_uuid have much lower cardinality and may be sparse (sometimes fk1_uuid=null or fk2_uuid=null).
In this case, I can either define only entity_uuid as the partition key or entity_uuid, fk1_uuid, fk2_uuid combination as the partition key.
And this is a LOOKUP-type of table, meaning we don't plan to do any aggregations/slice-dice based on this table. And the rows will be rotated out since we will be inserting with TTL defined for each row.
Can someone enlighten me:
What is the downside of having too many partition keys with very few
rows in each? Is there a hit/cost on the storage engine level?
My understanding is the cluster keys are ALWAYS sorted. Does that mean having text columns in a cluster will always incur tree
balancing cost?
Well you can tell where my heart lies by now. However, when all rows in a partition all TTL-ed out, that partition still lives, or is there a way they will be removed by the DB engine as well?
Thanks,
Bing
The major and possibly most significant difference between having big partitions and small partitions is the ability to do range scans. If you want to be able to do scan queries like
SELECT * FROM table_1 where entity_id = x and fk1_uuid > something
Then you'll need to have the clustering column for performance, otherwise this query would be difficult (a multi-get at best, full table scan at worst.) I've never heard of any cases where having too many partitions is a drag on performance but having too wide a partition (ie lots of clustering column values) can cause issues when you get into the 1B+ cell range.
In terms of the cost of clustering, it is basically free at write time (in memory sort is very very fast) but you can incur costs at read time as partitions become spread amongst various SSTables. Small partitions which are written once will not occur the merge penalty since they will most likely only exist in 1 SSTable.
TTL'd partitions will be removed but be sure to read up on GC_GRACE_SECONDS to see how Cassandra actually deals with removing data.
TL;DR
Everything is dependent on your read/write pattern
No Range Scans? No need for clustering keys
Yes Range Scans? Clustering keys a must