I am new to the timescale database. I was learning about chunks and how to create chunks based on time.
But there is another time/space chunking which is confusing me a lot. Please help me with below queries.
What is a dimension in a timescale DB?
What is space chunking and how it works?
Thanks in advance.
A dimension in TimescaleDB is associated with a column. Each hypertable requires to define at least a time dimension, which is a time column for the time series. Then a hypertable is divided into chunks, where each chunk contains data for a time interval of the time dimension. As result all new data usually arrives into the latets chunk, while other chunks contain older data.
Then, it is possible to define space dimensions on other columns, for example device column or/and location column. No interval is defined for space dimensions, instead a number of partitions is defined. So for the same time interval, several chunks will be created, which is equivalent to the number of partitions. Data are distributed by a hashing function on the values of the space dimension. For example, if 3 partitions are defined for a space dimension on device column and 12 different device values were present in the data, each space chunk will contain 4 different values with a hash function uniformly distributing the values.
Space dimensions are specifically useful for parallel I/O, when data are stored on several disks. Another scenario is multinode, i.e., distributed version of hypertable (beta feature, which coming to release in 2.0).
There are some complex usage cases when space partitioning will be also helpful.
You can read more in add_dimension docs, cloud KB about space partitioning
A note in the doc:
Supporting more than one additional dimension is currently experimental.
Related
AFAIK, in case of Relational Database on MPP hardware, the key to performance is a correct data distribution. While Dimensional Modeling is about query flexibility, you don't even know how the data will be queried (shuffled) in future.
For example, you have MPP Data Warehouse (Greenplum, Redshift, Synapse Analytics). For example, in 1-2 years, you expect your fact table will grow up to 10 billion of rows and you'll have 15-30 dimension tables of 10s millions of rows. How the data should be distributed accross DW nodes? Is there any common techniques? Like shard fact table and replicate dimension tables. Or should I minimize node amount in MPP DW?
I can bring specific use case, but I believe that the question arise from my misunderstanding of how Dimensional Modeling could be paired with scaling out.
One technique I’ve seen applied with success in the past is: segment the fact table (e.g., by mod’ing the date key), and distribute all dimensions across all nodes. That way all joins can be done locally.
Note that even with large dimensions, their total size on disk should be a small fraction of the total needed for the fact table.
after checking a lot of similar questions on stackoverflow, it seems that context will tell which way is the best to hold the data...
Short story, I add over 10,000 new rows of data in a very simple table containing only 3 columns. I will NEVER update the rows, only doing selects, grouping and making averages. I'm looking for the best way of storing this data to make the average calculations as fast as possible.
To put you in context, I'm analyzing a recorded audio file (Pink Noise playback in a sound mixing studio) using FFTs. The results for a single audio file is always in the same format: The frequency bin's ID (integer) and its value in decibels (float value). I'm want to store these values in a PostgreSQL DB.
Each bin (band) of frequencies (width = 8Hz) gets an amplitude in decibels. The first bin is ignored, so it goes like this (not actual dB values):
bin 1: 8Hz-16Hz, -85.0dB
bin 2: 16Hz-32Hz, -73.0dB
bin 3: 32Hz-40Hz, -65.0dB
...
bin 2499: 20,000Hz-20,008Hz, -49.0dB
The goal is to store an amplitude of each bin from 8Hz through 20,008Hz (1 bin covers 8Hz).
Many rows approach
For each analyzed audio file, there would be 2,499 rows of 3 columns: "Analysis UID", "Bin ID" and "dB".
For each studio (4), there is one recording daily that is to be appended in the database (that's 4 times 2,499 = 9,996 new rows per day).
After a recording in one studio, the new 2,499 rows are used to show a plot of the frequency response.
My concern is that we also need to make a plot of the averaged dB values of every bin in a single studio for 5-30 days, to see if the frequency response tends to change significantly over time (thus telling us that a calibration is needed in a studio).
I came up with the following data structure for the many rows approach:
"analysis" table:
analysisUID (serial)
studioUID (Foreign key)
analysisTimestamp
"analysis_results" table:
analysisUID (Foreign key)
freq_bin_id (integer)
amplitude_dB (float)
Is this the optimal way of storing data? A single table holding close to 10,000 new rows a day and making averages of 5 or more analysis, grouping by analysisUIDs and freq_bin_ids? That would give me 2,499 rows (each corresponding to a bin and giving me the averaged dB value).
Many columns approach:
I thought I could do it the other way around, breaking the frequency bins in 4 tables (Low, Med Low, Med High, High). Since Postgres documentation says the column limit is "250 - 1600 depending on column types", it would be realistic to make 4 tables containing around 625 columns (2,499 / 4) each representing a bin and containing the "dB" value, like so:
"low" table:
analysisUID (Foreign key)
freq_bin_id_1_amplitude_dB (float)
freq_bin_id_2_amplitude_dB (float)
...
freq_bin_id_625_amplitude_dB (float)
"med_low" table:
analysisUID (Foreign key)
freq_bin_id_626_amplitude_dB (float)
freq_bin_id_627_amplitude_dB (float)
...
freq_bin_id_1250_amplitude_dB (float)
etc...
Would the averages be computed faster if the server only has to Group by analysisUIDs and make averages of each column?
Rows are not going to be an issue, however, the way in which you insert said rows could be. If insert time is one of the primary concerns, then make sure you can bulk insert them OR go for a format with fewer rows.
You can potentially store all the data in a jsonb format, especially since you will not be doing any updates to the data-- it may be convenient to store it all in one table at a time, however the performance may be less.
In any case, since you're not updating the data, the (usually default) fillfactor of 100 is appropriate.
I would NOT use the "many column" approach, as the
amount of data you're talking about really isn't that much. Using your first example of 2 tables and few columns is very likely the optimal way to do your results.
It may be useful to index the following columns:
analysis_results.freq_bin_id
analysis.analysisTimestamp
As to breaking the data into different sections, it'll depend on what types of queries you're running. If you're looking at ALL freq bins, using multiple tables will just be a hassle and net you nothing.
If only querying at some freq_bin's at a time, it could theoretically help, however, you're basically doing table partitions and once you've moved into that land, you might as well make a partition for each frequency band.
If I were you, I'd create your first table structure, fill it with 30 days worth of data and query away. You may (as we often do) be overanalyzing the situation. Postgres can be very, very fast.
Remember, the raw data you're analyzing is something on the order of a few (5 or less) meg per day at an absolute maximum. Analyzing 150 mb of data is no sweat for a DB running with modern hardware if it's indexed and stored properly.
The optimizer is going to find the correct rows in the "smaller" table really, really fast and likely cache all of those, then go looking for the child rows, and it'll know exactly what ID's and ranges to search for. If your data is all inserted in chronological order, there's a good chance it'll read it all in very few reads with very few seeks.
My main concern is with the insert speed, as a doing 10,000 inserts can take a while if you're not doing bulk inserts.
Since the measurements seem well behaved, you could use an array, using the freq_bin as an index (Note: indices are 1-based in sql)
This has the additional advantage of the aray being stored in toasted storage, keeping the fysical table small.
CREATE TABLE herrie
( analysisUID serial NOT NULL PRIMARY KEY
, studioUID INTEGER NOT NULL REFERENCES studio(studioUID)
, analysisTimestamp TIMESTAMP NOT NULL
, decibels float[] -- array with 625 measurements
, UNIQUE (studioUID,analysisTimestamp)
);
So, I understand that in general one should use coalesce() when:
the number of partitions decreases due to a filter or some other operation that may result in reducing the original dataset (RDD, DF). coalesce() is useful for running operations more efficiently after filtering down a large dataset.
I also understand that it is less expensive than repartition as it reduces shuffling by moving data only if necessary. My problem is how to define the parameter that coalesce takes (idealPartionionNo). I am working on a project which was passed to me from another engineer and he was using the below calculation to compute the value of that parameter.
// DEFINE OPTIMAL PARTITION NUMBER
implicit val NO_OF_EXECUTOR_INSTANCES = sc.getConf.getInt("spark.executor.instances", 5)
implicit val NO_OF_EXECUTOR_CORES = sc.getConf.getInt("spark.executor.cores", 2)
val idealPartionionNo = NO_OF_EXECUTOR_INSTANCES * NO_OF_EXECUTOR_CORES * REPARTITION_FACTOR
This is then used with a partitioner object:
val partitioner = new HashPartitioner(idealPartionionNo)
but also used with:
RDD.filter(x=>x._3<30).coalesce(idealPartionionNo)
Is this the right approach? What is the main idea behind the idealPartionionNo value computation? What is the REPARTITION_FACTOR? How do I generally work to define that?
Also, since YARN is responsible for identifying the available executors on the fly is there a way of getting that number (AVAILABLE_EXECUTOR_INSTANCES) on the fly and use that for computing idealPartionionNo (i.e. replace NO_OF_EXECUTOR_INSTANCES with AVAILABLE_EXECUTOR_INSTANCES)?
Ideally, some actual examples of the form:
Here 's a dataset (size);
Here's a number of transformations and possible reuses of an RDD/DF.
Here is where you should repartition/coalesce.
Assume you have n executors with m cores and a partition factor equal to k
then:
The ideal number of partitions would be ==> ???
Also, if you can refer me to a nice blog that explains these I would really appreciate it.
In practice optimal number of partitions depends more on the data you have, transformations you use and overall configuration than the available resources.
If the number of partitions is too low you'll experience long GC pauses, different types of memory issues, and lastly suboptimal resource utilization.
If the number of partitions is too high then maintenance cost can easily exceed processing cost. Moreover, if you use non-distributed reducing operations (like reduce in contrast to treeReduce), a large number of partitions results in a higher load on the driver.
You can find a number of rules which suggest oversubscribing partitions compared to the number of cores (factor 2 or 3 seems to be common) or keeping partitions at a certain size but this doesn't take into account your own code:
If you allocate a lot you can expect long GC pauses and it is probably better to go with smaller partitions.
If a certain piece of code is expensive then your shuffle cost can be amortized by a higher concurrency.
If you have a filter you can adjust the number of partitions based on a discriminative power of the predicate (you make different decisions if you expect to retain 5% of the data and 99% of the data).
In my opinion:
With one-off jobs keep higher number partitions to stay on the safe side (slower is better than failing).
With reusable jobs start with conservative configuration then execute - monitor - adjust configuration - repeat.
Don't try to use fixed number of partitions based on the number of executors or cores. First understand your data and code, then adjust configuration to reflect your understanding.
Usually, it is relatively easy to determine the amount of raw data per partition for which your cluster exhibits stable behavior (in my experience it is somewhere in the range of few hundred megabytes, depending on the format, data structure you use to load data, and configuration). This is the "magic number" you're looking for.
Some things you have to remember in general:
Number of partitions doesn't necessarily reflect
data distribution. Any operation that requires shuffle (*byKey, join, RDD.partitionBy, Dataset.repartition) can result in non-uniform data distribution. Always monitor your jobs for symptoms of a significant data skew.
Number of partitions in general is not constant. Any operation with multiple dependencies (union, coGroup, join) can affect the number of partitions.
Your question is a valid one, but Spark partitioning optimization depends entirely on the computation you're running. You need to have a good reason to repartition/coalesce; if you're just counting an RDD (even if it has a huge number of sparsely populated partitions), then any repartition/coalesce step is just going to slow you down.
Repartition vs coalesce
The difference between repartition(n) (which is the same as coalesce(n, shuffle = true) and coalesce(n, shuffle = false) has to do with execution model. The shuffle model takes each partition in the original RDD, randomly sends its data around to all executors, and results in an RDD with the new (smaller or greater) number of partitions. The no-shuffle model creates a new RDD which loads multiple partitions as one task.
Let's consider this computation:
sc.textFile("massive_file.txt")
.filter(sparseFilterFunction) // leaves only 0.1% of the lines
.coalesce(numPartitions, shuffle = shuffle)
If shuffle is true, then the text file / filter computations happen in a number of tasks given by the defaults in textFile, and the tiny filtered results are shuffled. If shuffle is false, then the number of total tasks is at most numPartitions.
If numPartitions is 1, then the difference is quite stark. The shuffle model will process and filter the data in parallel, then send the 0.1% of filtered results to one executor for downstream DAG operations. The no-shuffle model will process and filter the data all on one core from the beginning.
Steps to take
Consider your downstream operations. If you're just using this dataset once, then you probably don't need to repartition at all. If you are saving the filtered RDD for later use (to disk, for example), then consider the tradeoffs above. It takes experience to become familiar with these models and when one performs better, so try both out and see how they perform!
As others have answered, there is no formula which calculates what you ask for. That said, You can make an educated guess on the first part and then fine tune it over time.
The first step is to make sure you have enough partitions. If you have NO_OF_EXECUTOR_INSTANCES executors and NO_OF_EXECUTOR_CORES cores per executor then you can process NO_OF_EXECUTOR_INSTANCES*NO_OF_EXECUTOR_CORES partitions at the same time (each would go to a specific core of a specific instance).
That said this assumes everything is divided equally between the cores and everything takes exactly the same time to process. This is rarely the case. There is a good chance that some of them would be finished before others either because of locallity (e.g. the data needs to come from a different node) or simply because they are not balanced (e.g. if you have data partitioned by root domain then partitions including google would probably be quite big). This is where the REPARTITION_FACTOR comes into play. The idea is that we "overbook" each core and therefore if one finishes very quickly and one finishes slowly we have the option of dividing the tasks between them. A factor of 2-3 is generally a good idea.
Now lets take a look at the size of a single partition. Lets say your entire data is X MB in size and you have N partitions. Each partition would be on average X/N MBs. If N is large relative to X then you might have very small average partition size (e.g. a few KB). In this case it is usually a good idea to lower N because the overhead of managing each partition becomes too high. On the other hand if the size is very large (e.g. a few GB) then you need to hold a lot of data at the same time which would cause issues such as garbage collection, high memory usage etc.
The optimal size is a good question but generally people seem to prefer partitions of 100-1000MB but in truth tens of MB probably would also be good.
Another thing you should note is when you do the calculation how your partitions change. For example, lets say you start with 1000 partitions of 100MB each but then filter the data so each partition becomes 1K then you should probably coalesce. Similar issues can happen when you do a groupby or join. In such cases both the size of the partition and the number of partitions change and might reach an undesirable size.
I am trying to create an estimate for how much space a table in Redshift is going to use, however, the only resources I found were in calculating the minimum table size:
https://aws.amazon.com/premiumsupport/knowledge-center/redshift-cluster-storage-space/
The purpose of this estimate is that I need to calculate how much space a table with the following dimensions is going to occupy without running out of space on Redshift (I.e. it will define how many nodes we end up using)
Rows : ~500 Billion (The exact number of rows is known)
Columns: 15 (The data types are known)
Any help in estimating this size would be greatly appreciated.
Thanks!
The article you reference (Why does a table in my Amazon Redshift cluster consume more disk storage space than expected?) does an excellent job of explaining how storage is consumed.
The main difficulty in predicting storage is predicting the efficiency of compression. Depending upon your data, Amazon Redshift will select an appropriate Compression Encoding that will reduce the storage space required by your data.
Compression also greatly improves the speed of Amazon Redshift queries by using Zone Maps, which identify the minimum and maximum value stored in each 1MB block. Highly compressed data will be stored on fewer blocks, thereby requiring less blocks to be read from disk during query execution.
The best way to estimate your storage space would be to load a subset of the data (eg 1 billion rows), allow Redshift to automatically select the compression types and then extrapolate to your full data size.
I had csv files of size 6GB and I tried using the import function on Matlab to load them but it failed due to memory issue. Is there a way to reduce the size of the files?
I think the no. of columns are causing the problem. I have a 133076 rows by 2329 columns. I had another file which is of the same no. of rows but only 12 rows and Matlab could handle that. However, once the columns increases, the files got really big.
Ulitmately, if I can read the data column wise so that I can have 2329 column vector of 133076, that will be great.
I am using Matlab 2014a
Numeric data are by default stored by Matlab in double precision format, which takes up 8 bytes per number. Data of size 133076 x 2329 therefore take up 2.3 GiB in memory. Do you have that much free memory? If not, reducing the file size won't help.
If the problem is not that the data themselves don't fit into memory, but is really about the process of reading such a large csv-file, then maybe using the syntax
M = csvread(filename,R1,C1,[R1 C1 R2 C2])
might help, which allows you to only read part of the data at one time. Read the data in chunks and assemble them in a (preallocated!) array.
If you do not have enough memory, another possibility is to read chunkwise and then convert each chunk to single precision before storing it. This reduces memory consumption by a factor of two.
And finally, if you don't process the data all at once, but can implement your algorithm such that it uses only a few rows or columns at a time, that same syntax may help you to avoid having all the data in memory at the same time.