I have multiple parquet files in S3 that are partitioned by date, like so:
s3://mybucket/myfolder/date=2022-01-01/file.parquet
s3://mybucket/myfolder/date=2022-01-02/file.parquet
and so on.
All of the files follow the same schema, except some which is why I am using the FILLRECORD (to fill the files with NULL values in case a column is not present). Now I want to load the content of all these files into an SQL temp table in redshift, like so:
DROP TABLE IF EXISTS table;
CREATE TEMP TABLE table
(
var1 bigint,
var2 bigint,
date timestamp
);
COPY table
FROM 's3://mybucket/myfolder/'
access_key_id 'id'secret_access_key 'key'
PARQUET FILLRECORD;
The problem is that the date column is not a column in the parquet files which is why the date column in the resulting table is NULL. I am trying to find a way to use the date to be inserted into the temp table.
Is there any way to do this?
I believe there are only 2 approaches to this:
Perform N COPY commands, one per S3 partition value, and populate the date column with the same information as the partition key value as a literal. A simple script can issue the SQL to Redshift. The issue with this is that you are issuing many COPY commands and if each partition in S3 has only 1 parquet file (or a few files) this will not take advantage of Redshift's parallelism.
Define the region of S3 with the partitioned parquet files as a Redshift partitioned external table and then INSERT INTO (SELECT * from );. The external table knows about the partition key and can insert this information into the local table. The down side is that you need to define the external schema and table and if this is a one time process, you will want to then tear these down after.
There are some other ways to attack this but none that are worth the effort or will be very slow.
I have come across Hive tables which I need to convert to Redshift/MySql equivalent.
I am having trouble understanding Hive query structure and would appreciate some help:
CREATE TABLE IF NOT EXISTS table_1 (
id BIGINT,
price DOUBLE,
asset string
)
PARTITIONED BY (
pt STRING
);
ALTER TABLE table_1 DROP IF EXISTS PARTITION (pt== '${yyyymmdd}');
INSERT OVERWRITE TABLE table_1 PARTITION (pt= '${yyyymmdd}')
select aa.id,aa.price,aa.symbol from
...
...
from
table_2 table
I am having trouble understanding the PARTITIONED BY clause. This, if I am understanding correctly, is different from MySQL table partitions, and is a Hive specific dynamic partition.
The partition does not define a column or a key, and partitions by the current date.
Does this mean that table_1 is partitioned by the date? Each day has a separate partition?
Then later on in the code there are notations similar to
inner join table_new table on table.pt = '${yyyymmdd}' and ...
In this context, does it mean only rows inserted on yyyymmdd are selected for the join?
Thank you.
Partition in Hive is a folder in HDFS by default with name key=value + metadata in the Hive metastore. You can alter partition location and create partition on top of any folder.
This PARTITIONED BY (pt STRING) defines partition column pt of type string, not date. Partition values are stored in the metadata. The pt column is not present in the table data files, it is only defined in PARTITIONED BY, all partition values are stored in the metadata. If you load partition dynamically, partition folder is being created with name pt='value'.
This sentence creates partition dynamically:
INSERT OVERWRITE TABLE table_1 PARTITION (pt)
select id, price, symbol
coln as pt --partition column should be the last one
from ...
And this sentence loads single STATIC partition:
INSERT OVERWRITE TABLE table_1 PARTITION (pt= '${yyyymmdd}')
select aa.id,aa.price,aa.symbol
from
No partition column is selected, partition value specified in the
PARTITION (pt= '${yyyymmdd}')
'${yyyymmdd}' here is a parameter with name yyyymmdd which is passed to the script using --hivevar like this:
hive --hivevar yyyymmdd=20200604 -f myscript.sql
You can pass ANY string as partition value in this case, though parameter name yyyymmdd suggests it's format.
BTW date format in hive is 'yyyy-MM-dd' Strings in 'yyyy-MM-dd' format can be implicitly converted to DATE.
I will try in one shot explain what is partitioning in Hive. First of all would be
WHEN TO USE TABLE PARTITIONING
Table partitioninig is good when:
Reading the entire dataset takes too long
Queries almost always filter on the partition columns
There are a reasonable number of different values for partition columns
Data generation of ETL process splits data by file or directory names
Partition column values are not in the data itself
Don't partition on columns with many unique values
Example: Partitioning customers by first name
CREATING PARTITIONED TABLES
To create a partitioned table, use the PARTITIONED BY clause in the CREATE TABLE statement.
The names and types of the partition columns must be specified
in the PARTITIONED BY clause, and only in the PARTITIONED BY clause.
They must not also appear in the list of all the other columns.
CREATE TABLE customers_by_country
(cust_id STRING, name STRING)
PARTITIONED BY (country STRING)
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t';
The example CREATE TABLE statement shown above creates the table customers_by_country,
which is partitioned by the STRING column named country.
Notice that the country column appears only in the PARTITIONED BY clause,
and not in the column list above it.
This example specifies only one partition column, but you can specify more than one by using
a comma-separated column list in the PARTITIONED BY clause.
Aside from these specific differences, this CREATE TABLE statement is the same
as the statement used to create an equivalent non-partitioned table.
Table partitioning is implemented in a way that is mostly transparent
to a user issuing queries with Hive.
A partition column is what’s known as a virtual column, because its values are not stored within the data files.
Following is the result of the DESCRIBE command on customers_by_country;
it displays the partition column country just as if it were a normal column within the table.
You can refer to partition columns in any of the usual clauses of a SELECT statement.
name type comment
cust_id string
name string
country string
You can load data in partitioned tables dynamically or statically
LOADING DATA WITH DYNAMIC PARTITION
One way to load data into a partitioned table is to use dynamic partitioning,
which automatically defines partitions when you load the data, using the values in the partition column.
(The other way is to manually define the partitions with Static Partitioning)
To use dynamic partitioning, you must load data using an INSERT statement.
In the INSERT statement, you must use the PARTITION clause to list the partition columns.
The data you are inserting must include values for the partition columns.
The partition columns must be the rightmost columns in the data you are inserting,
and they must be in the same order as they appear in the PARTITION clause.
INSERT OVERWRITE TABLE customers_by_country
PARTITION(country)
SELECT cust_id, name, country FROM customers;
The example shown above uses an INSERT … SELECT statement
to load data into the customers_by_country table with dynamic partitioning.
Notice that the partition column, country, is included
in the PARTITION clause and is specified last in the SELECT list.
When Hive executes this statement, it automatically creates partitions
for the country column and loads the data into these partitions based on the values in the country column.
The resulting data files in the partition subdirectories do not include values for the country column.
Since the country is known based on which subdirectory a data file is in,
it would be redundant to include country values in the data files as well.
Look at the contents of the customers_by_country directory.
It should now have one subdirectory for each value in the country column.
Look at the file in one of those directories.
Notice that the file contains the row for the customer from that country,
and no others; notice also that the country value is not included.
Note: Hive includes a safety feature that prevents users
from accidentally creating or overwriting a large number of partitions.
(See “Risks of Using Partitioning” for more about this.)
By default, Hive sets the property hive.exec.dynamic.partition.mode to strict.
This prevents you from using dynamic partitioning, though you can still use static partitions.
You can disable this safety feature in Hive by setting
the property hive.exec.dynamic.partition.mode to nonstrict:
SET hive.exec.dynamic.partition.mode=nonstrict;
Then you can use the INSERT statement to load the data dynamically.
Hive properties set in Beeline are for the current session only,
so the next time you start a Hive session this property will be set back to strict.
But you or your system administrator can configure properties permanently, if necessary.
When you run some SELECT queries on the partitioned table, if the table is big enough you can note significant difference in the time it takes to run.
Notice that you will not query the table any differently than you would query the customers table.
LOADING DATA WITH STATIC PARTITIONING
One way to load data into a partitioned table is to use static partitioning,
in which you manually define the different partitions.
With static partitioning, you create a partition manually, using an ALTER TABLE … ADD PARTITION statement,
and then load the data into the partition.
For example, this ALTER TABLE statement creates the partition for Pakistan (pk):
ALTER TABLE customers_by_country
ADD PARTITION (country='pk');
Notice how the partition column name, which is country, and the specific value that defines this partition,
which is pk, are both specified in the ADD PARTITION clause.
This creates a partition directory named country=pk inside the customers_by_country table directory.
After the partition for Pakistan is created, you can add data into the partition using an INSERT … SELECT statement:
INSERT OVERWRITE TABLE customers_by_country
PARTITION(country='pk')
SELECT cust_id, name FROM customers WHERE country='pk'
Notice how in the PARTITION clause, the partition column name, which is country,
and the specific value, which is pk, are both specified, just like in the ADD PARTITION command used to create the partition.
Also notice that in the SELECT statement, the partition column is not included in the SELECT list.
Finally, notice that the WHERE clause in the SELECT statement selects only customers from Pakistan.
With static partitioning, you need to repeat these two steps for each partition:
first create the partition, then add data.
You can actually use any method to load the data; you need not use an INSERT statement.
You could instead use hdfs dfs commands or a LOAD DATA INPATH command.
But however you load the data, it’s your responsibility to ensure that data is stored in the correct partition subdirectories.
For example, data for customers in Pakistan must be stored in the Pakistan partition subdirectory,
and data for customers in other countries must be stored in those countries’ partition subdirectories.
Static partitioning is most useful when the data being loaded
into the table is already divided into files based on the partition column,
or when the data grows in a manner that coincides with the partition column:
For example, suppose your company opens a new store in a different country,
like New Zealand ('nz'), and you're given a file of data for new customers, all from that country.
You could easily add a new partition and load that file into it.
RISKS OF USING PARTITIONING
A major risk when using partitioning is creating partitions that lead you into the small files problem.
When this happens, partitioning a table will actually worsen query performance
(the opposite of the goal when using partitioning) because it causes too many small files to be created.
This is more likely when using dynamic partitioning, but it could still
happen with static partitioning—for example if you added a new partition to a sales table
on a daily basis containing the sales from the previous day,
and each day’s data is not particularly big.
When choosing your partitions, you want to strike a happy balance between too many partitions
(causing the small files problem) and too few partitions (providing performance little benefit).
The partition column or columns should have a reasonable number of values
for the partitions—but what you should consider reasonable is difficult to quantify.
Using dynamic partitioning is particularly dangerous because if you're not careful,
it's easy to partition on a column with too many distinct values.
Imagine a use case where you are often looking for data that falls within
a time frame that you would specify in your query.
You might think that it's a good idea to partition on a column that pertains to time.
But a TIMESTAMP column could have the time to the nanosecond, so every row could have a unique value;
that would be a terrible choice for a partition column! Even to the minute or hour could create
far too many partitions, depending on the nature of your data;
partitioning by larger time units like day, month, or even year might be a better choice.
As another example, consider an employees table.
This has five columns: empl_id, first_name, last_name, salary, and office_id.
Before reading on, think for a moment, which of these might be reasonable for partitioning
The column empl_id is a unique identifier.
If that were your partition column, you would have a separate partition for each employee,
and each would have exactly one row.
In addition, it's not likely you'll be doing a lot of queries looking for a particular value,
or even a particular range of values. This is a poor choice.
The column first_name will not have one per employee, but there will likely be many columns that have only one row.
This is also true for last_name.
Also, like empl_id, it's not likely you'll need filter queries based on these columns. These are also poor choices.
The column salary also will have many divisions
(and even more so if your salaries go to the cent rather than to the dollar as our sample table does).
While it may be that you'll sometimes want to query on salary ranges,
it's not likely you'll want to use individual salaries.
So salary is a poor choice.
A more limited salary_grades specification, like the ones in the salary_grades table,
might be reasonable if your use case involves looking at the data by salary grade frequently.
The office_id column identifies the office where an employee works.
This will have a much smaller number of unique values, even if you have a large company with offices in many cities.
It's imaginable that your use case might be to frequently filter
your employee data based on office location, too. So this would be a good choice.
You also can use multiple columns and create nested partitions.
For example, a dataset of customers might include country and state_or_province columns.
You can partition by country and then partition those further by state_or_province, so customers from Ontario,
Canada would be in the country=ca/state_or_province=on/ partition directory.
This can be extremely helpful for large amounts of data that you want to access either by country or by state or province.
However, using multiple columns increases the danger of creating too many partitions, so you must take extra care when doing so.
The risk of creating too many partitions is why Hive includes the property
hive.exec.dynamic.partition.mode, set to strict by default, which must be reset to nonstrict before you can create a partition.
Rather than automatically and mechanically resetting that property when you're about to load data dynamically,
take it as an opportunity to think about the partitioning columns
and maybe check the number of unique values you would get when you load the data.
And that's all.
I am currently loading a hive table to pig relation using below code.
a = LOAD 'hive_db.hive_table' using org.apache.hive.hcatalog.pig.HCatLoader();
This step would get all the records from hive table into pig but for my current scenario I wouldn't need the whole table in pig. Is there way to filter out the unwanted records while I get the data from hive?
No you can't load partial table.However you can filter it after the load statement.You can use filter for specific partitions or filter out records based on column values in the table loaded.
Examples here
If your Hive table is partitioned, you can load only certain partitions by doing a FILTER statement immediately after your LOAD statement.
From the documentation:
If only some partitions of the specified table are needed, include a
partition filter statement immediately following the load statement in
the data flow. (In the script, however, a filter statement might not
immediately follow its load statement.) The filter statement can
include conditions on partition as well as non-partition columns.
A = LOAD 'tablename' USING org.apache.hive.hcatalog.pig.HCatLoader();
-- date is a partition column; age is not
B = filter A by date == '20100819' and age < 30;
The above will only load the partition date == '20100819'. This only works for partition columns.
How to rename a TABLE in Big query using StandardSQL or LegacySQL.
I'm trying with StandardSQL but it is giving following error,
RENAME TABLE dataset.old_table_name TO dataset.new_table_name;
Statement not supported: RenameStatement at [1:1]
Does it mean there is no any method(SQL QUERY) Which can rename a table?
I just want to change from non-partition table to partition-table
You can achieve this in two steps process
Step 1 - Export your table to Google Cloud Storage
Step 2 - Load file from GCS back to GBQ into new table with partitioned column
Both are free of charge
Still, have in mind some limitatins of partitioned tables - like number of partitions for example - it is 4000 per table as of today - https://cloud.google.com/bigquery/quotas#partitioned_tables
Currently it is not possible to rename table in Bigquery as explained in this document. You will have to create another table by following the steps given by Mikhail. Notice there is still some charge from table storage, but it is minimal. See this doc for detail information.
You can use the below query, it will create a new table with distinct records from old table with partition on given column.
create or replace table `dataset.new_table` PARTITION BY DATE(date_time_column) as select distinct * from `dataset.old_table`
I have a table created in hive stored in s3 location.
It has about 10 columns and is partitioned on 3 columns month, year and city , in the same order.
I am running a spark job that creates a dataframe(2 billion rows) and writes into this table.
val partitions:Seq[Column] = Seq(col("month"),col("year"),col("city"))
df.repartition(partitions: _*).selectExpr(cs.map(_.name): _*).write.mode("overwrite").insertInto(s"$tableName")
selectExpr(cs.map(_.name): _*) reorders the columns in the dataframe to align with the ordering in the table.
When i run the above command to insert into the table, I see there are so many staging files and multiple small files created under each city.
s3://s3Root/tableName/month/year/city/file1.csv
file2.csv
...
file200.csv
I am hoping to get a single file under each city per year per month.
To coalesce per partition.
Expected:
s3://s3Root/tableName/month/year/city/file.csv
Any help is appreciated.
to achieve one file by partition you should use
.partitionBy("")
val partitions:Seq[Column] = Seq(col("month"),col("year"),col("city"))
df.repartition(partitions: _*).selectExpr(cs.map(_.name): _*).write.partitionBy(partitions: _*).mode("overwrite").insertInto(s"$tableName")
I think that you could avoid to make the repartition before, if you do only the partitionBy,files will be partitioned by one per partition.