I have 10 million json files in s3 and I want load all the files in a PySpark DataFrame in databricks.
Below is the code that I have used:
schema = StructType([StructField("Col1",StringType(),True), ... , ..., ...])
df = spark.read.schema(schema).json("s3://data_bucket/json_files/")
After I counted the records I found that only 3858630 files are loaded in the dataframe. 6,141,370 files have not been loaded.
I don't know why not all the files are not being loaded.
I appreciate any help!
I would start with checking if those 6,141,370 files match the schema you specified exactly. If it doesn't, it could be nulled out.
I would chunk them into a few thousands of files and see if it works. If works, you can extend it to more files and merge the dataframes together.
Related
I'm using SPARK to read files in hdfs. There is a scenario, where we are getting files as chunks from legacy system in csv format.
ID1_FILENAMEA_1.csv
ID1_FILENAMEA_2.csv
ID1_FILENAMEA_3.csv
ID1_FILENAMEA_4.csv
ID2_FILENAMEA_1.csv
ID2_FILENAMEA_2.csv
ID2_FILENAMEA_3.csv
This files are loaded to FILENAMEA in HIVE using HiveWareHouse Connector, with few transformation like adding default values. Similarly we have around 70 tables. Hive tables are created in ORC format. Tables are partitioned on ID. Right now, I'm processing all these files one by one. It's taking much time.
I want to make this process much faster. Files will be in GBs.
Is there is any way to read all the FILENAMEA files at the same time and load it to HIVE tables.
You have two methods to read several CSV files in pyspark. If all CSV files are in the same directory and all have the same schema, you can read then at once by directly passing the path of directory as argument, as follow:
spark.read.csv('hdfs://path/to/directory')
If you have CSV files in different locations or CSV files in same directory but with other CSV/text files in it, you can pass them as string representing a list of path in .csv() method argument, as follow:
spark.read.csv('hdfs://path/to/filename1,hdfs://path/to/filename2')
You can have more information about how to read a CSV file with Spark here
If you need to build this list of paths from the list of files in HDFS directory, you can look at this answer, once you've created your list of paths, you can transform it to a string to pass to .csv() method with ','.join(your_file_list)
Using: spark.read.csv(["path1","path2","path3"...]) you can read multiple files from different paths. But that means you have first to make a list of the paths. A list not a string of comma-separated file paths
I have CSV files from multiple paths that are not parent directories in s3 bucket. All the tables have the same partition keys.
the directory of the s3:
table_name_1/partition_key_1 = <pk_1>/partition_key_2 = <pk_2>/file.csv
table_name_2/partition_key_1 = <pk_1>/partition_key_2 = <pk_2>/file.csv
...
I need to convert these csv files into parquet files and store them in another s3 bucket that has the same directory structure.
the directory of another s3:
table_name_1/partition_key_1 = <pk_1>/partition_key_2 = <pk_2>/file.parquet
table_name_2/partition_key_1 = <pk_1>/partition_key_2 = <pk_2>/file.parquet
...
I have a solution is iterating through the s3 bucket and find the CSV file and convert it to parquet and save to the another S3 path. I find this way is not efficient, because i have a loop and did the conversion one file by one file.
I want to utilize the spark library to improve the efficiency.
Then, I tried:
spark.read.csv('s3n://bucket_name/table_name_1/').write.partitionBy('partition_key_1', 'partition_key_2').parquet('s3n://another_bucket/table_name_1')
This way works good for each table, but to optimize it more, I want to take the table_name as a parameter, something like:
TABLE_NAMES = [table_name_1, table_name_2, ...]
spark.read.csv('s3n://bucket_name/{*TABLE_NAMES}/').write.partitionBy('partition_key_1', 'partition_key_2').parquet('s3n://another_bucket/{*TABLE_NAMES}')
Thanks
The mentioned question provides solutions for reading multiple files at once. The method spark.read.csv(...) accepts one or multiple paths as shown here. For reading the files you can apply the same logic. Although, when it comes to writing, Spark will merge all the given dataset/paths into one Dataframe. Therefore it is not possible to generate from one single dataframe multiple dataframes without applying a custom logic first. So to conclude, there is not such a method for extracting the initial dataframe directly into multiple directories i.e df.write.csv(*TABLE_NAMES).
The good news is that Spark provides a dedicated function namely input_file_name() which returns the file path of the current record. You can use it in combination with TABLE_NAMES to filter on the table name.
Here it is one possible untested PySpark solution:
from pyspark.sql.functions import input_file_name
TABLE_NAMES = [table_name_1, table_name_2, ...]
source_path = "s3n://bucket_name/"
input_paths = [f"{source_path}/{t}" for t in TABLE_NAMES]
all_df = spark.read.csv(*input_paths) \
.withColumn("file_name", input_file_name()) \
.cache()
dest_path = "s3n://another_bucket/"
def write_table(table_name: string) -> None:
all_df.where(all_df["file_name"].contains(table_name))
.write
.partitionBy('partition_key_1','partition_key_2')
.parquet(f"{dest_path}/{table_name}")
for t in TABLE_NAMES:
write_table(t)
Explanation:
We generate and store the input paths into input_paths. This will create paths such as: s3n://bucket_name/table1, s3n://bucket_name/table2 ... s3n://bucket_name/tableN.
Then we load all the paths into one dataframe in which we add a new column called file_name, this will hold the path of each row. Notice that we also use cache here, this is important since we have multiple len(TABLE_NAMES) actions in the following code. Using cache will prevent us from loading the datasource again and again.
Next we create the write_table which is responsible for saving the data for the given table. The next step is to filter based on the table name using all_df["file_name"].contains(table_name), this will return only the records that contain the value of the table_name in the file_name column. Finally we save the filtered data as you already did.
In the last step we call write_table for every item of TABLE_NAMES.
Related links
How to import multiple csv files in a single load?
Get HDFS file path in PySpark for files in sequence file format
I am having issues loading multiple files into a dataframe in Databricks. When I load a parquet file in an individual folder, it is fine, but the following error returns when I try to load multiple files in the dataframe:
DF = spark.read.parquet('S3 path/')
"org.apache.spark.sql.AnalysisException: Unable to infer schema for Parquet. It must be specified manually."
Per other StackOverflow answers, I added spark.sql.files.ignoreCorruptFiles true to the cluster configuration but it didn't seem to resolve the issue. Any other ideas?
I have a directory of CSV files. The files are named based on date similar to the image below:
I have many CSV files that go back to 2012.
So, I would like to read the CSV files that correspond to a certain date only. How is that could be possible in spark? In other words, I don't want my spark engine to bother and read all CSV files because my data is huge (TBs).
Any help is much appreciated!
You can specify a list of files to be processed when calling the load(paths) or csv(paths) methods from DataFrameReader.
So an option would be to list and filter files on the driver, then load only the "recent" files :
val files: Seq[String] = ???
spark.read.option("header","true").csv(files:_*)
Edit :
You can use this python code (not tested yet)
files=['foo','bar']
df=spark.read.csv(*files)
I am using apapche spark. I want to access multiple json files from spark on date basis. How can i pick multiple files i.e. i want to provide range that files ending with 1034.json up to files ending with 1434.json. I am trying this.
DataFrame df = sql.read().json("s3://..../..../.....-.....[1034*-1434*]");
But i am getting the following error
at java.util.regex.Pattern.error(Pattern.java:1924)
at java.util.regex.Pattern.range(Pattern.java:2594)
at java.util.regex.Pattern.clazz(Pattern.java:2507)
at java.util.regex.Pattern.sequence(Pattern.java:2030)
at java.util.regex.Pattern.expr(Pattern.java:1964)
at java.util.regex.Pattern.compile(Pattern.java:1665)
at java.util.regex.Pattern.<init>(Pattern.java:1337)
at java.util.regex.Pattern.compile(Pattern.java:1022)
at org.apache.hadoop.fs.GlobPattern.set(GlobPattern.java:156)
at org.apache.hadoop.fs.GlobPattern.<init>(GlobPattern.java:42)
at org.apache.hadoop.fs.GlobFilter.init(GlobFilter.java:67)
Please specify a way out.
You can read something like this.
sqlContext.read().json("s3n://bucket/filepath/*.json")
Also, you can use wildcards in the file path.
For example:
sqlContext.read().json("s3n://*/*/*-*[1034*-1434*]")