How to extract time from timestamp in pyspark? - apache-spark

I have a requirement to extract time from timestamp(this is a column in dataframe) using pyspark.
lets say this is the timestamp 2019-01-03T18:21:39 , I want to extract only time "18:21:39" such that it always appears in this manner "01:01:01"
df = spark.createDataFrame(["2020-06-17T00:44:30","2020-06-17T06:06:56","2020-06-17T15:04:34"],StringType()).toDF('datetime')
df=df.select(df['datetime'].cast(TimestampType()))
I tried like below but did not get the expected result
df1=df.withColumn('time',concat(hour(df['datetime']),lit(":"),minute(df['datetime']),lit(":"),second(df['datetime'])))
display(df1)
+-------------------+-------+
| datetime| time|
+-------------------+-------+
|2020-06-17 00:44:30|0:44:30|
|2020-06-17 06:06:56| 6:6:56|
|2020-06-17 15:04:34|15:4:34|
+-------------------+-------+
my results are like this 6:6:56 but i want them to be 06:06:56

Use the date_format function.
from pyspark.sql.types import StringType
df = spark \
.createDataFrame(["2020-06-17T00:44:30","2020-06-17T06:06:56","2020-06-17T15:04:34"], StringType()) \
.toDF('datetime')
from pyspark.sql.functions import date_format
q = df.withColumn('time', date_format('datetime', 'HH:mm:ss'))
>>> q.show()
+-------------------+--------+
| datetime| time|
+-------------------+--------+
|2020-06-17T00:44:30|00:44:30|
|2020-06-17T06:06:56|06:06:56|
|2020-06-17T15:04:34|15:04:34|
+-------------------+--------+

Related

Split column with JSON string to columns each containing one key-value pair from the string

I have a data frame that looks like this (one column named "value" with a JSON string in it). I send it to an Event Hub using Kafka API and then I want to read that data from the Event Hub and apply some transformations to it. The data is in received in binary format, as described in the Kafka documentation.
Here are a few columns in CSV format:
value
"{""id"":""e52f247c-f46c-4021-bc62-e28e56db1ad8"",""latitude"":""34.5016064725731"",""longitude"":""123.43996453687777""}"
"{""id"":""32782100-9b59-49c7-9d56-bb4dfc368a86"",""latitude"":""49.938541626415144"",""longitude"":""111.88360885971986""}"
"{""id"":""a72a600f-2b99-4c41-a388-9a24c00545c0"",""latitude"":""4.988768300413497"",""longitude"":""-141.92727675177588""}"
"{""id"":""5a5f056a-cdfd-4957-8e84-4d5271253509"",""latitude"":""41.802942545247134"",""longitude"":""90.45164573613573""}"
"{""id"":""d00d0926-46eb-45dd-9e35-ab765804340d"",""latitude"":""70.60161063520081"",""longitude"":""20.566520665122482""}"
"{""id"":""dda14397-6922-4bb6-9be3-a1546f08169d"",""latitude"":""68.400462882435"",""longitude"":""135.7167027587489""}"
"{""id"":""c7f13b8a-3468-4bc6-9db4-e0b1b34bf9ea"",""latitude"":""26.04757722355835"",""longitude"":""175.20227554031783""}"
"{""id"":""97f8f1cf-3aa0-49bb-b3d5-05b736e0c883"",""latitude"":""35.52624182094499"",""longitude"":""-164.18066699972852""}"
"{""id"":""6bed49bc-ee93-4ed9-893f-4f51c7b7f703"",""latitude"":""-24.319581484353847"",""longitude"":""85.27338980948076""}"
What I want to do is to apply a transformation and create a data frame with 3 columns one with id, one with latitude and one with longitude.
This is what I tried but the result is not what I expected:
from pyspark.sql.types import StructType
from pyspark.sql.functions import from_json
from pyspark.sql import functions as F
# df is the data frame received from Kafka
location_schema = StructType().add("id", "string").add("latitude", "float").add("longitude", "float")
string_df = df.selectExpr("CAST(value AS STRING)").withColumn("value", from_json(F.col("value"), location_schema))
string_df.printSchema()
string_df.show()
And this is the result:
It created a "value" column with a structure as a value. Any idea what to do to obtain 3 different columns, as I described?
Your df:
df = spark.createDataFrame(
[
(1, '{"id":"e52f247c-f46c-4021-bc62-e28e56db1ad8","latitude":"34.5016064725731","longitude":"123.43996453687777"}'),
(2, '{"id":"32782100-9b59-49c7-9d56-bb4dfc368a86","latitude":"49.938541626415144","longitude":"111.88360885971986"}'),
(3, '{"id":"a72a600f-2b99-4c41-a388-9a24c00545c0","latitude":"4.988768300413497","longitude":"-141.92727675177588"}'),
(4, '{"id":"5a5f056a-cdfd-4957-8e84-4d5271253509","latitude":"41.802942545247134","longitude":"90.45164573613573"}'),
(5, '{"id":"d00d0926-46eb-45dd-9e35-ab765804340d","latitude":"70.60161063520081","longitude":"20.566520665122482"}'),
(6, '{"id":"dda14397-6922-4bb6-9be3-a1546f08169d","latitude":"68.400462882435","longitude":"135.7167027587489"}'),
(7, '{"id":"c7f13b8a-3468-4bc6-9db4-e0b1b34bf9ea","latitude":"26.04757722355835","longitude":"175.20227554031783"}'),
(8, '{"id":"97f8f1cf-3aa0-49bb-b3d5-05b736e0c883","latitude":"35.52624182094499","longitude":"-164.18066699972852"}'),
(9, '{"id":"6bed49bc-ee93-4ed9-893f-4f51c7b7f703","latitude":"-24.319581484353847","longitude":"85.27338980948076"}')
],
['id', 'value']
).drop('id')
+--------------------------------------------------------------------------------------------------------------+
|value |
+--------------------------------------------------------------------------------------------------------------+
|{"id":"e52f247c-f46c-4021-bc62-e28e56db1ad8","latitude":"34.5016064725731","longitude":"123.43996453687777"} |
|{"id":"32782100-9b59-49c7-9d56-bb4dfc368a86","latitude":"49.938541626415144","longitude":"111.88360885971986"}|
|{"id":"a72a600f-2b99-4c41-a388-9a24c00545c0","latitude":"4.988768300413497","longitude":"-141.92727675177588"}|
|{"id":"5a5f056a-cdfd-4957-8e84-4d5271253509","latitude":"41.802942545247134","longitude":"90.45164573613573"} |
|{"id":"d00d0926-46eb-45dd-9e35-ab765804340d","latitude":"70.60161063520081","longitude":"20.566520665122482"} |
|{"id":"dda14397-6922-4bb6-9be3-a1546f08169d","latitude":"68.400462882435","longitude":"135.7167027587489"} |
|{"id":"c7f13b8a-3468-4bc6-9db4-e0b1b34bf9ea","latitude":"26.04757722355835","longitude":"175.20227554031783"} |
|{"id":"97f8f1cf-3aa0-49bb-b3d5-05b736e0c883","latitude":"35.52624182094499","longitude":"-164.18066699972852"}|
|{"id":"6bed49bc-ee93-4ed9-893f-4f51c7b7f703","latitude":"-24.319581484353847","longitude":"85.27338980948076"}|
+--------------------------------------------------------------------------------------------------------------+
Then:
from pyspark.sql import functions as F
from pyspark.sql.types import *
json_schema = StructType([
StructField("id", StringType(), True),
StructField("latitude", FloatType(), True),
StructField("longitude", FloatType(), True)
])
df\
.withColumn('json', F.from_json(F.col('value'), json_schema))\
.select(F.col('json').getItem('id').alias('id'),
F.col('json').getItem('latitude').alias('latitude'),
F.col('json').getItem('longitude').alias('longitude')
)\
.show(truncate=False)
+------------------------------------+-------------------+-------------------+
|id |latitude |longitude |
+------------------------------------+-------------------+-------------------+
|e52f247c-f46c-4021-bc62-e28e56db1ad8|34.5016064725731 |123.43996453687777 |
|32782100-9b59-49c7-9d56-bb4dfc368a86|49.938541626415144 |111.88360885971986 |
|a72a600f-2b99-4c41-a388-9a24c00545c0|4.988768300413497 |-141.92727675177588|
|5a5f056a-cdfd-4957-8e84-4d5271253509|41.802942545247134 |90.45164573613573 |
|d00d0926-46eb-45dd-9e35-ab765804340d|70.60161063520081 |20.566520665122482 |
|dda14397-6922-4bb6-9be3-a1546f08169d|68.400462882435 |135.7167027587489 |
|c7f13b8a-3468-4bc6-9db4-e0b1b34bf9ea|26.04757722355835 |175.20227554031783 |
|97f8f1cf-3aa0-49bb-b3d5-05b736e0c883|35.52624182094499 |-164.18066699972852|
|6bed49bc-ee93-4ed9-893f-4f51c7b7f703|-24.319581484353847|85.27338980948076 |
+------------------------------------+-------------------+-------------------+
If pattern remains unchanged then you can use regexp_replace()
>>> df = spark.read.option("header",False).option("inferSchema",True).csv("/dir1/dir2/Sample2.csv")
>>> df.show(truncate=False)
+-------------------------------------------------+------------------------------------+---------------------------------------+
|_c0 |_c1 |_c2 |
+-------------------------------------------------+------------------------------------+---------------------------------------+
|"{""id"":""e52f247c-f46c-4021-bc62-e28e56db1ad8""|""latitude"":""34.5016064725731"" |""longitude"":""123.43996453687777""}" |
|"{""id"":""32782100-9b59-49c7-9d56-bb4dfc368a86""|""latitude"":""49.938541626415144"" |""longitude"":""111.88360885971986""}" |
|"{""id"":""a72a600f-2b99-4c41-a388-9a24c00545c0""|""latitude"":""4.988768300413497"" |""longitude"":""-141.92727675177588""}"|
|"{""id"":""5a5f056a-cdfd-4957-8e84-4d5271253509""|""latitude"":""41.802942545247134"" |""longitude"":""90.45164573613573""}" |
|"{""id"":""d00d0926-46eb-45dd-9e35-ab765804340d""|""latitude"":""70.60161063520081"" |""longitude"":""20.566520665122482""}" |
|"{""id"":""dda14397-6922-4bb6-9be3-a1546f08169d""|""latitude"":""68.400462882435"" |""longitude"":""135.7167027587489""}" |
|"{""id"":""c7f13b8a-3468-4bc6-9db4-e0b1b34bf9ea""|""latitude"":""26.04757722355835"" |""longitude"":""175.20227554031783""}" |
|"{""id"":""97f8f1cf-3aa0-49bb-b3d5-05b736e0c883""|""latitude"":""35.52624182094499"" |""longitude"":""-164.18066699972852""}"|
|"{""id"":""6bed49bc-ee93-4ed9-893f-4f51c7b7f703""|""latitude"":""-24.319581484353847""|""longitude"":""85.27338980948076""}" |
+-------------------------------------------------+------------------------------------+---------------------------------------+
>>> df.withColumn("id",regexp_replace('_c0','\"\{\"\"id\"\":\"\"','')).withColumn("id",regexp_replace('id','\"\"','')).withColumn("latitude",regexp_replace('_c1','\"\"latitude\"\":\"\"','')).withColumn("latitude",regexp_replace('latitude','\"\"','')).withColumn("longitude",regexp_replace('_c2','\"\"longitude\"\":\"\"','')).withColumn("longitude",regexp_replace('longitude','\"\"\}\"','')).drop("_c0").drop("_c1").drop("_c2").show()
+--------------------+-------------------+-------------------+
| id| latitude| longitude|
+--------------------+-------------------+-------------------+
|e52f247c-f46c-402...| 34.5016064725731| 123.43996453687777|
|32782100-9b59-49c...| 49.938541626415144| 111.88360885971986|
|a72a600f-2b99-4c4...| 4.988768300413497|-141.92727675177588|
|5a5f056a-cdfd-495...| 41.802942545247134| 90.45164573613573|
|d00d0926-46eb-45d...| 70.60161063520081| 20.566520665122482|
|dda14397-6922-4bb...| 68.400462882435| 135.7167027587489|
|c7f13b8a-3468-4bc...| 26.04757722355835| 175.20227554031783|
|97f8f1cf-3aa0-49b...| 35.52624182094499|-164.18066699972852|
|6bed49bc-ee93-4ed...|-24.319581484353847| 85.27338980948076|
+--------------------+-------------------+-------------------+
You can use json_tuple to extract values from JSON string.
Input:
from pyspark.sql import functions as F
df = spark.createDataFrame(
[('{"id":"e52f247c-f46c-4021-bc62-e28e56db1ad8","latitude":"34.5016064725731","longitude":"123.43996453687777"}',)],
['value'])
Script:
cols = ['id', 'latitude', 'longitude']
df = df.select(F.json_tuple('value', *cols)).toDF(*cols)
df.show(truncate=0)
# +------------------------------------+----------------+------------------+
# |id |latitude |longitude |
# +------------------------------------+----------------+------------------+
# |e52f247c-f46c-4021-bc62-e28e56db1ad8|34.5016064725731|123.43996453687777|
# +------------------------------------+----------------+------------------+
If needed, cast to double:
.withColumn('latitude', F.col('latitude').cast('double'))
.withColumn('longitude', F.col('longitude').cast('double'))
It's easy to extract JSON string as columns using inline and from_json
df = spark.createDataFrame(
[('{"id":"e52f247c-f46c-4021-bc62-e28e56db1ad8","latitude":"34.5016064725731","longitude":"123.43996453687777"}',)],
['value'])
df = df.selectExpr(
"inline(array(from_json(value, 'struct<id:string, latitude:string, longitude:string>')))"
)
df.show(truncate=0)
# +------------------------------------+----------------+------------------+
# |id |latitude |longitude |
# +------------------------------------+----------------+------------------+
# |e52f247c-f46c-4021-bc62-e28e56db1ad8|34.5016064725731|123.43996453687777|
# +------------------------------------+----------------+------------------+
I used the sample data provided, created a dataframe called df and proceeded to use the same method as you.
The following is the image of the rows present inside df dataframe.
The fields are not displayed as required because of the their datatype. The values for latitude and longitude are present as string types in the dataframe df. But while creating the schema location_schema you have specified their type as float. Instead, try changing their type to string and later convert them to double type. The code looks as shown below:
location_schema = StructType().add("id", "string").add("latitude", "string").add("longitude", "string")
string_df = df.selectExpr('CAST(value AS STRING)').withColumn("value", from_json(F.col("value"), location_schema))
string_df.printSchema()
string_df.show(truncate=False)
Now using DataFrame.withColumn(), Column.withField() and cast() convert the string type fields latitude and longitude to Double Type.
string_df = string_df.withColumn("value", col("value").withField("latitude", col("value.latitude").cast(DoubleType())))\
.withColumn("value", col("value").withField("longitude", col("value.longitude").cast(DoubleType())))
string_df.printSchema()
string_df.show(truncate=False)
So, you can get the desired output as shown below.
Update:
To get separate columns you can simply use json_tuple() method. Refer to this official spark documentation:
pyspark.sql.functions.json_tuple — PySpark 3.3.0 documentation (apache.org)

Change the day of the date to a particular day

I basically have a requirement that needs a column that as the PeriodEndDate in. The period always ends on the 23rd of the month.
I need to take a date from a column in this case it is the last day of the month each day, and set the "day" of that date to be "23".
I have tried doing the following:
.withColumn("periodEndDate", change_day(jsonDF2.periodDate, sf.lit(23)))
cannot import name 'change_day' from 'pyspark.sql.functions'
You can use make_date
from pyspark.sql import functions as F
df = spark.createDataFrame([('2022-05-31',)], ['periodDate'])
df = df.withColumn('periodEndDate', F.expr("make_date(year(periodDate), month(periodDate), 23)"))
df.show()
# +----------+-------------+
# |periodDate|periodEndDate|
# +----------+-------------+
# |2022-05-31| 2022-05-23|
# +----------+-------------+
As far as I know, there is no function change_day however, you can make one using UDF. Pass a date and replace day.
Example:
from datetime import datetime
from pyspark.sql import SparkSession
from pyspark.sql.types import TimestampType
from pyspark.sql import functions as F
def change_day(date, day):
return date.replace(day=day)
change_day = F.udf(change_day, TimestampType())
spark = SparkSession.builder.getOrCreate()
df = spark.createDataFrame([{"date": datetime(2022, 1, 31)}])
df = df.withColumn("23day", change_day(F.col("date"), F.lit(23)))
df.show(20, False)
Result:
+-------------------+-------------------+
|date |23day |
+-------------------+-------------------+
|2022-01-31 00:00:00|2022-01-23 00:00:00|
+-------------------+-------------------+

spark : how to read csv with ";" as delimiter and "," as decimal separation?

I have a problem with a csv file witch contains decimal value like that "7,27431439586819e-05"
spark.read.option("header", "true")\
.option("delimiter", ";")\
.option("locale", "fr-FR")\ *doesnt work...*
.option("inferSchema", "true")\
.csv("file.csv").toPandas()
The comma doesn't seem to be a standard comma and i dont find an option to specify the symbol (.option('decimal',',') doesnt exist, .option('locale','fr-FR') doesnt work)
Do you have any idea ? I also tried re.sub("[^0-9]", ".") and then realize i had scientific value ('e-') so it doenst work neither.
Try using the regexp_replace() function from pyspark to replace "," with "." then cast into DoubleType().
import pandas as pd
import pyspark.sql.functions as F
from pyspark.sql.types import DoubleType
df = pd.DataFrame({"Name": ['a', 'b', 'c'], "Measures":["7,27431439586819e-05", "15,4689439586819e-01", "-2,97431439586819e02"]})
dfs = spark.createDataFrame(df)
dfs_transformed = dfs.withColumn('Measures', F.regexp_replace('Measures', ',', '.').cast(DoubleType()))
dfs_transformed.show()
And you should get correctly typed values:
+----+-------------------+
|Name| Measures|
+----+-------------------+
| a|7.27431439586819E-5|
| b| 1.54689439586819|
| c| -297.431439586819|
+----+-------------------+

How to insert value in a column using the lit in Spark dataframe?

I have Spark dataframe. I am trying to insert value in new column using lit, but the value is not being inserted.
Example:
I am trying below code:
df:
+--------------------+----------+---------+
| Programname|Projectnum| Drug|
+--------------------+----------+---------+
|Non-Oncology Phar...|SR0480-000|Invokamet|
+--------------------+----------+---------+
from pyspark.sql.functions import lit
df=df.withColumn("CDE_rec_crdt_dt", lit([str(x.CDE_rec_crdt_dt) for x in df_active.select('CDE_rec_crdt_dt').distinct().collect()][0]))
The value of -
[str(x.CDE_rec_crdt_dt) for x in df_active.select('CDE_rec_crdt_dt').distinct().collect()][0] ---'2020-12-03'
Desired output:
df:
+--------------------+----------+---------+----------------+
| Programname|Projectnum| Drug|CDE_rec_crdt_dt |
+--------------------+----------+---------+----------------+
|Non-Oncology Phar...|SR0480-000|Invokamet|2020-12-03 |
+--------------------+----------+---------+----------------+
val = str(df_active.select('CDE_rec_crdt_dt').distinct().collect()[0][0])
df = df.withColumn(
"CDE_rec_crdt_dt",
lit(val)
)

Spark order by second field to perform timeseries function

I have a csv with a timeseries:
timestamp, measure-name, value, type, quality
1503377580,x.x-2.A,0.5281250,Float,GOOD
1503377340,x.x-1.B,0.0000000,Float,GOOD
1503377400,x.x-1.B,0.0000000,Float,GOOD
The measure-name should be my partition key and I would like to calculate a moving average with pyspark, here my code (for instance) to calculate the max
def mysplit(line):
ll = line.split(",")
return (ll[1],float(ll[2]))
text_file.map(lambda line: mysplit(line)).reduceByKey(lambda a, b: max(a , b)).foreach(print)
However, for the average I would like to respect the timestamp ordering.
How to order by a second column?
You need to use a window function on pyspark dataframes:
First you should transform your rdd to a dataframe:
from pyspark.sql import HiveContext
hc = HiveContext(sc)
df = hc.createDataFrame(text_file.map(lambda l: l.split(','), ['timestamp', 'measure-name', 'value', 'type', 'quality'])
Or load it directly as a dataframe:
local:
import pandas as pd
df = hc.createDataFrame(pd.read_csv(path_to_csv, sep=",", header=0))
from hdfs:
df = hc.read.format("com.databricks.spark.csv").option("delimiter", ",").load(path_to_csv)
Then use a window function:
from pyspark.sql import Window
import pyspark.sql.functions as psf
w = Window.orderBy('timestamp')
df.withColumn('value_rol_mean', psf.mean('value').over(w))
+----------+------------+--------+-----+-------+-------------------+
| timestamp|measure_name| value| type|quality| value_rol_mean|
+----------+------------+--------+-----+-------+-------------------+
|1503377340| x.x-1.B| 0.0|Float| GOOD| 0.0|
|1503377400| x.x-1.B| 0.0|Float| GOOD| 0.0|
|1503377580| x.x-2.A|0.528125|Float| GOOD|0.17604166666666665|
+----------+------------+--------+-----+-------+-------------------+
in .orderByyou can order by as many columns as you want

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