I am trying to find the duplicate column value from dataframe in pyspark.
for example, I have a dataframe with single column 'A' with values like below:
==
A
==
1
1
2
3
4
5
5
I am expecting output like below(only duplicate values I need)
==
A
==
1
5
same answer as #Yuva but using the built-in functions :
df = sqlContext.createDataFrame([(1,),(1,),(2,),(3,),(4,),(5,),(5,)],('A',))
df.groupBy("A").count().where("count > 1").drop("count").show()
+---+
| A|
+---+
| 5|
| 1|
+---+
Can you try this, and see if this helps?
df = sqlContext.createDataFrame([(1,),(1,),(2,),(3,),(4,),(5,),(5,)],('A',))
df.createOrReplaceTempView(df_tbl)
spark.sql("select A, count(*) as COUNT from df_tbl group by a having COUNT > 1").show()
+---+-----+
| A|COUNT|
+---+-----+
| 5|2 |
| 1|2 |
+---+-----+
Related
I have assigned values to 4 variables in a conf or application.properties file,
A = 1
B = 2
C = 3
D = 4
I have a dataframe as follows,
+-----+
|name |
+-----+
| A |
| C |
| B |
| D |
| B |
+-----+
I want to add a new column that has the values assigned from the conf variables declared above for A,B,C,D respectively depending on the value in the name column.
Final Dataframe should have,
+----+----------+
|name|NAME_VALUE|
+----+----------+
| A | 1 |
| C | 3 |
| B | 2 |
| D | 4 |
| B | 2 |
+----+----------+
I tried lit function in .WITHCOLUMN with conf.getint($name), not accepting Column in lit func requires string, I have to hardcode the variable names in lit. Is there anyway for me to dynamically assign those respective conf variable names in LIT so it can automatically assign values to another column in spark scala?
For this moment i dont have any ideas how to do it as you intended with dynamic usage of vals names.
My proposition is to use a seq of tuples instead of multiple vals, in such case you can create some udf and try to map this value for each row, but you can also use join which i am showing in below example:
val data = Seq(("A"),("C"), ("B"), ("D"), ("B"))
val df = data.toDF("name")
val mappings = Seq(("A",1), ("B",2), ("C",3), ("D",4))
val mappingsDf = mappings.toDF("name", "value")
df.join(broadcast(mappingsDf), df("name") === mappingsDf("name"), "left")
.select(
df("name"),
mappingsDf("value")
).show
output is as expected:
+----+-----+
|name|value|
+----+-----+
| A| 1|
| C| 3|
| B| 2|
| D| 4|
| B| 2|
+----+-----+
This solution is pretty generic as your mapping are df here so you can hardcode them as showed in my example or load them from some csv or json easily with spark api
Due to broadcast join it should be quite efficient (you should remove this hint if you want to use big amount of mappings!)
I think its easy to understand and maintain as its not udf but only Spark api
I'm new to Pyspark and trying to transform data
Given dataframe
Col1
A=id1a A=id2a B=id1b C=id1c B=id2b
D=id1d A=id3a B=id3b C=id2c
A=id4a C=id3c
Required:
A B C
id1a id1b id1c
id2a id2b id2c
id3a id3b id3b
id4a null null
I have tried pivot, but that gives first value.
There might be a better way , however an approach is splitting the column on spaces to create array of the entries and then using higher order functions(spark 2.4+) to split on the '=' for each entry in the splitted array .Then explode and create 2 columns one with the id and one with the value. Then we can assign a row number to each partition and groupby then pivot:
import pyspark.sql.functions as F
df1 = (df.withColumn("Col1",F.split(F.col("Col1"),"\s+")).withColumn("Col1",
F.explode(F.expr("transform(Col1,x->split(x,'='))")))
.select(F.col("Col1")[0].alias("cols"),F.col("Col1")[1].alias("vals")))
from pyspark.sql import Window
w = Window.partitionBy("cols").orderBy("cols")
final = (df1.withColumn("Rnum",F.row_number().over(w)).groupBy("Rnum")
.pivot("cols").agg(F.first("vals")).orderBy("Rnum"))
final.show()
+----+----+----+----+----+
|Rnum| A| B| C| D|
+----+----+----+----+----+
| 1|id1a|id1b|id1c|id1d|
| 2|id2a|id2b|id2c|null|
| 3|id3a|id3b|id3c|null|
| 4|id4a|null|null|null|
+----+----+----+----+----+
this is how df1 looks like after the transformation:
df1.show()
+----+----+
|cols|vals|
+----+----+
| A|id1a|
| A|id2a|
| B|id1b|
| C|id1c|
| B|id2b|
| D|id1d|
| A|id3a|
| B|id3b|
| C|id2c|
| A|id4a|
| C|id3c|
+----+----+
May be I don't know the full picture, but the data format seems to be strange. If nothing can be done at the data source, then some collects, pivots and joins will be needed. Try this.
import pyspark.sql.functions as F
test = sqlContext.createDataFrame([('A=id1a A=id2a B=id1b C=id1c B=id2b',1),('D=id1d A=id3a B=id3b C=id2c',2),('A=id4a C=id3c',3)],schema=['col1','id'])
tst_spl = test.withColumn("item",(F.split('col1'," ")))
tst_xpl = tst_spl.select(F.explode("item"))
tst_map = tst_xpl.withColumn("key",F.split('col','=')[0]).withColumn("value",F.split('col','=')[1]).drop('col')
#%%
tst_pivot = tst_map.groupby(F.lit(1)).pivot('key').agg(F.collect_list(('value'))).drop('1')
#%%
tst_arr = [tst_pivot.select(F.posexplode(coln)).withColumnRenamed('col',coln) for coln in tst_pivot.columns]
tst_fin = reduce(lambda df1,df2:df1.join(df2,on='pos',how='full'),tst_arr).orderBy('pos')
tst_fin.show()
+---+----+----+----+----+
|pos| A| B| C| D|
+---+----+----+----+----+
| 0|id3a|id3b|id1c|id1d|
| 1|id4a|id1b|id2c|null|
| 2|id1a|id2b|id3c|null|
| 3|id2a|null|null|null|
+---+----+----+----+----
I am not sure if I am asking this correctly and maybe that is the reason why I didn't find the correct answer so far. Anyway, if it will be duplicate I will delete this question.
I have following data:
id | last_updated | count
__________________________
1 | 20190101 | 3
1 | 20190201 | 2
1 | 20190301 | 1
I want to group by this data by "id" column, get max value from "last_updated" and regarding "count" column I want keep value from row where "last_updated" has max value. So in that case result should be like that:
id | last_updated | count
__________________________
1 | 20190301 | 1
So I imagine it will look like that:
df
.groupBy("id")
.agg(max("last_updated"), ... ("count"))
Is there any function I can use to get "count" based on "last_updated" column.
I am using spark 2.4.0.
Thanks for any help
You have two options, the first the better as for my understanding
OPTION 1
Perform a window function over the ID, create a column with the max value over that window function. Then select where the desired column equals the max value and finally drop the column and rename the max column as desired
val w = Window.partitionBy("id")
df.withColumn("max", max("last_updated").over(w))
.where("max = last_updated")
.drop("last_updated")
.withColumnRenamed("max", "last_updated")
OPTION 2
You can perform a join with the original dataframe after grouping
df.groupBy("id")
.agg(max("last_updated").as("last_updated"))
.join(df, Seq("id", "last_updated"))
QUICK EXAMPLE
INPUT
df.show
+---+------------+-----+
| id|last_updated|count|
+---+------------+-----+
| 1| 20190101| 3|
| 1| 20190201| 2|
| 1| 20190301| 1|
+---+------------+-----+
OUTPUT
Option 1
import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions
val w = Window.partitionBy("id")
df.withColumn("max", max("last_updated").over(w))
.where("max = last_updated")
.drop("last_updated")
.withColumnRenamed("max", "last_updated")
+---+-----+------------+
| id|count|last_updated|
+---+-----+------------+
| 1| 1| 20190301|
+---+-----+------------+
Option 2
df.groupBy("id")
.agg(max("last_updated").as("last_updated")
.join(df, Seq("id", "last_updated")).show
+---+-----------------+----------+
| id| last_updated| count |
+---+-----------------+----------+
| 1| 20190301| 1|
+---+-----------------+----------+
I have a PySpark dataframe with a column contains Python list
id value
1 [1,2,3]
2 [1,2]
I want to remove all rows with len of the list in value column is less than 3.
So I tried:
df.filter(len(df.value) >= 3)
and indeed it does not work.
How can I filter the dataframe by the length of the inside data?
Refer to this link -
size() - It returns the length of the array or map stored in the column.
from pyspark.sql.functions import size
myValues = [(1,[1,2,3]),(2,[1,2])]
df = sqlContext.createDataFrame(myValues,['id','value'])
df.show()
+----+---------+
| id| value|
+--------------+
| 1| [1,2,3]|
| 2| [1,2]|
+----+---------+
df = df.filter(size(df.value) >= 3).show()
+----+---------+
| id| value|
+--------------+
| 1| [1,2,3]|
+----+---------+
This question already has answers here:
How to select the first row of each group?
(9 answers)
Closed 5 years ago.
I have a dataframe df as mentioned below:
**customers** **product** **val_id** **rule_name** **rule_id** **priority**
1 A 1 ABC 123 1
3 Z r ERF 789 2
2 B X ABC 123 2
2 B X DEF 456 3
1 A 1 DEF 456 2
I want to create a new dataframe df2, which will have only unique customer ids, but as rule_name and rule_id columns are different for same customer in data, so I want to pick those records which has highest priority for the same customer, so my final outcome should be:
**customers** **product** **val_id** **rule_name** **rule_id** **priority**
1 A 1 ABC 123 1
3 Z r ERF 789 2
2 B X ABC 123 2
Can anyone please help me to achieve it using Spark scala. Any help will be appericiated.
You basically want to select rows with extreme values in a column. This is a really common issue, so there's even a whole tag greatest-n-per-group. Also see this question SQL Select only rows with Max Value on a Column which has a nice answer.
Here's an example for your specific case.
Note that this could select multiple rows for a customer, if there are multiple rows for that customer with the same (minimum) priority value.
This example is in pyspark, but it should be straightforward to translate to Scala
# find best priority for each customer. this DF has only two columns.
cusPriDF = df.groupBy("customers").agg( F.min(df["priority"]).alias("priority") )
# now join back to choose only those rows and get all columns back
bestRowsDF = df.join(cusPriDF, on=["customers","priority"], how="inner")
To create df2 you have to first order df by priority and then find unique customers by id. Like this:
val columns = df.schema.map(_.name).filterNot(_ == "customers").map(col => first(col).as(col))
val df2 = df.orderBy("priority").groupBy("customers").agg(columns.head, columns.tail:_*).show
It would give you expected output:
+----------+--------+-------+----------+--------+---------+
| customers| product| val_id| rule_name| rule_id| priority|
+----------+--------+-------+----------+--------+---------+
| 1| A| 1| ABC| 123| 1|
| 3| Z| r| ERF| 789| 2|
| 2| B| X| ABC| 123| 2|
+----------+--------+-------+----------+--------+---------+
Corey beat me to it, but here's the Scala version:
val df = Seq(
(1,"A","1","ABC",123,1),
(3,"Z","r","ERF",789,2),
(2,"B","X","ABC",123,2),
(2,"B","X","DEF",456,3),
(1,"A","1","DEF",456,2)).toDF("customers","product","val_id","rule_name","rule_id","priority")
val priorities = df.groupBy("customers").agg( min(df.col("priority")).alias("priority"))
val top_rows = df.join(priorities, Seq("customers","priority"), "inner")
+---------+--------+-------+------+---------+-------+
|customers|priority|product|val_id|rule_name|rule_id|
+---------+--------+-------+------+---------+-------+
| 1| 1| A| 1| ABC| 123|
| 3| 2| Z| r| ERF| 789|
| 2| 2| B| X| ABC| 123|
+---------+--------+-------+------+---------+-------+
You will have to use min aggregation on priority column grouping the dataframe by customers and then inner join the original dataframe with the aggregated dataframe and select the required columns.
val aggregatedDF = dataframe.groupBy("customers").agg(max("priority").as("priority_1"))
.withColumnRenamed("customers", "customers_1")
val finalDF = dataframe.join(aggregatedDF, dataframe("customers") === aggregatedDF("customers_1") && dataframe("priority") === aggregatedDF("priority_1"))
finalDF.select("customers", "product", "val_id", "rule_name", "rule_id", "priority").show
you should have the desired result