I have two dataframes that need to be joined in a particular way I am struggling with.
dataframe 1:
+--------------------+---------+----------------+
| asset_domain| eid| oid|
+--------------------+---------+----------------+
| test-domain...| 126656| 126656|
| nebraska.aaa.com| 335660| 335660|
| netflix.com| 460| 460|
+--------------------+---------+----------------+
dataframe 2:
+--------------------+--------------------+---------+--------------+----+----+------------+
| asset| asset_domain|dns_count| ip| ev|post|form_present|
+--------------------+--------------------+---------+--------------+----+----+------------+
| sub1.test-domain...| test-domain...| 6354| 11.11.111.111| 1| 1| null|
| netflix.com| netflix.com| 3836| 22.22.222.222|null|null| null|
+--------------------+--------------------+---------+--------------+----+----+------------+
desired result:
+--------------------+---------+-------------+----+----+------------+---------+----------------+
| asset|dns_count| ip| ev|post|form_present| eid| oid|
+--------------------+---------+-------------+----+----+------------+---------+----------------+
| netflix.com| 3836|22.22.222.222|null|null| null| 460| 460|
| sub1.test-domain...| 5924|111.11.111.11| 1| 1| null| 126656| 126656|
| nebraska.aaa.com| null| null|null|null| null| 335660| 335660|
+--------------------+---------+-------------+----+----+------------+---------+----------------+
Basically – it should join df1 and df2 on asset_domain but if that doesn't exist in df2, then the resulting asset should be the asset_domain from df1.
I tried df = df2.join(df1, ["asset_domain"], "right").drop("asset_domain") but that obviously leaves null in the asset column for nebraska.aaa.com since it does not have a matching domain in df2. How do I go about adding those to the asset column for this particular case?
you can use coalesce function after join to create asset column.
df2.join(df1, ["asset_domain"], "right").select(coalesce("asset","asset_domain").alias("asset"),"dns_count","ip","ev","post","form_present","eid","oid").orderBy("asset").show()
#+----------------+---------+-------------+----+----+------------+------+------+
#| asset|dns_count| ip| ev|post|form_present| eid| oid|
#+----------------+---------+-------------+----+----+------------+------+------+
#|nebraska.aaa.com| null| null|null|null| null|335660|335660|
#| netflix.com| 3836|22.22.222.222|null|null| None| 460| 460|
#|sub1.test-domain| 6354|11.11.111.111| 1| 1| null|126656|126656|
#+----------------+---------+-------------+----+----+------------+------+------+
After the join you can use the isNull() function
import pyspark.sql.functions as F
tst1 = sqlContext.createDataFrame([('netflix',1),('amazon',2)],schema=("asset_domain",'xtra1'))
tst2= sqlContext.createDataFrame([('netflix','yahoo',1),('amazon','yahoo',2),('flipkart',None,2)],schema=("asset_domain","asset",'xtra'))
tst_j = tst1.join(tst2,on='asset_domain',how='right')
#%%
tst_res = tst_j.withColumn("asset",F.when(F.col('asset').isNull(),F.col('asset_domain')).otherwise(F.col('asset')))
Related
I would like to get the first and last row of each partition in spark (I'm using pyspark). How do I go about this?
In my code I repartition my dataset based on a key column using:
mydf.repartition(keyColumn).sortWithinPartitions(sortKey)
Is there a way to get the first row and last row for each partition?
Thanks
I would highly advise against working with partitions directly. Spark does a lot of DAG optimisation, so when you try executing specific functionality on each partition, all your assumptions about the partitions and their distribution might be completely false.
You seem to however have a keyColumn and sortKey, so then I'd just suggest to do the following:
import pyspark
import pyspark.sql.functions as f
w_asc = pyspark.sql.Window.partitionBy(keyColumn).orderBy(f.asc(sortKey))
w_desc = pyspark.sql.Window.partitionBy(keyColumn).orderBy(f.desc(sortKey))
res_df = mydf. \
withColumn("rn_asc", f.row_number().over(w_asc)). \
withColumn("rn_desc", f.row_number().over(w_desc)). \
where("rn_asc = 1 or rn_desc = 1")
The resulting dataframe will have 2 additional columns, where rn_asc=1 indicates the first row and rn_desc=1 indicates the last row.
Scala: I think the repartition is not by come key column but it requires the integer how may partition you want to set. I made a way to select the first and last row by using the Window function of the spark.
First, this is my test data.
+---+-----+
| id|value|
+---+-----+
| 1| 1|
| 1| 2|
| 1| 3|
| 1| 4|
| 2| 1|
| 2| 2|
| 2| 3|
| 3| 1|
| 3| 3|
| 3| 5|
+---+-----+
Then, I use the Window function twice, because I cannot know the last row easily but the reverse is quite easy.
import org.apache.spark.sql.expressions.Window
val a = Window.partitionBy("id").orderBy("value")
val d = Window.partitionBy("id").orderBy(col("value").desc)
val df = spark.read.option("header", "true").csv("test.csv")
df.withColumn("marker", when(rank.over(a) === 1, "Y").otherwise("N"))
.withColumn("marker", when(rank.over(d) === 1, "Y").otherwise(col("marker")))
.filter(col("marker") === "Y")
.drop("marker").show
The final result is then,
+---+-----+
| id|value|
+---+-----+
| 3| 5|
| 3| 1|
| 1| 4|
| 1| 1|
| 2| 3|
| 2| 1|
+---+-----+
Here is another approach using mapPartitions from RDD API. We iterate over the elements of each partition until we reach the end. I would expect this iteration to be very fast since we skip all the elements of the partition except the two edges. Here is the code:
df = spark.createDataFrame([
["Tom", "a"],
["Dick", "b"],
["Harry", "c"],
["Elvis", "d"],
["Elton", "e"],
["Sandra", "f"]
], ["name", "toy"])
def get_first_last(it):
first = last = next(it)
for last in it:
pass
# Attention: if first equals last by reference return only one!
if first is last:
return [first]
return [first, last]
# coalesce here is just for demonstration
first_last_rdd = df.coalesce(2).rdd.mapPartitions(get_first_last)
spark.createDataFrame(first_last_rdd, ["name", "toy"]).show()
# +------+---+
# | name|toy|
# +------+---+
# | Tom| a|
# | Harry| c|
# | Elvis| d|
# |Sandra| f|
# +------+---+
PS: Odd positions will contain the first partition element and the even ones the last item. Also note that the number of results will be (numPartitions * 2) - numPartitionsWithOneItem which I expect to be relatively small therefore you shouldn't bother about the cost of the new createDataFrame statement.
I'm trying to group by an ID column in a pyspark dataframe and sum a column depending on the value of another column.
To illustrate, consider the following dummy dataframe:
+-----+-------+---------+
| ID| type| amount|
+-----+-------+---------+
| 1| a| 55|
| 2| b| 1455|
| 2| a| 20|
| 2| b| 100|
| 3| null| 230|
+-----+-------+---------+
My desired output is:
+-----+--------+----------+----------+
| ID| sales| sales_a| sales_b|
+-----+--------+----------+----------+
| 1| 55| 55| 0|
| 2| 1575| 20| 1555|
| 3| 230| 0| 0|
+-----+--------+----------+----------+
So basically, sales will be the sum of amount, while sales_a and sales_b are the sum of amount when type is a or b respectively.
For sales, I know this could be done like this:
from pyspark.sql import functions as F
df = df.groupBy("ID").agg(F.sum("amount").alias("sales"))
For the others, I'm guessing F.when would be useful but I'm not sure how to go about it.
You could create two columns before the aggregation based off of the value of type.
df.withColumn("sales_a", F.when(col("type") == "a", col("amount"))) \
.withColumn("sales_b", F.when(col("type") == "b", col("amount"))) \
.groupBy("ID") \
.agg(F.sum("amount").alias("sales"),
F.sum("sales_a").alias("sales_a"),
F.sum("sales_b").alias("sales_b"))
from pyspark.sql import functions as F
df = df.groupBy("ID").agg(F.sum("amount").alias("sales"))
dfPivot = df.filter("type is not null").groupBy("ID").pivot("type").agg(F.sum("amount").alias("sales"))
res = df.join(dfPivot, df.id== dfPivot.id,how='left')
Then replace null with 0.
This is generic solution will work irrespective of values in type column.. so if type c is added in dataframe then it will create column _c
Related question: How to drop columns which have same values in all rows via pandas or spark dataframe?
So I have a pyspark dataframe, and I want to drop the columns where all values are the same in all rows while keeping other columns intact.
However the answers in the above question are only for pandas. Is there a solution for pyspark dataframe?
Thanks
You can apply the countDistinct() aggregation function on each column to get count of distinct values per column. Column with count=1 means it has only 1 value in all rows.
# apply countDistinct on each column
col_counts = df.agg(*(countDistinct(col(c)).alias(c) for c in df.columns)).collect()[0].asDict()
# select the cols with count=1 in an array
cols_to_drop = [col for col in df.columns if col_counts[col] == 1 ]
# drop the selected column
df.drop(*cols_to_drop).show()
You can use approx_count_distinct function (link) to count the number of distinct elements in a column. In case there is just one distinct, the remove the corresponding column.
Creating the DataFrame
from pyspark.sql.functions import approx_count_distinct
myValues = [(1,2,2,0),(2,2,2,0),(3,2,2,0),(4,2,2,0),(3,1,2,0)]
df = sqlContext.createDataFrame(myValues,['value1','value2','value3','value4'])
df.show()
+------+------+------+------+
|value1|value2|value3|value4|
+------+------+------+------+
| 1| 2| 2| 0|
| 2| 2| 2| 0|
| 3| 2| 2| 0|
| 4| 2| 2| 0|
| 3| 1| 2| 0|
+------+------+------+------+
Couting number of distinct elements and converting it into dictionary.
count_distinct_df=df.select([approx_count_distinct(x).alias("{0}".format(x)) for x in df.columns])
count_distinct_df.show()
+------+------+------+------+
|value1|value2|value3|value4|
+------+------+------+------+
| 4| 2| 1| 1|
+------+------+------+------+
dict_of_columns = count_distinct_df.toPandas().to_dict(orient='list')
dict_of_columns
{'value1': [4], 'value2': [2], 'value3': [1], 'value4': [1]}
#Storing those keys in the list which have just 1 distinct key.
distinct_columns=[k for k,v in dict_of_columns.items() if v == [1]]
distinct_columns
['value3', 'value4']
Drop the columns having distinct values
df=df.drop(*distinct_columns)
df.show()
+------+------+
|value1|value2|
+------+------+
| 1| 2|
| 2| 2|
| 3| 2|
| 4| 2|
| 3| 1|
+------+------+
This question already has answers here:
How to avoid duplicate columns after join?
(10 answers)
Closed 4 years ago.
I want to use join with 3 dataframe, but there are some columns we don't need or have some duplicate name with other dataframes, so I want to drop some columns like below:
result_df = (aa_df.join(bb_df, 'id', 'left')
.join(cc_df, 'id', 'left')
.withColumnRenamed(bb_df.status, 'user_status'))
Please note that status column is in two dataframes, i.e. aa_df and bb_df.
The above doesn't work. I also tried to use withColumn, but the new column is created, and the old column is still existed.
If you are trying to rename the status column of bb_df dataframe then you can do so while joining as
result_df = aa_df.join(bb_df.withColumnRenamed('status', 'user_status'),'id', 'left').join(cc_df, 'id', 'left')
I want to use join with 3 dataframe, but there are some columns we don't need or have some duplicate name with other dataframes
That's a fine use case for aliasing a Dataset using alias or as operators.
alias(alias: String): Dataset[T] or alias(alias: Symbol): Dataset[T]
Returns a new Dataset with an alias set. Same as as.
as(alias: String): Dataset[T] or as(alias: Symbol): Dataset[T]
Returns a new Dataset with an alias set.
(And honestly I did only now see the Symbol-based variants.)
NOTE There are two as operators, as for aliasing and as for type mapping. Consult the Dataset API.
After you've aliases a Dataset, you can reference columns using [alias].[columnName] format. This is particularly handy with joins and star column dereferencing using *.
val ds1 = spark.range(5)
scala> ds1.as('one).select($"one.*").show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
val ds2 = spark.range(10)
// Using joins with aliased datasets
// where clause is in a longer form to demo how ot reference columns by alias
scala> ds1.as('one).join(ds2.as('two)).where($"one.id" === $"two.id").show
+---+---+
| id| id|
+---+---+
| 0| 0|
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
+---+---+
so I want to drop some columns like below
My general recommendation is not to drop columns, but select what you want to include in the result. That makes life more predictable as you know what you get (not what you don't). I was told that our brains work by positives which could also make a point for select.
So, as you asked and I showed in the above example, the result has two columns of the same name id. The question is how to have only one.
There are at least two answers with using the variant of join operator with the join columns or condition included (as you did show in your question), but that would not answer your real question about "dropping unwanted columns", would it?
Given I prefer select (over drop), I'd do the following to have a single id column:
val q = ds1.as('one)
.join(ds2.as('two))
.where($"one.id" === $"two.id")
.select("one.*") // <-- select columns from "one" dataset
scala> q.show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
Regardless of the reasons why you asked the question (which could also be answered with the points I raised above), let me answer the (burning) question how to use withColumnRenamed when there are two matching columns (after join).
Let's assume you ended up with the following query and so you've got two id columns (per join side).
val q = ds1.as('one)
.join(ds2.as('two))
.where($"one.id" === $"two.id")
scala> q.show
+---+---+
| id| id|
+---+---+
| 0| 0|
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
+---+---+
withColumnRenamed won't work for this use case since it does not accept aliased column names.
scala> q.withColumnRenamed("one.id", "one_id").show
+---+---+
| id| id|
+---+---+
| 0| 0|
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
+---+---+
You could select the columns you're interested in as follows:
scala> q.select("one.id").show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
scala> q.select("two.*").show
+---+
| id|
+---+
| 0|
| 1|
| 2|
| 3|
| 4|
+---+
Please see the docs : withColumnRenamed()
You need to pass the name of the existing column and the new name to the function. Both of these should be strings.
result_df = aa_df.join(bb_df,'id', 'left').join(cc_df, 'id', 'left').withColumnRenamed('status', 'user_status')
If you have 'status' columns in 2 dataframes, you can use them in the join as aa_df.join(bb_df, ['id','status'], 'left') assuming aa_df and bb_df have the common column. This way you will not end up having 2 'status' columns.
I have several data frames in spark with partly similar schema (header) int the beginning and different columns (custom) in the end.
case class First(header1:String, header2:String, header3:Int, custom1:String)
case class Second(header1:String, header2:String, header3:Int, custom1:String, custom5:String)
case class Third(header1:String, header2:String, header3:Int, custom2:String, custom3:Int, custom4:Double)
val first = Seq(First("A", "Ba1", 1, "custom1"), First("A", "Ba2", 2, "custom2")).toDS
val second = Seq(Second("B", "Bb1", 1, "custom12", "custom5"), Second("B", "Bb2", 22, "custom12", "custom55")).toDS
val third = Seq(Third("A", "Bc1", 1, "custom2", 22, 44.4)).toDS
This could look like:
+-------+-------+-------+-------+
|header1|header2|header3|custom1|
+-------+-------+-------+-------+
| A| Ba1| 1|custom1|
| A| Ba2| 2|custom2|
+-------+-------+-------+-------+
+-------+-------+-------+--------+--------+
|header1|header2|header3| custom1| custom5|
+-------+-------+-------+--------+--------+
| B| Bb1| 1|custom12| custom5|
| B| Bb2| 22|custom12|custom55|
+-------+-------+-------+--------+--------+
+-------+-------+-------+-------+-------+-------+
|header1|header2|header3|custom2|custom3|custom4|
+-------+-------+-------+-------+-------+-------+
| A| Bc1| 1|custom2| 22| 44.4|
+-------+-------+-------+-------+-------+-------+
How can I merge the schema to basically concatenate all the dataframes into a single schema
case class All(header1:String, header2:String, header3:Int, custom1:Option[String], custom3:Option[String],
custom4: Option[Double], custom5:Option[String], type:String)
where some columns which are not present will be nullable?
Output should should look like this in case of the first record from data frame named first
+-------+-------+-------+-------+-------+-------+-------+-------+
|header1|header2|header3|custom1|custom2|custom3|custom4|custom5|
+-------+-------+-------+-------+-------+-------+-------+-------+
| A| B| 1|custom1|Nan |Nan | Nan| Nan. |
+-------+-------+-------+-------+-------+-------+-------+-------+
I was thinking about joining the data frames via the header columns, however,only some (lets say header1) would hold the same (actually joinable) values and the others (header2,3) would hold different values i.e.
first
.join(second, Seq("header1", "header2", "header3"), "LEFT")
.join(third, Seq("header1", "header2", "header3"), "LEFT")
.show
resulting in
+-------+-------+-------+-------+-------+-------+-------+-------+-------+
|header1|header2|header3|custom1|custom1|custom5|custom2|custom3|custom4|
+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| A| Ba1| 1|custom1| null| null| null| null| null|
| A| Ba2| 2|custom2| null| null| null| null| null|
+-------+-------+-------+-------+-------+-------+-------+-------+-------+
is not correct as I just want to pd.Concat(axis=0) the dataFrames i.e. am lacking most of the records.
Also it would be lacking a type column identifying the original data frame i.e. first, second, third
edit
I think a classical full outer join is the solution
first
.join(second, Seq("header1", "header2", "header3"), "fullouter")
.join(third, Seq("header1", "header2", "header3"), "fullouter")
.show
yields:
+-------+-------+-------+-------+--------+--------+-------+-------+-------+
|header1|header2|header3|custom1| custom1| custom5|custom2|custom3|custom4|
+-------+-------+-------+-------+--------+--------+-------+-------+-------+
| A| Ba1| 1|custom1| null| null| null| null| null|
| A| Ba2| 2|custom2| null| null| null| null| null|
| A| Bb1| 1| null|custom12| custom5| null| null| null|
| A| Bb2| 22| null|custom12|custom55| null| null| null|
| A| Bc1| 1| null| null| null|custom2| 22| 44.4|
+-------+-------+-------+-------+--------+--------+-------+-------+-------+
As you see, actually there will never be a real join, rows are concatenated. Is there a simpler operation to achieve the same functionality?
This answer is not optimal, as custom1 is a duplicate name. I rather would want to see a single custom1 column (with no null values if there is a second one to fill).
Check out my comment to similar question. Basically you need to union all the frames. To make similar schema you need to use dataframe.withColumn(ColumnName, expr("null")) expression:
import org.apache.spark.sql.functions._
val first1 = first.withColumn("custom5", expr("null"))
.withColumn("custom4", expr("null"))
val second2 = second.withColumn("custom4", expr("null"))
val result = first1.unionAll(second2).unionAll(third)
Please test the SQL Union approach if it provides the desired result.
SELECT header1,
header2,
header3,
custom1,
To_char(NULL) "custom2",
To_char(NULL) "custom3",
To_number(NULL) "custom4",
To_char(NULL) "custom5"
FROM table1
UNION
SELECT header1,
header2,
header3,
custom1,
To_char(NULL) "custom2",
To_char(NULL) "custom3",
To_number(NULL) "custom4",
custom5
FROM table2
UNION
SELECT header1,
header2,
header3,
To_char(NULL) "custom1",
custom2,
custom3,
custom4,
To_char(NULL) "custom5"
FROM table3;
If you are writing files to HDFS then you can achieve this by setting following property Spark.sql.parquet.mergeSchema to TRUE and write files to HDFS location.
It automatically updates the schema and returns all columns.
You can achieve this using below ways
withColumn and union
Specify schema before itself and perform union
spark.conf.set("spark.sql.parquet.mergeSchema","true")
eb = spark.read.format("csv").schema(schem).option("path","/retail/ebay.csv").load()
eb.printSchema()
eb.write.format("parquet").mode("append").save("/retail/parquet_test")
from pyspark.sql.functions import lit
eb1 = eb.withColumn("dummy",lit(35))
eb1.printSchema()
eb1.write.format("parquet").mode("append").save("/retail/parquet_test")
eb2 = spark.read.parquet("/srinchin/parquet_test")
eb2.printSchema()