When you join two DFs with similar column names:
df = df1.join(df2, df1['id'] == df2['id'])
Join works fine but you can't call the id column because it is ambiguous and you would get the following exception:
pyspark.sql.utils.AnalysisException: "Reference 'id' is ambiguous,
could be: id#5691, id#5918.;"
This makes id not usable anymore...
The following function solves the problem:
def join(df1, df2, cond, how='left'):
df = df1.join(df2, cond, how=how)
repeated_columns = [c for c in df1.columns if c in df2.columns]
for col in repeated_columns:
df = df.drop(df2[col])
return df
What I don't like about it is that I have to iterate over the column names and delete them why by one. This looks really clunky...
Do you know of any other solution that will either join and remove duplicates more elegantly or delete multiple columns without iterating over each of them?
If the join columns at both data frames have the same names and you only need equi join, you can specify the join columns as a list, in which case the result will only keep one of the join columns:
df1.show()
+---+----+
| id|val1|
+---+----+
| 1| 2|
| 2| 3|
| 4| 4|
| 5| 5|
+---+----+
df2.show()
+---+----+
| id|val2|
+---+----+
| 1| 2|
| 1| 3|
| 2| 4|
| 3| 5|
+---+----+
df1.join(df2, ['id']).show()
+---+----+----+
| id|val1|val2|
+---+----+----+
| 1| 2| 2|
| 1| 2| 3|
| 2| 3| 4|
+---+----+----+
Otherwise you need to give the join data frames alias and refer to the duplicated columns by the alias later:
df1.alias("a").join(
df2.alias("b"), df1['id'] == df2['id']
).select("a.id", "a.val1", "b.val2").show()
+---+----+----+
| id|val1|val2|
+---+----+----+
| 1| 2| 2|
| 1| 2| 3|
| 2| 3| 4|
+---+----+----+
df.join(other, on, how) when on is a column name string, or a list of column names strings, the returned dataframe will prevent duplicate columns.
when on is a join expression, it will result in duplicate columns. We can use .drop(df.a) to drop duplicate columns. Example:
cond = [df.a == other.a, df.b == other.bb, df.c == other.ccc]
# result will have duplicate column a
result = df.join(other, cond, 'inner').drop(df.a)
Assuming 'a' is a dataframe with column 'id' and 'b' is another dataframe with column 'id'
I use the following two methods to remove duplicates:
Method 1: Using String Join Expression as opposed to boolean expression. This automatically remove a duplicate column for you
a.join(b, 'id')
Method 2: Renaming the column before the join and dropping it after
b.withColumnRenamed('id', 'b_id')
joinexpr = a['id'] == b['b_id']
a.join(b, joinexpr).drop('b_id)
The code below works with Spark 1.6.0 and above.
salespeople_df.show()
+---+------+-----+
|Num| Name|Store|
+---+------+-----+
| 1| Henry| 100|
| 2| Karen| 100|
| 3| Paul| 101|
| 4| Jimmy| 102|
| 5|Janice| 103|
+---+------+-----+
storeaddress_df.show()
+-----+--------------------+
|Store| Address|
+-----+--------------------+
| 100| 64 E Illinos Ave|
| 101| 74 Grand Pl|
| 102| 2298 Hwy 7|
| 103|No address available|
+-----+--------------------+
Assuming -in this example- that the name of the shared column is the same:
joined=salespeople_df.join(storeaddress_df, ['Store'])
joined.orderBy('Num', ascending=True).show()
+-----+---+------+--------------------+
|Store|Num| Name| Address|
+-----+---+------+--------------------+
| 100| 1| Henry| 64 E Illinos Ave|
| 100| 2| Karen| 64 E Illinos Ave|
| 101| 3| Paul| 74 Grand Pl|
| 102| 4| Jimmy| 2298 Hwy 7|
| 103| 5|Janice|No address available|
+-----+---+------+--------------------+
.join will prevent the duplication of the shared column.
Let's assume that you want to remove the column Num in this example, you can just use .drop('colname')
joined=joined.drop('Num')
joined.show()
+-----+------+--------------------+
|Store| Name| Address|
+-----+------+--------------------+
| 103|Janice|No address available|
| 100| Henry| 64 E Illinos Ave|
| 100| Karen| 64 E Illinos Ave|
| 101| Paul| 74 Grand Pl|
| 102| Jimmy| 2298 Hwy 7|
+-----+------+--------------------+
After I've joined multiple tables together, I run them through a simple function to drop columns in the DF if it encounters duplicates while walking from left to right. Alternatively, you could rename these columns too.
Where Names is a table with columns ['Id', 'Name', 'DateId', 'Description'] and Dates is a table with columns ['Id', 'Date', 'Description'], the columns Id and Description will be duplicated after being joined.
Names = sparkSession.sql("SELECT * FROM Names")
Dates = sparkSession.sql("SELECT * FROM Dates")
NamesAndDates = Names.join(Dates, Names.DateId == Dates.Id, "inner")
NamesAndDates = dropDupeDfCols(NamesAndDates)
NamesAndDates.saveAsTable("...", format="parquet", mode="overwrite", path="...")
Where dropDupeDfCols is defined as:
def dropDupeDfCols(df):
newcols = []
dupcols = []
for i in range(len(df.columns)):
if df.columns[i] not in newcols:
newcols.append(df.columns[i])
else:
dupcols.append(i)
df = df.toDF(*[str(i) for i in range(len(df.columns))])
for dupcol in dupcols:
df = df.drop(str(dupcol))
return df.toDF(*newcols)
The resulting data frame will contain columns ['Id', 'Name', 'DateId', 'Description', 'Date'].
In my case I had a dataframe with multiple duplicate columns after joins and I was trying to same that dataframe in csv format, but due to duplicate column I was getting error. I followed below steps to drop duplicate columns. Code is in scala
1) Rename all the duplicate columns and make new dataframe
2) make separate list for all the renamed columns
3) Make new dataframe with all columns (including renamed - step 1)
4) drop all the renamed column
private def removeDuplicateColumns(dataFrame:DataFrame): DataFrame = {
var allColumns: mutable.MutableList[String] = mutable.MutableList()
val dup_Columns: mutable.MutableList[String] = mutable.MutableList()
dataFrame.columns.foreach((i: String) =>{
if(allColumns.contains(i))
if(allColumns.contains(i))
{allColumns += "dup_" + i
dup_Columns += "dup_" +i
}else{
allColumns += i
}println(i)
})
val columnSeq = allColumns.toSeq
val df = dataFrame.toDF(columnSeq:_*)
val unDF = df.drop(dup_Columns:_*)
unDF
}
to call the above function use below code and pass your dataframe which contains duplicate columns
val uniColDF = removeDuplicateColumns(df)
Here is simple solution for remove duplicate column
final_result=df1.join(df2,(df1['subjectid']==df2['subjectid']),"left").drop(df1['subjectid'])
If you join on a list or string, dup cols are automatically]1 removed
This is a scala solution, you could translate the same idea into any language
// get a list of duplicate columns or use a list/seq
// of columns you would like to join on (note that this list
// should include columns for which you do not want duplicates)
val duplicateCols = df1.columns.intersect(df2.columns)
// no duplicate columns in resulting DF
df1.join(df2, duplicateCols.distinct.toSet)
Spark SQL version of this answer:
df1.createOrReplaceTempView("t1")
df2.createOrReplaceTempView("t2")
spark.sql("select * from t1 inner join t2 using (id)").show()
# +---+----+----+
# | id|val1|val2|
# +---+----+----+
# | 1| 2| 2|
# | 1| 2| 3|
# | 2| 3| 4|
# +---+----+----+
This works for me when multiple columns used to join and need to drop more than one column which are not string type.
final_data = mdf1.alias("a").join(df3.alias("b")
(mdf1.unique_product_id==df3.unique_product_id) &
(mdf1.year_week==df3.year_week) ,"left" ).select("a.*","b.promotion_id")
Give a.* to select all columns from one table and from the other table choose specific columns.
Related
I'm trying to compare two data frames with have same number of columns i.e. 4 columns with id as key column in both data frames
df1 = spark.read.csv("/path/to/data1.csv")
df2 = spark.read.csv("/path/to/data2.csv")
Now I want to append new column to DF2 i.e. column_names which is the list of the columns with different values than df1
df2.withColumn("column_names",udf())
DF1
+------+---------+--------+------+
| id | |name | sal | Address |
+------+---------+--------+------+
| 1| ABC | 5000 | US |
| 2| DEF | 4000 | UK |
| 3| GHI | 3000 | JPN |
| 4| JKL | 4500 | CHN |
+------+---------+--------+------+
DF2:
+------+---------+--------+------+
| id | |name | sal | Address |
+------+---------+--------+------+
| 1| ABC | 5000 | US |
| 2| DEF | 4000 | CAN |
| 3| GHI | 3500 | JPN |
| 4| JKL_M | 4800 | CHN |
+------+---------+--------+------+
Now I want DF3
DF3:
+------+---------+--------+------+--------------+
| id | |name | sal | Address | column_names |
+------+---------+--------+------+--------------+
| 1| ABC | 5000 | US | [] |
| 2| DEF | 4000 | CAN | [address] |
| 3| GHI | 3500 | JPN | [sal] |
| 4| JKL_M | 4800 | CHN | [name,sal] |
+------+---------+--------+------+--------------+
I saw this SO question, How to compare two dataframe and print columns that are different in scala. Tried that, however the result is different.
I'm thinking of going with a UDF function by passing row from each dataframe to udf and compare column by column and return column list. However for that both the data frames should be in sorted order so that same id rows will be sent to udf. Sorting is costly operation here. Any solution?
Assuming that we can use id to join these two datasets I don't think that there is a need for UDF. This could be solved just by using inner join, array and array_remove functions among others.
First let's create the two datasets:
df1 = spark.createDataFrame([
[1, "ABC", 5000, "US"],
[2, "DEF", 4000, "UK"],
[3, "GHI", 3000, "JPN"],
[4, "JKL", 4500, "CHN"]
], ["id", "name", "sal", "Address"])
df2 = spark.createDataFrame([
[1, "ABC", 5000, "US"],
[2, "DEF", 4000, "CAN"],
[3, "GHI", 3500, "JPN"],
[4, "JKL_M", 4800, "CHN"]
], ["id", "name", "sal", "Address"])
First we do an inner join between the two datasets then we generate the condition df1[col] != df2[col] for each column except id. When the columns aren't equal we return the column name otherwise an empty string. The list of conditions will consist the items of an array from which finally we remove the empty items:
from pyspark.sql.functions import col, array, when, array_remove
# get conditions for all columns except id
conditions_ = [when(df1[c]!=df2[c], lit(c)).otherwise("") for c in df1.columns if c != 'id']
select_expr =[
col("id"),
*[df2[c] for c in df2.columns if c != 'id'],
array_remove(array(*conditions_), "").alias("column_names")
]
df1.join(df2, "id").select(*select_expr).show()
# +---+-----+----+-------+------------+
# | id| name| sal|Address|column_names|
# +---+-----+----+-------+------------+
# | 1| ABC|5000| US| []|
# | 3| GHI|3500| JPN| [sal]|
# | 2| DEF|4000| CAN| [Address]|
# | 4|JKL_M|4800| CHN| [name, sal]|
# +---+-----+----+-------+------------+
Here is your solution with UDF, I have changed first dataframe name dynamically so that it will be not ambiguous during check. Go through below code and let me know in case any concerns.
>>> from pyspark.sql.functions import *
>>> df.show()
+---+----+----+-------+
| id|name| sal|Address|
+---+----+----+-------+
| 1| ABC|5000| US|
| 2| DEF|4000| UK|
| 3| GHI|3000| JPN|
| 4| JKL|4500| CHN|
+---+----+----+-------+
>>> df1.show()
+---+----+----+-------+
| id|name| sal|Address|
+---+----+----+-------+
| 1| ABC|5000| US|
| 2| DEF|4000| CAN|
| 3| GHI|3500| JPN|
| 4|JKLM|4800| CHN|
+---+----+----+-------+
>>> df2 = df.select([col(c).alias("x_"+c) for c in df.columns])
>>> df3 = df1.join(df2, col("id") == col("x_id"), "left")
//udf declaration
>>> def CheckMatch(Column,r):
... check=''
... ColList=Column.split(",")
... for cc in ColList:
... if(r[cc] != r["x_" + cc]):
... check=check + "," + cc
... return check.replace(',','',1).split(",")
>>> CheckMatchUDF = udf(CheckMatch)
//final column that required to select
>>> finalCol = df1.columns
>>> finalCol.insert(len(finalCol), "column_names")
>>> df3.withColumn("column_names", CheckMatchUDF(lit(','.join(df1.columns)),struct([df3[x] for x in df3.columns])))
.select(finalCol)
.show()
+---+----+----+-------+------------+
| id|name| sal|Address|column_names|
+---+----+----+-------+------------+
| 1| ABC|5000| US| []|
| 2| DEF|4000| CAN| [Address]|
| 3| GHI|3500| JPN| [sal]|
| 4|JKLM|4800| CHN| [name, sal]|
+---+----+----+-------+------------+
Python: PySpark version of my previous scala code.
import pyspark.sql.functions as f
df1 = spark.read.option("header", "true").csv("test1.csv")
df2 = spark.read.option("header", "true").csv("test2.csv")
columns = df1.columns
df3 = df1.alias("d1").join(df2.alias("d2"), f.col("d1.id") == f.col("d2.id"), "left")
for name in columns:
df3 = df3.withColumn(name + "_temp", f.when(f.col("d1." + name) != f.col("d2." + name), f.lit(name)))
df3.withColumn("column_names", f.concat_ws(",", *map(lambda name: f.col(name + "_temp"), columns))).select("d1.*", "column_names").show()
Scala: Here is my best approach for your problem.
val df1 = spark.read.option("header", "true").csv("test1.csv")
val df2 = spark.read.option("header", "true").csv("test2.csv")
val columns = df1.columns
val df3 = df1.alias("d1").join(df2.alias("d2"), col("d1.id") === col("d2.id"), "left")
columns.foldLeft(df3) {(df, name) => df.withColumn(name + "_temp", when(col("d1." + name) =!= col("d2." + name), lit(name)))}
.withColumn("column_names", concat_ws(",", columns.map(name => col(name + "_temp")): _*))
.show(false)
First, I join two dataframe into df3 and used the columns from df1. By folding left to the df3 with temp columns that have the value for column name when df1 and df2 has the same id and other column values.
After that, concat_ws for those column names and the null's are gone away and only the column names are left.
+---+----+----+-------+------------+
|id |name|sal |Address|column_names|
+---+----+----+-------+------------+
|1 |ABC |5000|US | |
|2 |DEF |4000|UK |Address |
|3 |GHI |3000|JPN |sal |
|4 |JKL |4500|CHN |name,sal |
+---+----+----+-------+------------+
The only thing different from your expected result is that the output is not a list but string.
p.s. I forgot to use PySpark but this is the normal spark, sorry.
You can get that query build for you in PySpark and Scala by the spark-extension package.
It provides the diff transformation that does exactly that.
from gresearch.spark.diff import *
options = DiffOptions().with_change_column('changes')
df1.diff_with_options(df2, options, 'id').show()
+----+-----------+---+---------+----------+--------+---------+------------+-------------+
|diff| changes| id|left_name|right_name|left_sal|right_sal|left_Address|right_Address|
+----+-----------+---+---------+----------+--------+---------+------------+-------------+
| N| []| 1| ABC| ABC| 5000| 5000| US| US|
| C| [Address]| 2| DEF| DEF| 4000| 4000| UK| CAN|
| C| [sal]| 3| GHI| GHI| 3000| 3500| JPN| JPN|
| C|[name, sal]| 4| JKL| JKL_M| 4500| 4800| CHN| CHN|
+----+-----------+---+---------+----------+--------+---------+------------+-------------+
While this is a simple example, diffing DataFrames can become complicated when wide schemas, insertions, deletions and null values are involved. That package is well-tested, so you don't have to worry about getting that query right yourself.
There is a wonderful package for pyspark that compares two dataframes. The name of the package is datacompy
https://capitalone.github.io/datacompy/
example code:
import datacompy as dc
comparison = dc.SparkCompare(spark, base_df=df1, compare_df=df2, join_columns=common_keys, match_rates=True)
comparison.report()
The above code will generate a summary report, and the one below it will give you the mismatches.
comparison.rows_both_mismatch.display()
There are also more fearures that you can explore.
I have a table which has columns [col1, col2, col3 .... col9].
I want to merge all the columns data into one column as col in python?
from pyspark.sql.functions import concat
values = [('A','B','C','D'),('E','F','G','H'),('I','J','K','L')]
df = sqlContext.createDataFrame(values,['col1','col2','col3','col4'])
df.show()
+----+----+----+----+
|col1|col2|col3|col4|
+----+----+----+----+
| A| B| C| D|
| E| F| G| H|
| I| J| K| L|
+----+----+----+----+
req_column = ['col1','col2','col3','col4']
df = df.withColumn('concatenated_cols',concat(*req_column))
df.show()
+----+----+----+----+-----------------+
|col1|col2|col3|col4|concatenated_cols|
+----+----+----+----+-----------------+
| A| B| C| D| ABCD|
| E| F| G| H| EFGH|
| I| J| K| L| IJKL|
+----+----+----+----+-----------------+
using Spark SQL
new_df=sqlContext.sql("SELECT CONCAT(col1,col2,col3,col3) FROM df")
Using Non Spark SQL way you can use Concat function
new_df = df.withColumn('joined_column', concat(col('col1'),col('col2'),col('col3'),col('col4'))
In Spark(pySpark) for reasons, there is no edit of existing data. What you can do is create a new column. Please check the following link.
How do I add a new column to a Spark DataFrame (using PySpark)?
Using a UDF function, you can aggregate/combine all those values in a row and return you as a single value.
Few cautions, please look out for following data issues while aggregation
Null values
Type mismatches
String Encoding issues
My dataframe is this and I want to split my data frame by colon (:)
+------------------+
|Name:Roll_no:Class|
+------------------+
| #ab:cd#:23:C|
| #sd:ps#:34:A|
| #ra:kh#:14:H|
| #ku:pa#:36:S|
| #ra:sh#:50:P|
+------------------+
and I want my dataframe like:
+-----+-------+-----+
| Name|Roll_no|Class|
+-----+-------+-----+
|ab:cd| 23| C|
|sd:ps| 34| A|
|ra:kh| 14| H|
|ku:pa| 36| S|
|ra:sh| 50| P|
+-----+-------+-----+
If need split by last 2 : use Series.str.rsplit, then set columns by split column name and last remove first and last # by indexing:
col = 'Name:Roll_no:Class'
df1 = df[col].str.rsplit(':', n=2, expand=True)
df1.columns = col.split(':')
df1['Name'] = df1['Name'].str[1:-1]
#if only first and last value
#df1['Name'] = df1['Name'].str.strip('#')
print (df1)
Name Roll_no Class
0 ab:cd 23 C
1 sd:ps 34 A
2 ra:kh 14 H
3 ku:pa 36 S
4 ra:sh 50 P
Use read_csv() sep=':' and quotechar='#'
str = """Name:Roll_no:Class
#ab:cd#:23:C
#sd:ps#:34:A
#ra:kh#:14:H
#ku:pa#:36:S
#ra:sh#:50:P"""
df = pd.read_csv(pd.io.common.StringIO(str), sep=':', quotechar='#')
>>> df
Name Roll_no Class
#0 ab:cd 23 C
#1 sd:ps 34 A
#2 ra:kh 14 H
#3 ku:pa 36 S
#4 ra:sh 50 P
This is how you could do this in pyspark:
Specify the separator and the quote on read
If you're reading the data from a file, you can use spark.read_csv with the following arguments:
df = spark.read.csv("path/to/file", sep=":", quote="#", header=True)
df.show()
#+-----+-------+-----+
#| Name|Roll_no|Class|
#+-----+-------+-----+
#|ab:cd| 23| C|
#|sd:ps| 34| A|
#|ra:kh| 14| H|
#|ku:pa| 36| S|
#|ra:sh| 50| P|
#+-----+-------+-----+
Use Regular Expressions
If you're unable to change the way the data is read and you're starting with the DataFrame shown in the question, you can use regular expressions to get the desired output.
First get the new column names by splitting the existing column name on ":"
new_columns = df.columns[0].split(":")
print(new_columns)
#['Name', 'Roll_no', 'Class']
For the Name column you need to extract the data between the #s. For the other two columns, you need to remove the strings between the #s (and the following ":") and use pyspark.sql.functions.split to extract the components
from pyspark.sql.functions import regexp_extract, regexp_replace, split
df.withColumn(new_columns[0], regexp_extract(df.columns[0], r"(?<=#).+(?=#)", 0))\
.withColumn(new_columns[1], split(regexp_replace(df.columns[0], "#.+#:", ""), ":")[0])\
.withColumn(new_columns[2], split(regexp_replace(df.columns[0], "#.+#:", ""), ":")[1])\
.select(*new_columns)\
.show()
#+-----+-------+-----+
#| Name|Roll_no|Class|
#+-----+-------+-----+
#|ab:cd| 23| C|
#|sd:ps| 34| A|
#|ra:kh| 14| H|
#|ku:pa| 36| S|
#|ra:sh| 50| P|
#+-----+-------+-----+
I have a dataframe with many columns. My aim is to produce a dataframe thats lists each column name, along with the number of null values in that column.
Example:
+-------------+-------------+
| Column_Name | NULL_Values |
+-------------+-------------+
| Column_1 | 15 |
| Column_2 | 56 |
| Column_3 | 18 |
| ... | ... |
+-------------+-------------+
I have managed to get the number of null values for ONE column like so:
df.agg(F.count(F.when(F.isnull(c), c)).alias('NULL_Count'))
where c is a column in the dataframe. However, it does not show the name of the column. The output is:
+------------+
| NULL_Count |
+------------+
| 15 |
+------------+
Any ideas?
You can use a list comprehension to loop over all of your columns in the agg, and use alias to rename the output column:
import pyspark.sql.functions as F
df_agg = df.agg(*[F.count(F.when(F.isnull(c), c)).alias(c) for c in df.columns])
However, this will return the results in one row as shown below:
df_agg.show()
#+--------+--------+--------+
#|Column_1|Column_2|Column_3|
#+--------+--------+--------+
#| 15| 56| 18|
#+--------+--------+--------+
If you wanted the results in one column instead, you could union each column from df_agg using functools.reduce as follows:
from functools import reduce
df_agg_col = reduce(
lambda a, b: a.union(b),
(
df_agg.select(F.lit(c).alias("Column_Name"), F.col(c).alias("NULL_Count"))
for c in df_agg.columns
)
)
df_agg_col.show()
#+-----------+----------+
#|Column_Name|NULL_Count|
#+-----------+----------+
#| Column_1| 15|
#| Column_2| 56|
#| Column_3| 18|
#+-----------+----------+
Or you can skip the intermediate step of creating df_agg and do:
df_agg_col = reduce(
lambda a, b: a.union(b),
(
df.agg(
F.count(F.when(F.isnull(c), c)).alias('NULL_Count')
).select(F.lit(c).alias("Column_Name"), "NULL_Count")
for c in df.columns
)
)
Scala alternative could be
case class Test(id:Int, weight:Option[Int], age:Int, gender: Option[String])
val df1 = Seq(Test(1, Some(100), 23, Some("Male")), Test(2, None, 25, None), Test(3, None, 33, Some("Female"))).toDF()
df1.show()
+---+------+---+------+
| id|weight|age|gender|
+---+------+---+------+
| 1| 100| 23| Male|
| 2| null| 25| null|
| 3| null| 33|Female|
+---+------+---+------+
val s = df1.columns.map(c => sum(col(c).isNull.cast("integer")).alias(c))
val df2 = df1.agg(s.head, s.tail:_*)
val t = df2.columns.map(c => df2.select(lit(c).alias("col_name"), col(c).alias("null_count")))
val df_agg_col = t.reduce((df1, df2) => df1.union(df2))
df_agg_col.show()
I have two spark dataframes:
Dataframe A:
|col_1 | col_2 | ... | col_n |
|val_1 | val_2 | ... | val_n |
and dataframe B:
|col_1 | col_2 | ... | col_m |
|val_1 | val_2 | ... | val_m |
Dataframe B can contain duplicate, updated and new rows from dataframe A. I want to write an operation in spark where I can create a new dataframe containing the rows from dataframe A and the updated and new rows from dataframe B.
I started by creating a hash column containing only the columns that are not updatable. This is the unique id. So let's say col1 and col2 can change value (can be updated), but col3,..,coln are unique. I have created a hash function as hash(col3,..,coln):
A=A.withColumn("hash", hash(*[col(colname) for colname in unique_cols_A]))
B=B.withColumn("hash", hash(*[col(colname) for colname in unique_cols_B]))
Now I want to write some spark code that basically selects the rows from B that have the hash not in A (so new rows and updated rows) and join them into a new dataframe together with the rows from A. How can I achieve this in pyspark?
Edit:
Dataframe B can have extra columns from dataframe A, so a union is not possible.
Sample example
Dataframe A:
+-----+-----+
|col_1|col_2|
+-----+-----+
| a| www|
| b| eee|
| c| rrr|
+-----+-----+
Dataframe B:
+-----+-----+-----+
|col_1|col_2|col_3|
+-----+-----+-----+
| a| wew| 1|
| d| yyy| 2|
| c| rer| 3|
+-----+-----+-----+
Result:
Dataframe C:
+-----+-----+-----+
|col_1|col_2|col_3|
+-----+-----+-----+
| a| wew| 1|
| b| eee| null|
| c| rer| 3|
| d| yyy| 2|
+-----+-----+-----+
This is closely related to update a dataframe column with new values, except that you also want to add the rows from DataFrame B. One approach would be to first do what is outlined in the linked question and then union the result with DataFrame B and drop duplicates.
For example:
dfA.alias('a').join(dfB.alias('b'), on=['col_1'], how='left')\
.select(
'col_1',
f.when(
~f.isnull(f.col('b.col_2')),
f.col('b.col_2')
).otherwise(f.col('a.col_2')).alias('col_2'),
'b.col_3'
)\
.union(dfB)\
.dropDuplicates()\
.sort('col_1')\
.show()
#+-----+-----+-----+
#|col_1|col_2|col_3|
#+-----+-----+-----+
#| a| wew| 1|
#| b| eee| null|
#| c| rer| 3|
#| d| yyy| 2|
#+-----+-----+-----+
Or more generically using a list comprehension if you have a lot of columns to replace and you don't want to hard code them all:
cols_to_update = ['col_2']
dfA.alias('a').join(dfB.alias('b'), on=['col_1'], how='left')\
.select(
*[
['col_1'] +
[
f.when(
~f.isnull(f.col('b.{}'.format(c))),
f.col('b.{}'.format(c))
).otherwise(f.col('a.{}'.format(c))).alias(c)
for c in cols_to_update
] +
['b.col_3']
]
)\
.union(dfB)\
.dropDuplicates()\
.sort('col_1')\
.show()
I would opt for different solution, which I believe is less verbose, more generic and does not involve column listing. I would first identify subset of dfA that will be updated (replaceDf) by performing inner join based on keyCols (list). Then I would subtract this replaceDF from dfA and union it with dfB.
replaceDf = dfA.alias('a').join(dfB.alias('b'), on=keyCols, how='inner').select('a.*')
resultDf = dfA.subtract(replaceDf).union(dfB).show()
Even though there will be different columns in dfA and dfB, you can still overcome this with obtaining list of columns from both DataFrames and finding their union. Then I would
prepare select query (instead of "select.('a.')*") so that I would just list columns from dfA that exist in dfB + "null as colname" for those that do not exist in dfB.
If you want to keep only unique values, and require strictly correct results, then union followed by dropDupilcates should do the trick:
columns_which_dont_change = [...]
old_df.union(new_df).dropDuplicates(subset=columns_which_dont_change)