Related
I have some dataframe in Pyspark:
from pyspark.sql import SQLContext, SparkSession
spark = SparkSession.builder.getOrCreate()
sqlcontext = SQLContext(spark)
df = sqlcontext.createDataFrame([['a'],['b'],['c'],['d'],['e']], ['id'])
df.show()
+---+
| id|
+---+
| a|
| b|
| c|
| d|
| e|
+---+
And I have a list of lists:
l = [[1,1], [2,2], [3,3], [4,4], [5,5]]
Is it possible to append this list as a column to df? Namely, the first element of l should appear next to the first row of df, the second element of l next to the second row of df, etc. It should look like this:
+----+---+--+
| id| l|
+----+---+--+
| a| [1,1]|
| b| [2,2]|
| c| [3,3]|
| d| [4,4]|
| e| [5,5]|
+----+---+--+
UDF's are generally slow but a more efficient way without using any UDF's would be:
import pyspark.sql.functions as F
ldf = spark.createDataFrame(l, schema = "array<int>")
df1 = df.withColumn("m_id", F.monotonically_increasing_id())
df2 = ldf.withColumn("m_id", F.monotonically_increasing_id())
df3 = df2.join(df1, "m_id", "outer").drop("m_id")
df3.select("id", "value").show()
+---+------+
| id| value|
+---+------+
| a|[1, 1]|
| b|[2, 2]|
| d|[4, 4]|
| c|[3, 3]|
| e|[5, 5]|
+---+------+
Assuming that you are going to have same amount of rows in your df and items in your list (df.count==len(l)).
You can add a row_id (to specify the order) to your df, and based on that, access to the item on your list (l).
from pyspark.sql.functions import row_number, lit
from pyspark.sql.window import *
df = df.withColumn("row_num", row_number().over(Window().orderBy(lit('A'))))
df.show()
Above code will look like:
+---+-------+
| id|row_num|
+---+-------+
| 1| 1|
| 2| 2|
| 3| 3|
| 4| 4|
| 5| 5|
+---+-------+
Then, you can just iterate your df and access the specified index in your list:
def map_df(row):
return (row.id, l[row.row_num-1])
new_df = df.rdd.map(map_df).toDF(["id", "l"])
new_df.show()
Output:
+---+------+
| id| l|
+---+------+
| 1|[1, 1]|
| 2|[2, 2]|
| 3|[3, 3]|
| 4|[4, 4]|
| 5|[5, 5]|
+---+------+
Thanks to Cesar's answer, I figured out how to do it without making the dataframe an RDD and coming back. It would be something like this:
from pyspark.sql import SQLContext, SparkSession
from pyspark.sql.functions import row_number, lit, udf
from pyspark.sql.window import Window
from pyspark.sql.types import ArrayType, FloatType, IntegerType
spark = SparkSession.builder.getOrCreate()
sqlcontext = SQLContext(spark)
df = sqlcontext.createDataFrame([['a'],['b'],['c'],['d'],['e']], ['id'])
df = df.withColumn("row_num", row_number().over(Window().orderBy(lit('A'))))
new_col = [[1.,1.], [2.,2.], [3.,3.], [4.,4.], [5.,5.]]
map_list_to_column = udf(lambda row_num: new_col[row_num -1], ArrayType(FloatType()))
df.withColumn('new_col', map_list_to_column(df.row_num)).drop('row_num').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)
I want to create a sample single-column DataFrame, but the following code is not working:
df = spark.createDataFrame(["10","11","13"], ("age"))
## ValueError
## ...
## ValueError: Could not parse datatype: age
The expected result:
age
10
11
13
the following code is not working
With single element you need a schema as type
spark.createDataFrame(["10","11","13"], "string").toDF("age")
or DataType:
from pyspark.sql.types import StringType
spark.createDataFrame(["10","11","13"], StringType()).toDF("age")
With name elements should be tuples and schema as sequence:
spark.createDataFrame([("10", ), ("11", ), ("13", )], ["age"])
Well .. There is some pretty easy method for creating sample dataframe in PySpark
>>> df = sc.parallelize([[1,2,3], [2,3,4]]).toDF()
>>> df.show()
+---+---+---+
| _1| _2| _3|
+---+---+---+
| 1| 2| 3|
| 2| 3| 4|
+---+---+---+
to create with some column names
>>> df1 = sc.parallelize([[1,2,3], [2,3,4]]).toDF(("a", "b", "c"))
>>> df1.show()
+---+---+---+
| a| b| c|
+---+---+---+
| 1| 2| 3|
| 2| 3| 4|
+---+---+---+
In this way, no need to define schema too.Hope this is the simplest way
from pyspark.sql import SparkSession
spark = SparkSession.builder.getOrCreate()
df = spark.createDataFrame([{"a": "x", "b": "y", "c": "3"}])
Output: (no need to define schema)
+---+---+---+
| a | b | c |
+---+---+---+
| x| y| 3|
+---+---+---+
For pandas + pyspark users, if you've already installed pandas in the cluster, you can do this simply:
# create pandas dataframe
df = pd.DataFrame({'col1':[1,2,3], 'col2':['a','b','c']})
# convert to spark dataframe
df = spark.createDataFrame(df)
Local Spark Setup
import findspark
findspark.init()
import pyspark
spark = (pyspark
.sql
.SparkSession
.builder
.master("local")
.getOrCreate())
See my farsante lib for creating a DataFrame with fake data:
import farsante
df = farsante.quick_pyspark_df(['first_name', 'last_name'], 7)
df.show()
+----------+---------+
|first_name|last_name|
+----------+---------+
| Tommy| Hess|
| Arthur| Melendez|
| Clemente| Blair|
| Wesley| Conrad|
| Willis| Dunlap|
| Bruna| Sellers|
| Tonda| Schwartz|
+----------+---------+
Here's how to explicitly specify the schema when creating the PySpark DataFrame:
df = spark.createDataFrame(
[(10,), (11,), (13,)],
StructType([StructField("some_int", IntegerType(), True)]))
df.show()
+--------+
|some_int|
+--------+
| 10|
| 11|
| 13|
+--------+
You can also try something like this -
from pyspark.sql import SQLContext
sqlContext = SQLContext(sc) # sc is the spark context
sample = sqlContext.createDataFrame(
[
('qwe', 23), # enter your data here
('rty',34),
('yui',56),
],
['abc', 'def'] # the row header/column labels should be entered here
There are several ways to create a DataFrame, PySpark Create DataFrame is one of the first steps you learn while working on PySpark
I assume you already have data, columns, and an RDD.
1) df = rdd.toDF()
2) df = rdd.toDF(columns) //Assigns column names
3) df = spark.createDataFrame(rdd).toDF(*columns)
4) df = spark.createDataFrame(data).toDF(*columns)
5) df = spark.createDataFrame(rowData,columns)
Besides these, you can find several examples on pyspark create dataframe
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.
I have a pyspark 2.0.1. I'm trying to groupby my data frame & retrieve the value for all the fields from my data frame. I found that
z=data1.groupby('country').agg(F.collect_list('names'))
will give me values for country & names attribute & for names attribute it will give column header as collect_list(names). But for my job I have dataframe with around 15 columns & I will run a loop & will change the groupby field each time inside loop & need the output for all of the remaining fields.Can you please suggest me how to do it using collect_list() or any other pyspark functions?
I tried this code too
from pyspark.sql import functions as F
fieldnames=data1.schema.names
names1= list()
for item in names:
if item != 'names':
names1.append(item)
z=data1.groupby('names').agg(F.collect_list(names1))
z.show()
but got error message
Py4JError: An error occurred while calling z:org.apache.spark.sql.functions.collect_list. Trace: py4j.Py4JException: Method collect_list([class java.util.ArrayList]) does not exist
Use struct to combine the columns before calling groupBy
suppose you have a dataframe
df = spark.createDataFrame(sc.parallelize([(0,1,2),(0,4,5),(1,7,8),(1,8,7)])).toDF("a","b","c")
df = df.select("a", f.struct(["b","c"]).alias("newcol"))
df.show()
+---+------+
| a|newcol|
+---+------+
| 0| [1,2]|
| 0| [4,5]|
| 1| [7,8]|
| 1| [8,7]|
+---+------+
df = df.groupBy("a").agg(f.collect_list("newcol").alias("collected_col"))
df.show()
+---+--------------+
| a| collected_col|
+---+--------------+
| 0|[[1,2], [4,5]]|
| 1|[[7,8], [8,7]]|
+---+--------------+
Aggregation operation can be done only on single columns.
After aggregation, You can collect the result and iterate over it to separate the combined columns generate the index dict. or you can write a
udf to separate the combined columns.
from pyspark.sql.types import *
def foo(x):
x1 = [y[0] for y in x]
x2 = [y[1] for y in x]
return(x1,x2)
st = StructType([StructField("b", ArrayType(LongType())), StructField("c", ArrayType(LongType()))])
udf_foo = udf(foo, st)
df = df.withColumn("ncol",
udf_foo("collected_col")).select("a",
col("ncol").getItem("b").alias("b"),
col("ncol").getItem("c").alias("c"))
df.show()
+---+------+------+
| a| b| c|
+---+------+------+
| 0|[1, 4]|[2, 5]|
| 1|[7, 8]|[8, 7]|
+---+------+------+
Actually we can do it in pyspark 2.2 .
First we need create a constant column ("Temp"), groupBy with that column ("Temp") and apply agg by pass iterable *exprs in which expression of collect_list exits.
Below is the code:
import pyspark.sql.functions as ftions
import functools as ftools
def groupColumnData(df, columns):
df = df.withColumn("Temp", ftions.lit(1))
exprs = [ftions.collect_list(colName) for colName in columns]
df = df.groupby('Temp').agg(*exprs)
df = df.drop("Temp")
df = df.toDF(*columns)
return df
Input Data:
df.show()
+---+---+---+
| a| b| c|
+---+---+---+
| 0| 1| 2|
| 0| 4| 5|
| 1| 7| 8|
| 1| 8| 7|
+---+---+---+
Output Data:
df.show()
+------------+------------+------------+
| a| b| c|
+------------+------------+------------+
|[0, 0, 1, 1]|[1, 4, 7, 8]|[2, 5, 8, 7]|
+------------+------------+------------+
in spark 2.4.4 and python 3.7 (I guess its also relevant for previous spark and python version) --
My suggestion is a based on pauli's answer,
instead of creating the struct and then using the agg function, create the struct inside collect_list:
df = spark.createDataFrame([(0,1,2),(0,4,5),(1,7,8),(1,8,7)]).toDF("a","b","c")
df.groupBy("a").agg(collect_list(struct(["b","c"])).alias("res")).show()
result :
+---+-----------------+
| a|res |
+---+-----------------+
| 0|[[1, 2], [4, 5]] |
| 1|[[7, 8], [8, 7]] |
+---+-----------------+
I just use Concat_ws function it's perfectly fine.
> from pyspark.sql.functions import * df =
> spark.createDataFrame([(0,1,2),(0,4,5),(1,7,8),(1,8,7)]).toDF("a","b","c")
> df.groupBy('a').agg(collect_list(concat_ws(',','b','c'))).alias('r').show()