I am working with PySpark and loading a csv
file. I have a column with numbers in European format, which means that comma replaces the dot and vice versa.
For example: I have 2.416,67
instead of 2,416.67
.
My data in .csv file looks like this -
ID; Revenue
21; 2.645,45
23; 31.147,05
.
.
55; 1.009,11
In pandas, such a file can easily be read by specifying decimal=','
and thousands='.'
options inside pd.read_csv()
to read European formats.
Pandas code:
import pandas as pd
df=pd.read_csv("filepath/revenues.csv",sep=';',decimal=',',thousands='.')
I don't know how can this be done in PySpark.
PySpark code:
from pyspark.sql.types import StructType, StructField, FloatType, StringType
schema = StructType([
StructField("ID", StringType(), True),
StructField("Revenue", FloatType(), True)
])
df=spark.read.csv("filepath/revenues.csv",sep=';',encoding='UTF-8', schema=schema, header=True)
Can anyone suggest as to how we can load such a file in PySpark using the above mentioned .csv()
function?
You won't be able to read it as a float because the format of the data. You need to read it as a string, clean it up and then cast to float:
from pyspark.sql.functions import regexp_replace
from pyspark.sql.types import FloatType
df = spark.read.option("headers", "true").option("inferSchema", "true").csv("my_csv.csv", sep=";")
df = df.withColumn('revenue', regexp_replace('revenue', '\\.', ''))
df = df.withColumn('revenue', regexp_replace('revenue', ',', '.'))
df = df.withColumn('revenue', df['revenue'].cast("float"))
You can probably just chain these all together too:
df = spark.read.option("headers", "true").option("inferSchema", "true").csv("my_csv.csv", sep=";")
df = (
df
.withColumn('revenue', regexp_replace('revenue', '\\.', ''))
.withColumn('revenue', regexp_replace('revenue', ',', '.'))
.withColumn('revenue', df['revenue'].cast("float"))
)
Please note this I haven't tested this so there may be a typo or two in there.