问题
My input dataframe looks like the below
from pyspark.sql import SparkSession
spark = SparkSession.builder.appName("Basics").getOrCreate()
df=spark.createDataFrame(data=[('Alice',4.300,None),('Bob',float('nan'),897)],schema=['name','High','Low'])
+-----+----+----+
| name|High| Low|
+-----+----+----+
|Alice| 4.3|null|
| Bob| NaN| 897|
+-----+----+----+
Expected Output if divided by 10.0
+-----+----+----+
| name|High| Low|
+-----+----+----+
|Alice| 0.43|null|
| Bob| NaN| 89.7|
+-----+----+----+
回答1:
I don't know about any library function that could do this, but this snippet seems to do job just fine:
CONSTANT = 10.0
for field in df.schema.fields:
if str(field.dataType) in ['DoubleType', 'FloatType', 'LongType', 'IntegerType', 'DecimalType']:
name = str(field.name)
df = df.withColumn(name, col(name)/CONSTANT)
df.show()
outputs:
+-----+----+----+
| name|High| Low|
+-----+----+----+
|Alice|0.43|null|
| Bob| NaN|89.7|
+-----+----+----+
回答2:
The below code should solve your problem in a time efficient manner
from pyspark.sql.functions import col
allowed_types = ['DoubleType', 'FloatType', 'LongType', 'IntegerType', 'DecimalType']
df = df.select(*[(col(field.name)/10).name(field.name) if str(field.dataType) in allowed_types else col(field.name) for field in df.schema.fields]
Using "withColumn" iteratively might not be a good idea when the number of columns is large.
This is because PySpark dataframes are immutable, so essentially we will be creating a new DataFrame for each column casted using withColumn, which will be a very slow process.
This is where the above code comes in handy.
来源:https://stackoverflow.com/questions/44807818/how-divide-or-multiply-every-non-string-columns-of-a-pyspark-dataframe-with-a-fl