(Py)Spark DataFrame 中的映射值

Question

在Pandas中，可以这样操作：

mapping = {
    'a': 'The letter A',
    'b': 'The letter B',
    'c': 'The third letter'
}

x = pd.Series(['a', 'b', 'a', c']).map(mapping)

并获得类似

的东西

pd.Series([
    'The letter A',
    'The letter B',
    'The letter A',
    'The third letter'
])

天真地，我可以在 PySpark DataFrame 中使用类似

的东西来实现这一点

import pyspark.sql.functions as F
import pyspark.sql.functions as T

def _map_values_str(value, mapping, default=None):
    """ Apply a mapping, assuming the result is a string """
    return mapping.get(value, default)

map_values_str = F.udf(_map_values_str, T.StringType())

mapping = {
    'a': 'The letter A',
    'b': 'The letter B',
    'c': 'The third letter'
}

data = spark.createDataFrame([('a',), ('b',), ('a',), ('c',)], schema=['letters'])
data = data.withColumn('letters_mapped', map_values_str(F.col('letters'), mapping))

但根据我的经验，像这样的 UDF 在大型数据集上往往会有些慢。有没有更有效的方法？

Answer 1

我认为在这种情况下，您可以将 dict 转换为 DataFrame 并简单地使用 join:

import pyspark.sql.functions as F

mapping = {
    'a': 'The letter A',
    'b': 'The letter B',
    'c': 'The third letter'
}
# Convert so Spark DataFrame
mapping_df = spark.sparkContext.parallelize([(k,)+(v,) for k,v in mapping.items()]).toDF(['letters','val'])

data = spark.createDataFrame([('a',), ('b',), ('a',), ('c',)], schema=['letters'])
data = data.join(mapping_df.withColumnRenamed('val','letters_mapped'),'letters','left')
data.show()

输出：

+-------+----------------+
|letters|  letters_mapped|
+-------+----------------+
|      c|The third letter|
|      b|    The letter B|
|      a|    The letter A|
|      a|    The letter A|
+-------+----------------+

希望对您有所帮助！

(Py)Spark DataFrame 中的映射值

Mapping values in a (Py)Spark DataFrame

python

pyspark

pyspark-sql