我想将向量列展开为数据帧中的普通列。 .transform创建单独的列,但是数据类型或“可为空”有问题,当我尝试进行.show时会出错,请参见下面的示例代码。如何解决该问题?
from pyspark.sql.types import *
from pyspark.ml.feature import VectorAssembler
from pyspark.sql.functions import udf
spark = SparkSession\
.builder\
.config("spark.driver.maxResultSize", "40g") \
.config('spark.sql.shuffle.partitions', '2001') \
.getOrCreate()
data = [(0.2, 53.3, 0.2, 53.3),
(1.1, 43.3, 0.3, 51.3),
(2.6, 22.4, 0.4, 43.3),
(3.7, 25.6, 0.2, 23.4)]
df = spark.createDataFrame(data, ['A','B','C','D'])
df.show(3)
df.printSchema()
vecAssembler = VectorAssembler(inputCols=['C','D'], outputCol="features")
new_df = vecAssembler.transform(df)
new_df.printSchema()
new_df.show(3)
split1_udf = udf(lambda value: value[0], DoubleType())
split2_udf = udf(lambda value: value[1], DoubleType())
new_df = new_df.withColumn('c1', split1_udf('features')).withColumn('c2', split2_udf('features'))
new_df.printSchema()
new_df.show(3)
答案 0 :(得分:0)
功能列的类型为pyspark.ml.linalg.DenseVector
,特征矢量元素的类型为numpy.float64
。
要将numpy dtypes
转换为本地python
,请输入value.item()
split1_udf = udf(lambda value: value[0].item(), DoubleType())
split2_udf = udf(lambda value: value[1].item(), DoubleType())
使用此修复程序将产生以下输出
+---+----+---+----+----------+---+----+
| A| B| C| D| features| c1| c2|
+---+----+---+----+----------+---+----+
|0.2|53.3|0.2|53.3|[0.2,53.3]|0.2|53.3|
|1.1|43.3|0.3|51.3|[0.3,51.3]|0.3|51.3|
|2.6|22.4|0.4|43.3|[0.4,43.3]|0.4|43.3|
|3.7|25.6|0.2|23.4|[0.2,23.4]|0.2|23.4|
+---+----+---+----+----------+---+----+
答案 1 :(得分:0)
我不知道UDF的问题是什么。但是我在下面找到了另一个解决方案。
data = [(0.2, 53.3, 0.2, 53.3),
(1.1, 43.3, 0.3, 51.3),
(2.6, 22.4, 0.4, 43.3),
(3.7, 25.6, 0.2, 23.4)]
df = spark.createDataFrame(data, ['A','B','C','D'])
vecAssembler = VectorAssembler(inputCols=['C','D'], outputCol="features")
new_df = vecAssembler.transform(df)
def extract(row):
return (row.A, row.B,row.C,row.D,) + tuple(row.features.toArray().tolist())
extracted_df = new_df.rdd.map(extract).toDF(['A','B','C','D', 'col1', 'col2'])
extracted_df.show()