PySpark行式功能组合

时间：2016-04-12 21:58:14

标签： python apache-spark pyspark apache-spark-sql

作为一个简化的例子，我有一个带有“col1，col2”列的数据框“df”，我想在将函数应用到每一列之后计算一个行的最大值：

def f(x):
    return (x+1)

max_udf=udf(lambda x,y: max(x,y), IntegerType())
f_udf=udf(f, IntegerType())

df2=df.withColumn("result", max_udf(f_udf(df.col1),f_udf(df.col2)))

所以如果df：

col1   col2
1      2
3      0

然后

DF2：

col1   col2  result
1      2     3
3      0     4

以上似乎不起作用并产生“无法评估表达式：PythonUDF＃f ......”

我绝对肯定“f_udf”在我的桌子上工作得很好，主要问题在于max_udf。

如果不创建额外的列或使用基本的map / reduce，有没有办法完全使用数据帧和udfs完成上述操作？我该如何修改“max_udf”？

我也试过了：

max_udf=udf(max, IntegerType())

产生相同的错误。

我也确认以下工作：

df2=(df.withColumn("temp1", f_udf(df.col1))
       .withColumn("temp2", f_udf(df.col2))

df2=df2.withColumn("result", max_udf(df2.temp1,df2.temp2))

为什么我不能一次性完成这些？

我希望看到一个可以概括为任何函数“f_udf”和“max_udf”的答案。

4 个答案:

答案 0 :(得分：33)

我遇到了类似的问题，并在this stackoverflow question

的答案中找到了解决方案

要将多列或整行传递给UDF，请使用struct：

from pyspark.sql.functions import udf, struct
from pyspark.sql.types import IntegerType

df = sqlContext.createDataFrame([(None, None), (1, None), (None, 2)], ("a", "b"))

count_empty_columns = udf(lambda row: len([x for x in row if x == None]), IntegerType())

new_df = df.withColumn("null_count", count_empty_columns(struct([df[x] for x in df.columns])))

new_df.show()

返回：

+----+----+----------+
|   a|   b|null_count|
+----+----+----------+
|null|null|         2|
|   1|null|         1|
|null|   2|         1|
+----+----+----------+

答案 1 :(得分：6)

UserDefinedFunction在接受UDF作为参数时抛出错误。

您可以像下面一样修改max_udf以使其正常工作。 

df = sc.parallelize([(1, 2), (3, 0)]).toDF(["col1", "col2"])

max_udf = udf(lambda x, y: max(x + 1, y + 1), IntegerType())

df2 = df.withColumn("result", max_udf(df.col1, df.col2))

或

def f_udf(x):
    return (x + 1)

max_udf = udf(lambda x, y: max(x, y), IntegerType())
## f_udf=udf(f, IntegerType())

df2 = df.withColumn("result", max_udf(f_udf(df.col1), f_udf(df.col2)))

注意：

当且仅当内部函数（此处为f_udf）生成有效的SQL表达式时，第二种方法才有效。

此处可行，因为f_udf(df.col1)和f_udf(df.col2)在传递给Column<b'(col1 + 1)'>之前分别被评估为Column<b'(col2 + 1)'>和max_udf。它不适用于任意函数。

如果我们尝试这样的例子，它就行不通了：

from math import exp

df.withColumn("result", max_udf(exp(df.col1), exp(df.col2)))

答案 2 :(得分：0)

下面是一个有用的代码，专门通过仅调用顶级业务规则而创建的任何新列，该业务规则与技术性和繁琐的Spark东西完全隔离（无需花费$，并且不再依赖于Databricks库）。 我的建议是，在您的组织中，尝试使生活简单，整洁，以获取顶级数据用户的利益：

def createColumnFromRule(df, columnName, ruleClass, ruleName, inputColumns=None, inputValues=None, columnType=None):
    from pyspark.sql import functions as F
    from pyspark.sql import types as T
    def _getSparkClassType(shortType):
        defaultSparkClassType = "StringType"
        typesMapping = {
            "bigint"    : "LongType",
            "binary"    : "BinaryType",
            "boolean"   : "BooleanType",
            "byte"      : "ByteType",
            "date"      : "DateType",
            "decimal"   : "DecimalType",
            "double"    : "DoubleType",
            "float"     : "FloatType",
            "int"       : "IntegerType",
            "integer"   : "IntegerType",
            "long"      : "LongType",
            "numeric"   : "NumericType",
            "string"    : defaultSparkClassType,
            "timestamp" : "TimestampType"
        }
        sparkClassType = None
        try:
            sparkClassType = typesMapping[shortType]
        except:
            sparkClassType = defaultSparkClassType
        return sparkClassType
    if (columnType != None): sparkClassType = _getSparkClassType(columnType)
    else: sparkClassType = "StringType"
    aUdf = eval("F.udf(ruleClass." + ruleName + ", T." + sparkClassType + "())")
    columns = None
    values = None
    if (inputColumns != None): columns = F.struct([df[column] for column in inputColumns])
    if (inputValues != None): values = F.struct([F.lit(value) for value in inputValues])
    # Call the rule
    if (inputColumns != None and inputValues != None): df = df.withColumn(columnName, aUdf(columns, values))
    elif (inputColumns != None): df = df.withColumn(columnName, aUdf(columns, F.lit(None)))
    elif (inputValues != None): df = df.withColumn(columnName, aUdf(F.lit(None), values))
    # Create a Null column otherwise
    else:
        if (columnType != None):
            df = df.withColumn(columnName, F.lit(None).cast(columnType))
        else:
            df = df.withColumn(columnName, F.lit(None))
    # Return the resulting dataframe
    return df

用法示例：

# Define your business rule (you can get columns and values)
class CustomerRisk:
    def churnRisk(self, columns=None, values=None):
        isChurnRisk = False
        # ... Rule implementation starts here
        if (values != None):
            if (values[0] == "FORCE_CHURN=true"): isChurnRisk = True
        if (isChurnRisk == False and columns != None):
            if (columns["AGE"]) <= 25): isChurnRisk = True
        # ...
        return isChurnRisk

# Execute the rule, it will create your new column in one line of code, that's all, easy isn't ?
# And look how to pass columns and values, it's really easy !
df = createColumnFromRule(df, columnName="CHURN_RISK", ruleClass=CustomerRisk(), ruleName="churnRisk", columnType="boolean", inputColumns=["NAME", "AGE", "ADDRESS"], inputValues=["FORCE_CHURN=true", "CHURN_RISK=100%"])

答案 3 :(得分：0)

处理此问题的最佳方法是转义 pyspark.sql.DataFrame 表示并通过 pyspark.RDDs 和 [pyspark.RDD.map()](https://spark.apache.org/docs/latest/api/python/reference/api/pyspark.RDD.map.html#pyspark.RDD.map) 使用 pyspark.sql.Row.asDict()。

import typing

# Save yourself some pain and always import these things: functions as F and types as T
import pyspark.sql.functions as F
import pyspark.sql.types as T
from pyspark.sql import Row, SparkSession, SQLContext


spark = (
    SparkSession.builder.appName("Stack Overflow Example")
    .getOrCreate()
)
sc = spark.sparkContext

# sqlContet is needed sometimes to create DataFrames from RDDs
sqlContext = SQLContext(sc)

df = sc.parallelize([Row(**{"a": "hello", "b": 1, "c": 2}), Row(**{"a": "goodbye", "b": 2, "c": 1})]).toDF(["a", "b", "c"])


def to_string(record:dict) -> Row:
    """Create a readable string representation of the record"""
    
    record["readable"] = f'Word: {record["a"]} A: {record["b"]} B: {record["c"]}'
    return Row(**record)


# Apply the function with a map after converting the Row to a dict
readable_rdd = df.rdd.map(lambda x: x.asDict()).map(to_string)

# Test the function without running the entire DataFrame through it
print(readable_rdd.first())

# This results in: Row(a='hello', b=1, c=2, readable='Word: hello A: 1 B: 2')

# Sometimes you can use `toDF()` to get a dataframe
readable_df = readable_rdd.toDF()

readable_df.show()

# +-------+---+---+--------------------+
# |      a|  b|  c|            readable|
# +-------+---+---+--------------------+
# |  hello|  1|  2|Word: hello A: 1 ...|
# |goodbye|  2|  1|Word: goodbye A: ...|
# +-------+---+---+--------------------+

# Sometimes you have to use createDataFrame with a specified schema
schema = T.StructType(
    [
        T.StructField("a", T.StringType(), True),
        T.StructField("b", T.IntegerType(), True),
        T.StructField("c", T.StringType(), True),
        T.StructField("readable", T.StringType(), True),
    ]
)

# This is more reliable, you should use it in production!
readable_df = sqlContext.createDataFrame(readable_rdd, schema)

readable_df.show()

# +-------+---+---+--------------------+
# |      a|  b|  c|            readable|
# +-------+---+---+--------------------+
# |  hello|  1|  2|Word: hello A: 1 ...|
# |goodbye|  2|  1|Word: goodbye A: ...|
# +-------+---+---+--------------------+

有时 RDD.map() 函数无法使用某些 Python 库，因为映射器会被序列化，因此您需要将数据划分为足够的分区以占用集群的所有核心，然后使用 pyspark.RDD.mapPartition() 进行处理一次整个分区（只是一个 Iterable 的 dicts）。这使您可以一次实例化一个开销较大的对象（例如 spaCy Language model），并且一次将其应用于一条记录，而无需重新创建它。

def to_string_partition(partition:typing.Iterable[dict]) -> typing.Iterable[Row]:
    """Add a readable string form to an entire partition"""
    # Instantiate expensive objects here
    
    # Apply these objects' methods here
    for record in partition:
        record["readable"] = f'Word: {record["a"]} A: {record["b"]} B: {record["c"]}'
        yield Row(**record)


readable_rdd = df.rdd.map(lambda x: x.asDict()).mapPartitions(to_string_partition)

print(readable_rdd.first())

# Row(a='hello', b=1, c=2, readable='Word: hello A: 1 B: 2')

# mapPartitions are more likely to require a specified schema
schema = T.StructType(
    [
        T.StructField("a", T.StringType(), True),
        T.StructField("b", T.IntegerType(), True),
        T.StructField("c", T.StringType(), True),
        T.StructField("readable", T.StringType(), True),
    ]
)

# This is more reliable, you should use it in production!
readable_df = sqlContext.createDataFrame(readable_rdd, schema)

readable_df.show()

# +-------+---+---+--------------------+
# |      a|  b|  c|            readable|
# +-------+---+---+--------------------+
# |  hello|  1|  2|Word: hello A: 1 ...|
# |goodbye|  2|  1|Word: goodbye A: ...|
# +-------+---+---+--------------------+

DataFrame API 很好，因为它们允许类似 SQL 的操作更快，但有时您需要没有任何限制的直接 Python 的强大功能，学习使用 RDD 将极大地有益于您的分析实践。例如，您可以对记录进行分组，然后在 RAM 中评估整个组，只要它适合 - 您可以通过更改分区键和限制工人/增加他们的 RAM 来安排。

import numpy as np


def median_b(x):
    """Process a group and determine the median value"""
    
    key = x[0]
    values = x[1]
    
    # Get the median value
    m = np.median([record["b"] for record in values])

    # Return a Row of the median for each group
    return Row(**{"a": key, "median_b": m})


median_b_rdd = df.rdd.map(lambda x: x.asDict()).groupBy(lambda x: x["a"]).map(median_b)
median_b_rdd.first()

# Row(a='hello', median_b=1.0)
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