Question

我正在尝试在pyspark中加入2个数据帧。我的问题是我想要我的＆＃34;内部加入＆＃34;无论NULL如何都给它一个通行证。我可以看到，在scala中，我有一个＆lt; =＆gt; 的替代品。但是，＆lt; =＆gt; 在pyspark中无效。

userLeft = sc.parallelize([
Row(id=u'1', 
    first_name=u'Steve', 
    last_name=u'Kent', 
    email=u's.kent@email.com'),
Row(id=u'2', 
    first_name=u'Margaret', 
    last_name=u'Peace', 
    email=u'marge.peace@email.com'),
Row(id=u'3', 
    first_name=None, 
    last_name=u'hh', 
    email=u'marge.hh@email.com')]).toDF()

userRight = sc.parallelize([
Row(id=u'2', 
    first_name=u'Margaret', 
    last_name=u'Peace', 
    email=u'marge.peace@email.com'),
Row(id=u'3', 
    first_name=None, 
    last_name=u'hh', 
    email=u'marge.hh@email.com')]).toDF()

目前的工作版本：

userLeft.join(userRight, (userLeft.last_name==userRight.last_name) & (userLeft.first_name==userRight.first_name)).show()

当前结果：

    +--------------------+----------+---+---------+--------------------+----------+---+---------+
|               email|first_name| id|last_name|               email|first_name| id|last_name|
    +--------------------+----------+---+---------+--------------------+----------+---+---------+ 
    |marge.peace@email...|  Margaret|  2|    Peace|marge.peace@email...|  Margaret|  2|    Peace|
    +--------------------+----------+---+---------+--------------------+----------+---+---------+

预期结果：

    +--------------------+----------+---+---------+--------------------+----------+---+---------+
|               email|first_name| id|last_name|               email|first_name| id|last_name|
+--------------------+----------+---+---------+--------------------+----------+---+---------+
|  marge.hh@email.com|      null|  3|       hh|  marge.hh@email.com|      null|  3|       hh|
|marge.peace@email...|  Margaret|  2|    Peace|marge.peace@email...|  Margaret|  2|    Peace|
+--------------------+----------+---+---------+--------------------+----------+---+---------+

Answer 1

对于 PYSPARK <2.3.0，您仍然可以使用这样的表达式列来构建 <=> 运算符：

import pyspark.sql.functions as F
df1.alias("df1").join(df2.alias("df2"), on = F.expr('df1.column <=> df2.column'))

对于 PYSPARK> = 2.3.0 ，您可以使用 Column.eqNullSafe 或 IS NOT DISTINCT FROM 作为回答here

Answer 2

使用其他值而不是null：

userLeft = userLeft.na.fill("unknown")
userRight = userRight.na.fill("unknown")

userLeft.join(userRight, ["last_name", "first_name"])

    +---------+----------+--------------------+---+--------------------+---+
    |last_name|first_name|               email| id|               email| id|
    +---------+----------+--------------------+---+--------------------+---+
    |    Peace|  Margaret|marge.peace@email...|  2|marge.peace@email...|  2|
    |       hh|   unknown|  marge.hh@email.com|  3|  marge.hh@email.com|  3|
    +---------+----------+--------------------+---+--------------------+---+

PySpark：在联接中处理NULL

2 个答案: