如何在 Spark DataFrame 中添加常量列? [英] How to add a constant column in a Spark DataFrame?

问题描述

我想在 DataFrame 中添加一个具有任意值的列(每行都相同).当我使用 withColumn 时出现错误，如下所示:

I want to add a column in a DataFrame with some arbitrary value (that is the same for each row). I get an error when I use withColumn as follows:

dt.withColumn('new_column', 10).head(5)

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-50-a6d0257ca2be> in <module>()
      1 dt = (messages
      2     .select(messages.fromuserid, messages.messagetype, floor(messages.datetime/(1000*60*5)).alias("dt")))
----> 3 dt.withColumn('new_column', 10).head(5)

/Users/evanzamir/spark-1.4.1/python/pyspark/sql/dataframe.pyc in withColumn(self, colName, col)
   1166         [Row(age=2, name=u'Alice', age2=4), Row(age=5, name=u'Bob', age2=7)]
   1167         """
-> 1168         return self.select('*', col.alias(colName))
   1169 
   1170     @ignore_unicode_prefix

AttributeError: 'int' object has no attribute 'alias'

似乎我可以通过添加和减去其他列之一(因此它们添加为零)然后添加我想要的数字(在这种情况下为 10)来诱使函数按照我想要的方式工作:

It seems that I can trick the function into working as I want by adding and subtracting one of the other columns (so they add to zero) and then adding the number I want (10 in this case):

dt.withColumn('new_column', dt.messagetype - dt.messagetype + 10).head(5)

[Row(fromuserid=425, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=47019141, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=49746356, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=93506471, messagetype=1, dt=4809600.0, new_column=10),
 Row(fromuserid=80488242, messagetype=1, dt=4809600.0, new_column=10)]

这太hacky了，对吧?我认为有更合法的方法可以做到这一点?

This is supremely hacky, right? I assume there is a more legit way to do this?

推荐答案

Spark 2.2+

Spark 2.2 引入了 typedLit 以支持 Seq、Map 和 Tuples(SPARK-19254) 并且应该支持以下调用(Scala):

Spark 2.2 introduces typedLit to support Seq, Map, and Tuples (SPARK-19254) and following calls should be supported (Scala):

import org.apache.spark.sql.functions.typedLit

df.withColumn("some_array", typedLit(Seq(1, 2, 3)))
df.withColumn("some_struct", typedLit(("foo", 1, 0.3)))
df.withColumn("some_map", typedLit(Map("key1" -> 1, "key2" -> 2)))

Spark 1.3+ (lit), 1.4+ (array, struct>)、2.0+(map):

Spark 1.3+ (lit), 1.4+ (array, struct), 2.0+ (map):

DataFrame.withColumn 的第二个参数应该是 Column 所以你必须使用文字:

The second argument for DataFrame.withColumn should be a Column so you have to use a literal:

from pyspark.sql.functions import lit

df.withColumn('new_column', lit(10))

如果你需要复杂的列，你可以使用像 array 这样的块来构建它们:

If you need complex columns you can build these using blocks like array:

from pyspark.sql.functions import array, create_map, struct

df.withColumn("some_array", array(lit(1), lit(2), lit(3)))
df.withColumn("some_struct", struct(lit("foo"), lit(1), lit(.3)))
df.withColumn("some_map", create_map(lit("key1"), lit(1), lit("key2"), lit(2)))

在 Scala 中可以使用完全相同的方法.

Exactly the same methods can be used in Scala.

import org.apache.spark.sql.functions.{array, lit, map, struct}

df.withColumn("new_column", lit(10))
df.withColumn("map", map(lit("key1"), lit(1), lit("key2"), lit(2)))

要为 structs 提供名称，请​​在每个字段上使用 alias:

To provide names for structs use either alias on each field:

df.withColumn(
    "some_struct",
    struct(lit("foo").alias("x"), lit(1).alias("y"), lit(0.3).alias("z"))
 )

或 cast 对整个对象

df.withColumn(
    "some_struct", 
    struct(lit("foo"), lit(1), lit(0.3)).cast("struct<x: string, y: integer, z: double>")
 )

虽然速度较慢，但​​也可以使用 UDF.

It is also possible, although slower, to use an UDF.

注意:

可以使用相同的构造将常量参数传递给 UDF 或 SQL 函数.

The same constructs can be used to pass constant arguments to UDFs or SQL functions.

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