将数组传递给 Spark Lit 函数 [英] Passing Array to Spark Lit function

问题描述

假设我有一个包含数字 1-10 的 numpy 数组 a:
[1 2 3 4 5 6 7 8 9 10]

我还有一个 Spark 数据框，我想向其中添加我的 numpy 数组 a.我认为一列文字可以完成这项工作.这不起作用:

df = df.withColumn(NewColumn", F.lit(a))

<块引用>

不支持的文字类型类 java.util.ArrayList

但这有效:

df = df.withColumn(NewColumn", F.lit(a[0]))

怎么做?

之前的示例 DF:

<头>

col1
a b c d e f g h i j

预期结果:

<头>

col1	新列
a b c d e f g h i j	1 2 3 4 5 6 7 8 9 10

解决方案

Spark array

中的列表解析

a = [1,2,3,4,5,6,7,8,9,10]df = spark.createDataFrame([['a b c d e f g h i j '],], ['col1'])df = df.withColumn(NewColumn", F.array([F.lit(x) for x in a]))df.show(截断=假)df.printSchema()# +--------------------+-------------------------------+# |col1 |新列|# +--------------------+-------------------------------+# |a b c d e f g h i j |[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]|# +--------------------+-------------------------------+#  根# |-- col1: string (nullable = true)# |-- NewColumn: 数组 (nullable = false)# ||-- 元素:整数(containsNull = false)

@pault 评论 (Python 2.7):

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您可以使用 map 隐藏循环:
df.withColumn(NewColumn", F.array(map(F.lit, a)))

@abegehr 添加了 Python 3 版本:

<块引用>

df.withColumn(NewColumn", F.array(*map(F.lit, a)))

Spark 的 udf

#定义UDFdef arrayUdf():返回一个callArrayUdf = F.udf(arrayUdf, T.ArrayType(T.IntegerType()))# 调用UDFdf = df.withColumn(NewColumn", callArrayUdf())

输出是一样的.

Let's say I have a numpy array a that contains the numbers 1-10:
[1 2 3 4 5 6 7 8 9 10]

I also have a Spark dataframe to which I want to add my numpy array a. I figure that a column of literals will do the job. This doesn't work:

df = df.withColumn("NewColumn", F.lit(a))

Unsupported literal type class java.util.ArrayList

But this works:

df = df.withColumn("NewColumn", F.lit(a[0]))

How to do it?

Example DF before:

col1
a b c d e f g h i j

Expected result:

col1	NewColumn
a b c d e f g h i j	1 2 3 4 5 6 7 8 9 10

解决方案

List comprehension inside Spark's array

a = [1,2,3,4,5,6,7,8,9,10]
df = spark.createDataFrame([['a b c d e f g h i j '],], ['col1'])
df = df.withColumn("NewColumn", F.array([F.lit(x) for x in a]))

df.show(truncate=False)
df.printSchema()
#  +--------------------+-------------------------------+
#  |col1                |NewColumn                      |
#  +--------------------+-------------------------------+
#  |a b c d e f g h i j |[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]|
#  +--------------------+-------------------------------+
#  root
#   |-- col1: string (nullable = true)
#   |-- NewColumn: array (nullable = false)
#   |    |-- element: integer (containsNull = false)

@pault commented (Python 2.7):

You can hide the loop using map:
df.withColumn("NewColumn", F.array(map(F.lit, a)))

@ abegehr added Python 3 version:

df.withColumn("NewColumn", F.array(*map(F.lit, a)))

Spark's udf

# Defining UDF
def arrayUdf():
    return a
callArrayUdf = F.udf(arrayUdf, T.ArrayType(T.IntegerType()))

# Calling UDF
df = df.withColumn("NewColumn", callArrayUdf())

Output is the same.

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