Pyspark：将多个数组列拆分为行

Question

我有一个数据框架，它有一行和几列。有些列是单个值，有些是列表。所有列表列的长度相同。我想将每个列表列拆分为单独的行，同时保留任何非列表列。

样本DF：

from pyspark import Row
from pyspark.sql import SQLContext
from pyspark.sql.functions import explode

sqlc = SQLContext(sc)

df = sqlc.createDataFrame([Row(a=1, b=[1,2,3],c=[7,8,9], d='foo')])
# +---+---------+---------+---+
# |  a|        b|        c|  d|
# +---+---------+---------+---+
# |  1|[1, 2, 3]|[7, 8, 9]|foo|
# +---+---------+---------+---+

我想要什么：

+---+---+----+------+
|  a|  b|  c |    d |
+---+---+----+------+
|  1|  1|  7 |  foo |
|  1|  2|  8 |  foo |
|  1|  3|  9 |  foo |
+---+---+----+------+

如果我只有一个列表列，那么只需执行< code>explode就很容易了:

df_exploded = df.withColumn('b', explode('b'))
# >>> df_exploded.show()
# +---+---+---------+---+
# |  a|  b|        c|  d|
# +---+---+---------+---+
# |  1|  1|[7, 8, 9]|foo|
# |  1|  2|[7, 8, 9]|foo|
# |  1|  3|[7, 8, 9]|foo|
# +---+---+---------+---+

然而，如果我尝试分解＜code＞c＜/code＞列，我得到的数据帧长度为我想要的平方：

df_exploded_again = df_exploded.withColumn('c', explode('c'))
# >>> df_exploded_again.show()
# +---+---+---+---+
# |  a|  b|  c|  d|
# +---+---+---+---+
# |  1|  1|  7|foo|
# |  1|  1|  8|foo|
# |  1|  1|  9|foo|
# |  1|  2|  7|foo|
# |  1|  2|  8|foo|
# |  1|  2|  9|foo|
# |  1|  3|  7|foo|
# |  1|  3|  8|foo|
# |  1|  3|  9|foo|
# +---+---+---+---+

我想要的是-对于每一列，取该列中数组的第n个元素，并将其添加到新行。我尝试过将爆炸映射到数据帧中的所有列，但似乎也不起作用：

df_split = df.rdd.map(lambda col: df.withColumn(col, explode(col))).toDF()

Answer 1

df.withColumn("bc", arrays_zip("b","c"))
  .select("a", explode("bc").alias("tbc"))
  .select("a", col"tbc.b", "tbc.c").show()

< code >从pyspark.sql.functions导入arrays_zip

1. 创建一列 bc，它是 b 列和 c 列的array_zip
2. 分解 bc 以获得结构 tbc
3. 选择所需的列 a、b 和 c（全部根据需要分解）。

> df.withColumn("bc", arrays_zip("b","c")).select("a", explode("bc").alias("tbc")).select("a", "tbc.b", col("tbc.c")).show()
+---+---+---+
|  a|  b|  c|
+---+---+---+
|  1|  1|  7|
|  1|  2|  8|
|  1|  3|  9|
+---+---+---+

Answer 2

您需要使用 flatMap，而不是映射，因为您希望从每个输入行中生成多个输出行。

from pyspark.sql import Row
def dualExplode(r):
    rowDict = r.asDict()
    bList = rowDict.pop('b')
    cList = rowDict.pop('c')
    for b,c in zip(bList, cList):
        newDict = dict(rowDict)
        newDict['b'] = b
        newDict['c'] = c
        yield Row(**newDict)

df_split = sqlContext.createDataFrame(df.rdd.flatMap(dualExplode))

Answer 3

火花

您可以将zip_ udf 替换为arrays_zip函数

from pyspark.sql.functions import arrays_zip, col, explode

(df
    .withColumn("tmp", arrays_zip("b", "c"))
    .withColumn("tmp", explode("tmp"))
    .select("a", col("tmp.b"), col("tmp.c"), "d"))

火花

使用< code >数据帧和UDF:

from pyspark.sql.types import ArrayType, StructType, StructField, IntegerType
from pyspark.sql.functions import col, udf, explode

zip_ = udf(
  lambda x, y: list(zip(x, y)),
  ArrayType(StructType([
      # Adjust types to reflect data types
      StructField("first", IntegerType()),
      StructField("second", IntegerType())
  ]))
)

(df
    .withColumn("tmp", zip_("b", "c"))
    # UDF output cannot be directly passed to explode
    .withColumn("tmp", explode("tmp"))
    .select("a", col("tmp.first").alias("b"), col("tmp.second").alias("c"), "d"))

使用RDD：

(df
    .rdd
    .flatMap(lambda row: [(row.a, b, c, row.d) for b, c in zip(row.b, row.c)])
    .toDF(["a", "b", "c", "d"]))

由于 Python 通信开销，这两种解决方案都效率低下。如果数据大小是固定的，您可以执行如下操作：

from functools import reduce
from pyspark.sql import DataFrame

# Length of array
n = 3

# For legacy Python you'll need a separate function
# in place of method accessor 
reduce(
    DataFrame.unionAll, 
    (df.select("a", col("b").getItem(i), col("c").getItem(i), "d")
        for i in range(n))
).toDF("a", "b", "c", "d")

甚至：

from pyspark.sql.functions import array, struct

# SQL level zip of arrays of known size
# followed by explode
tmp = explode(array(*[
    struct(col("b").getItem(i).alias("b"), col("c").getItem(i).alias("c"))
    for i in range(n)
]))

(df
    .withColumn("tmp", tmp)
    .select("a", col("tmp").getItem("b"), col("tmp").getItem("c"), "d"))

与UDF或RDD相比，这应该要快得多。泛化以支持任意数量的列：

# This uses keyword only arguments
# If you use legacy Python you'll have to change signature
# Body of the function can stay the same
def zip_and_explode(*colnames, n):
    return explode(array(*[
        struct(*[col(c).getItem(i).alias(c) for c in colnames])
        for i in range(n)
    ]))

df.withColumn("tmp", zip_and_explode("b", "c", n=3))