如何在Pyspark中逐元素连接两个ArrayType（StringType（））列？

Question

我在spark数据帧中有两个ArrayType(StringType())列，我想将这两个列逐元素连接：

输入：

+-------------+-------------+
|col1         |col2         |
+-------------+-------------+
|['a','b']    |['c','d']    |
|['a','b','c']|['e','f','g']|
+-------------+-------------+

输出：

+-------------+-------------+----------------+
|col1         |col2         |col3            |
+-------------+-------------+----------------+
|['a','b']    |['c','d']    |['ac', 'bd']    |
|['a','b','c']|['e','f','g']|['ae','bf','cg']|
+-------------+----------- -+----------------+

谢谢。

Answer 1

对于Spark 2.4+，您可以使用transform函数，如下所示：

col3_expr = "transform(col1, (x, i) -> concat(x, col2[i]))"
df.withColumn("col3", expr(col3_expr)).show()

transform函数将第一数组列col1作为参数，对其元素进行迭代，并应用lambda函数(x, i) -> concat(x, col2[i])，其中x是实际元素，而{{1 }}其索引用于从数组i中获取相应的元素。

礼物：

col2

或更高级的zip_with函数更简单：

+------+------+--------+
|  col1|  col2|    col3|
+------+------+--------+
|[a, b]|[c, d]|[ac, bd]|
+------+------+--------+

Answer 2

它不会真正扩展，但是您可以在每个数组中获得0th和1st条目，然后说col3是a[0] + b[0]然后是{{1} }。 将所有4个条目设置为单独的值，然后将它们组合输出。

Answer 3

这是一个通用答案。只需查看结果即可。 2个相等大小的数组，因此两个元素均为n个元素。

from pyspark.sql.functions import *
from pyspark.sql.types import *

# Gen in this case numeric data, etc. 3 rows with 2 arrays of varying length, but both the same length as in your example
df = spark.createDataFrame([   ( list([x,x+1,4, x+100]), 4, list([x+100,x+200,999,x+500])   ) for x in range(3)], ['array1', 'value1', 'array2'] )    
num_array_elements = len(df.select("array1").first()[0])

# concat
df2 = df.select(([ concat(col("array1")[i], col("array2")[i]) for i in range(num_array_elements)]))
df2.withColumn("res", array(df2.schema.names)).show(truncate=False)

返回：

Answer 4

这里是替代答案，可以用于更新的非原创问题。使用array和array_except演示此类方法的用法。接受的答案更加优雅。

from pyspark.sql.functions import *
from pyspark.sql.types import *

# Arbitrary max number of elements to apply array over, need not broadcast such a small amount of data afaik.
max_entries = 5 

# Gen in this case numeric data, etc. 3 rows with 2 arrays of varying length,but per row constant length. 
dfA = spark.createDataFrame([   ( list([x,x+1,4, x+100]), 4, list([x+100,x+200,999,x+500])   ) for x in range(3)], ['array1', 'value1', 'array2'] ).withColumn("s",size(col("array1")))    
dfB = spark.createDataFrame([   ( list([x,x+1]), 4, list([x+100,x+200])   ) for x in range(5)], ['array1', 'value1', 'array2'] ).withColumn("s",size(col("array1"))) 
df = dfA.union(dfB)

# concat the array elements which are variable in size up to a max amount.
df2 = df.select(( [concat(col("array1")[i], col("array2")[i]) for i in range(max_entries)]))
df3 = df2.withColumn("res", array(df2.schema.names))

# Get results but strip out null entires from array.
df3.select(array_except(df3.res, array(lit(None)))).show(truncate=False)

无法获取要传递到范围内的列的s值。

这将返回：

+------------------------------+
|array_except(res, array(NULL))|
+------------------------------+
|[0100, 1200, 4999, 100500]    |
|[1101, 2201, 4999, 101501]    |
|[2102, 3202, 4999, 102502]    |
|[0100, 1200]                  |
|[1101, 2201]                  |
|[2102, 3202]                  |
|[3103, 4203]                  |
|[4104, 5204]                  |
+------------------------------+