Question

我正在使用n列的PySpark DataFrame。我有一组m列（m

例如：

输入：PySpark DataFrame包含：

col_1 = [1,2,3], col_2 = [2,1,4], col_3 = [3,2,5]

输出：

col_4 = max(col1, col_2, col_3) = [3,2,5]

pandas中有类似this问题中解释的内容。

有没有办法在PySpark中执行此操作，还是应该将我的PySpark df转换为Pandas df然后执行操作？

Answer 1

您可以减少在列表列表中使用SQL表达式：

from pyspark.sql.functions import max as max_, col, when
from functools import reduce

def row_max(*cols):
    return reduce(
        lambda x, y: when(x > y, x).otherwise(y),
        [col(c) if isinstance(c, str) else c for c in cols]
    )

df = (sc.parallelize([(1, 2, 3), (2, 1, 2), (3, 4, 5)])
    .toDF(["a", "b", "c"]))

df.select(row_max("a", "b", "c").alias("max")))

Spark 1.5+还提供least，greatest

from pyspark.sql.functions import greatest

df.select(greatest("a", "b", "c"))

如果你想保留最大名称你可以使用`structs：

from pyspark.sql.functions import struct, lit

def row_max_with_name(*cols):
    cols_ = [struct(col(c).alias("value"), lit(c).alias("col")) for c in cols]
    return greatest(*cols_).alias("greatest({0})".format(",".join(cols)))

 maxs = df.select(row_max_with_name("a", "b", "c").alias("maxs"))

最后，您可以使用上面的选项来查找选择＆＃34; top＆＃34;柱：

from pyspark.sql.functions import max

((_, c), ) = (maxs
    .groupBy(col("maxs")["col"].alias("col"))
    .count()
    .agg(max(struct(col("count"), col("col"))))
    .first())

df.select(c)

Answer 2

您还可以使用内置least的pyspark：

from pyspark.sql.functions import least, col
df = df.withColumn('min', least(col('c1'), col('c2'), col('c3')))

Answer 3

Scala解决方案：

df = sc.parallelize(Seq((10, 10, 1 ), (200, 2, 20), (3, 30, 300), (400, 40, 4))).toDF("c1", "c2", "c3"))  

df.rdd.map(row=>List[String](row(0).toString,row(1).toString,row(2).toString)).map(x=>(x(0),x(1),x(2),x.min)).toDF("c1","c2","c3","min").show

+---+---+---+---+  
| c1| c2| c3|min|  
+---+---+---+---+  
| 10| 10|  1|  1|    
|200|  2| 20|  2|  
|  3| 30|300|  3|  
|400| 40|  4|  4|  
+---+---+---+---+

Answer 4

另一种简单的方法。我们假设下面的df是您的数据框

df = sc.parallelize([(10, 10, 1 ), (200, 2, 20), (3, 30, 300), (400, 40, 4)]).toDF(["c1", "c2", "c3"])
df.show()

+---+---+---+
| c1| c2| c3|
+---+---+---+
| 10| 10|  1|
|200|  2| 20|
|  3| 30|300|
|400| 40|  4|
+---+---+---+

您可以按照以下方式处理上述df以获得结果

from pyspark.sql.functions import lit, min

df.select( lit('c1').alias('cn1'), min(df.c1).alias('c1'),
           lit('c2').alias('cn2'), min(df.c2).alias('c2'),
           lit('c3').alias('cn3'), min(df.c3).alias('c3')
          )\
         .rdd.flatMap(lambda r: [ (r.cn1, r.c1), (r.cn2, r.c2), (r.cn3, r.c3)])\
         .toDF(['Columnn', 'Min']).show()

+-------+---+
|Columnn|Min|
+-------+---+
|     c1|  3|
|     c2|  2|
|     c3|  1|
+-------+---+

Answer 5

我们可以使用greatest

创建DataFrame

df = spark.createDataFrame(
    [[1,2,3], [2,1,2], [3,4,5]], 
    ['col_1','col_2','col_3']
)
df.show()
+-----+-----+-----+
|col_1|col_2|col_3|
+-----+-----+-----+
|    1|    2|    3|
|    2|    1|    2|
|    3|    4|    5|
+-----+-----+-----+

解决方案

from pyspark.sql.functions import greatest
df2 = df.withColumn('max_by_rows', greatest('col_1','col_2','col_3'))

#Only if you need col
#from pyspark.sql.functions import col
#df2 = df.withColumn('max',greatest(col('col_1'),col('col_2'),col('col_3')))
df2.show()

+-----+-----+-----+-----------+
|col_1|col_2|col_3|max_by_rows|
+-----+-----+-----+-----------+
|    1|    2|    3|          3|
|    2|    1|    2|          2|
|    3|    4|    5|          5|
+-----+-----+-----+-----------+

比较Pyspark中的列

5 个答案: