Question

我有一个返回不可变列表的方法。我想向其中添加元素，这就是为什么必须将其转换为可变列表的原因。目前，我正在从不可变列表中创建一个新的ArrayList，如下所示：

final List<someDTO> mutableList = new ArrayList<>(someDTO.getImmutableList());

是否有更好的方法，例如使用某些集合复制方法，java流或类似方法？

Answer 1

说实话，这是最好的，但是另一个变种是：

scala> val df = Seq(("ServiceCent4","AP-1-IOO-PPP","241.206.155.172","06-12-18:17:42:34",162,53,1544098354885L)).toDF("COL1","COL2","COL3","EventTime","COL4","COL5","COL6")
df: org.apache.spark.sql.DataFrame = [COL1: string, COL2: string ... 5 more fields]

scala> df.show(1,false)
+------------+------------+---------------+-----------------+----+----+-------------+
|COL1        |COL2        |COL3           |EventTime        |COL4|COL5|COL6         |
+------------+------------+---------------+-----------------+----+----+-------------+
|ServiceCent4|AP-1-IOO-PPP|241.206.155.172|06-12-18:17:42:34|162 |53  |1544098354885|
+------------+------------+---------------+-----------------+----+----+-------------+
only showing top 1 row

scala> df.printSchema
root
 |-- COL1: string (nullable = true)
 |-- COL2: string (nullable = true)
 |-- COL3: string (nullable = true)
 |-- EventTime: string (nullable = true)
 |-- COL4: integer (nullable = false)
 |-- COL5: integer (nullable = false)
 |-- COL6: long (nullable = false)


scala> val schema = df.schema
schema: org.apache.spark.sql.types.StructType = StructType(StructField(COL1,StringType,true), StructField(COL2,StringType,true), StructField(COL3,StringType,true), StructField(EventTime,StringType,true), StructField(COL4,IntegerType,false), StructField(COL5,IntegerType,false), StructField(COL6,LongType,false))

scala> val df2 = df.columns.foldLeft(df){ (acc,r) => acc.withColumn(r,col(r).cast("string")) }
df2: org.apache.spark.sql.DataFrame = [COL1: string, COL2: string ... 5 more fields]

scala> df2.printSchema
root
 |-- COL1: string (nullable = true)
 |-- COL2: string (nullable = true)
 |-- COL3: string (nullable = true)
 |-- EventTime: string (nullable = true)
 |-- COL4: string (nullable = false)
 |-- COL5: string (nullable = false)
 |-- COL6: string (nullable = false)


scala> val x = df2.flatMap(r => (0 until schema.length).map { idx => ((idx, r.getString(idx)), 1l) } )
x: org.apache.spark.sql.Dataset[((Int, String), Long)] = [_1: struct<_1: int, _2: string>, _2: bigint]

scala> x.show(5,false)
+---------------------+---+
|_1                   |_2 |
+---------------------+---+
|[0,ServiceCent4]     |1  |
|[1,AP-1-IOO-PPP]     |1  |
|[2,241.206.155.172]  |1  |
|[3,06-12-18:17:42:34]|1  |
|[4,162]              |1  |
+---------------------+---+
only showing top 5 rows


scala>

Answer 2

实际上没有更好的方法。

有人认为Collectors.toList带有静态导入的流使其看起来更好，但这纯粹是美观的。

List<someDTO> mutableList = someDTO.getImmutableList().stream().collect(toList());

将不可变转换为可变列表Java

2 个答案: