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這篇文章主要介紹“Apache下Flink transformation的用法”，在日常操作中，相信很多人在Apache下Flink transformation的用法問題上存在疑惑，小編查閱了各式資料，整理出簡單好用的操作方法，希望對大家解答”Apache下Flink transformation的用法”的疑惑有所幫助！接下來，請跟著小編一起來學習吧！
新建一個Object
object DataSetTransformationApp {
def main(args: Array[String]): Unit = {
val environment = ExecutionEnvironment.getExecutionEnvironment
}
def mapFunction(env: ExecutionEnvironment): Unit = {
val data = env.fromCollection(List(1,2,3,4,5,6,7,8,9,10))
}
}這里的數(shù)據(jù)源是一個1到10的list集合。Map的原理是：假設data數(shù)據(jù)集中有N個元素，將每一個元素進行轉(zhuǎn)化：
data.map { x => x.toInt }好比：y=f(x)
// 對data中的每一個元素都去做一個+1操作 data.map((x:Int) => x + 1 ).print()
然后對每一個元素都做一個+1操作。
簡單寫法：
如果這個里面只有一個元素，就可以直接寫成下面形式：
data.map((x) => x + 1).print()
更簡潔的寫法：
data.map(x => x + 1).print()
更簡潔的方法：
data.map(_ + 1).print()
輸出結(jié)果：
2 3 4 5 6 7 8 9 10 11
public static void main(String[] args) throws Exception {
ExecutionEnvironment executionEnvironment = ExecutionEnvironment.getExecutionEnvironment();
mapFunction(executionEnvironment);
}
public static void mapFunction(ExecutionEnvironment executionEnvironment) throws Exception {
List<String> list = new ArrayList<>();
for (int i = 1; i <= 10; i++) {
list.add(i + "");
}
DataSource<String> data = executionEnvironment.fromCollection(list);
data.map(new MapFunction<String, Integer>() {
public Integer map(String input) {
return Integer.parseInt(input) + 1;
}
}).print();
}因為我們定義的List是一個String的泛型，因此MapFunction的泛型是<String, Integer>，第一個參數(shù)表示輸入的類型，第二個參數(shù)表示輸出是一個Integer類型。
將每個元素執(zhí)行+1操作，并取出大于5的元素。
def filterFunction(env: ExecutionEnvironment): Unit = {
val data = env.fromCollection(List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))
data.map(_ + 1).filter(_ > 5).print()
}filter只會返回滿足條件的記錄。
public static void filterFunction(ExecutionEnvironment env) throws Exception {
List<Integer> list = new ArrayList<>();
for (int i = 1; i <= 10; i++) {
list.add(i);
}
DataSource<Integer> data = env.fromCollection(list);
data.map(new MapFunction<Integer, Integer>() {
public Integer map(Integer input) {
return input + 1;
}
}).filter(new FilterFunction<Integer>() {
@Override
public boolean filter(Integer input) throws Exception {
return input > 5;
}
}).print();
}map function 與 MapPartition function有什么區(qū)別？
需求：DataSource 中有100個元素,把結(jié)果存儲在數(shù)據(jù)庫中
如果使用map function ，那么實現(xiàn)方法如下：
// DataSource 中有100個元素,把結(jié)果存儲在數(shù)據(jù)庫中
def mapPartitionFunction(env: ExecutionEnvironment): Unit = {
val students = new ListBuffer[String]
for (i <- 1 to 100) {
students.append("Student" + i)
}
val data = env.fromCollection(students)
data.map(x=>{
// 每一個元素要存儲到數(shù)據(jù)庫中去，肯定需要先獲取到connection
val connection = DBUtils.getConnection()
println(connection + " ... ")
// TODO .... 保存數(shù)據(jù)到DB
DBUtils.returnConnection(connection)
}).print()
}打印結(jié)果，將會打印100個獲取DBUtils.getConnection()的請求。如果數(shù)據(jù)量增多，顯然不停的獲取連接是不現(xiàn)實的。
因此MapPartition就應運而生了，轉(zhuǎn)換一個分區(qū)里面的數(shù)據(jù)，也就是說一個分區(qū)中的數(shù)據(jù)調(diào)用一次。
因此要首先設置分區(qū)：
val data = env.fromCollection(students).setParallelism(4)
設置4個分區(qū)，也就是并行度，然后使用mapPartition來處理：
data.mapPartition(x => {
val connection = DBUtils.getConnection()
println(connection + " ... ")
// TODO .... 保存數(shù)據(jù)到DB
DBUtils.returnConnection(connection)
x
}).print()那么就會的到4次連接請求，每一個分區(qū)獲取一個connection。
public static void mapPartitionFunction(ExecutionEnvironment env) throws Exception {
List<String> list = new ArrayList<>();
for (int i = 1; i <= 100; i++) {
list.add("student:" + i);
}
DataSource<String> data = env.fromCollection(list);
/*data.map(new MapFunction<String, String>() {
@Override
public String map(String input) throws Exception {
String connection = DBUtils.getConnection();
System.out.println("connection = [" + connection + "]");
DBUtils.returnConnection(connection);
return input;
}
}).print();*/
data.mapPartition(new MapPartitionFunction<String, Object>() {
@Override
public void mapPartition(Iterable<String> values, Collector<Object> out) throws Exception {
String connection = DBUtils.getConnection();
System.out.println("connection = [" + connection + "]");
DBUtils.returnConnection(connection);
}
}).print();
}first表示獲取前幾個，groupBy表示分組，sortGroup表示分組內(nèi)排序
def firstFunction(env:ExecutionEnvironment): Unit = {
val info = ListBuffer[(Int, String)]()
info.append((1, "hadoop"))
info.append((1, "spark"))
info.append((1, "flink"))
info.append((2, "java"))
info.append((2, "springboot"))
info.append((3, "linux"))
info.append((4, "vue"))
val data = env.fromCollection(info)
data.first(3).print()
//輸出:(1,hadoop)
//(1,spark)
//(1,flink)
data.groupBy(0).first(2).print()//根據(jù)第一個字段分組，每個分組獲取前兩個數(shù)據(jù)
//(3,linux)
//(1,hadoop)
//(1,spark)
//(2,java)
//(2,springboot)
//(4,vue)
data.groupBy(0).sortGroup(1, Order.ASCENDING).first(2).print() //根據(jù)第一個字段分組，然后在分組內(nèi)根據(jù)第二個字段升序排序，并取出前兩個數(shù)據(jù)
//輸出(3,linux)
//(1,flink)
//(1,hadoop)
//(2,java)
//(2,springboot)
//(4,vue)
} public static void firstFunction(ExecutionEnvironment env) throws Exception {
List<Tuple2<Integer, String>> info = new ArrayList<>();
info.add(new Tuple2<>(1, "hadoop"));
info.add(new Tuple2<>(1, "spark"));
info.add(new Tuple2<>(1, "flink"));
info.add(new Tuple2<>(2, "java"));
info.add(new Tuple2<>(2, "springboot"));
info.add(new Tuple2<>(3, "linux"));
info.add(new Tuple2<>(4, "vue"));
DataSource<Tuple2<Integer, String>> data = env.fromCollection(info);
data.first(3).print();
data.groupBy(0).first(2).print();
data.groupBy(0).sortGroup(1, Order.ASCENDING).first(2).print();
}獲取一個元素，然后產(chǎn)生0個、1個或多個元素
def flatMapFunction(env: ExecutionEnvironment): Unit = {
val info = ListBuffer[(String)]()
info.append("hadoop,spark");
info.append("hadoop,flink");
info.append("flink,flink");
val data = env.fromCollection(info)
data.flatMap(_.split(",")).print()
}輸出：
hadoop spark hadoop flink flink flink
FlatMap將每個元素都用逗號分割，然后變成多個。
經(jīng)典例子：
data.flatMap(_.split(",")).map((_,1)).groupBy(0).sum(1).print()將每個元素用逗號分割，然后每個元素做map，然后根據(jù)第一個字段分組，然后根據(jù)第二個字段求和。
輸出結(jié)果如下：
(hadoop,2) (flink,3) (spark,1)
同樣實現(xiàn)一個經(jīng)典案例wordcount
public static void flatMapFunction(ExecutionEnvironment env) throws Exception {
List<String> info = new ArrayList<>();
info.add("hadoop,spark");
info.add("hadoop,flink");
info.add("flink,flink");
DataSource<String> data = env.fromCollection(info);
data.flatMap(new FlatMapFunction<String, String>() {
@Override
public void flatMap(String input, Collector<String> out) throws Exception {
String[] splits = input.split(",");
for(String split: splits) {
//發(fā)送出去
out.collect(split);
}
}
}).map(new MapFunction<String, Tuple2<String, Integer>>() {
@Override
public Tuple2<String, Integer> map(String value) throws Exception {
return new Tuple2<>(value,1);
}
}).groupBy(0).sum(1).print();
}去重操作
def distinctFunction(env: ExecutionEnvironment): Unit = {
val info = ListBuffer[(String)]()
info.append("hadoop,spark");
info.append("hadoop,flink");
info.append("flink,flink");
val data = env.fromCollection(info)
data.flatMap(_.split(",")).distinct().print()
}這樣就將每一個元素都做了去重操作。輸出如下：
hadoop flink spark
public static void distinctFunction(ExecutionEnvironment env) throws Exception {
List<String> info = new ArrayList<>();
info.add("hadoop,spark");
info.add("hadoop,flink");
info.add("flink,flink");
DataSource<String> data = env.fromCollection(info);
data.flatMap(new FlatMapFunction<String, String>() {
@Override
public void flatMap(String input, Collector<String> out) throws Exception {
String[] splits = input.split(",");
for(String split: splits) {
//發(fā)送出去
out.collect(split);
}
}
}).distinct().print();
}Joins two data sets by creating all pairs of elements that are equal on their keys. Optionally uses a JoinFunction to turn the pair of elements into a single element, or a FlatJoinFunction to turn the pair of elements into arbitrarily many (including none) elements. See the keys section to learn how to define join keys.
result = input1.join(input2) .where(0) // key of the first input (tuple field 0) .equalTo(1); // key of the second input (tuple field 1)
表示第一個tuple input1中的第0個字段，與第二個tuple input2中的第一個字段進行join。
def joinFunction(env: ExecutionEnvironment): Unit = {
val info1 = ListBuffer[(Int, String)]() //編號 名字
info1.append((1, "hadoop"))
info1.append((2, "spark"))
info1.append((3, "flink"))
info1.append((4, "java"))
val info2 = ListBuffer[(Int, String)]() //編號 城市
info2.append((1, "北京"))
info2.append((2, "上海"))
info2.append((3, "深圳"))
info2.append((5, "廣州"))
val data1 = env.fromCollection(info1)
val data2 = env.fromCollection(info2)
data1.join(data2).where(0).equalTo(0).apply((first, second)=>{
(first._1, first._2, second._2)
}).print()
}輸出結(jié)果如下：
(3,flink,深圳) (1,hadoop,北京) (2,spark,上海)
public static void joinFunction(ExecutionEnvironment env) throws Exception {
List<Tuple2<Integer, String>> info1 = new ArrayList<>(); //編號 名字
info1.add(new Tuple2<>(1, "hadoop"));
info1.add(new Tuple2<>(2, "spark"));
info1.add(new Tuple2<>(3, "flink"));
info1.add(new Tuple2<>(4, "java"));
List<Tuple2<Integer, String>> info2 = new ArrayList<>(); //編號 城市
info2.add(new Tuple2<>(1, "北京"));
info2.add(new Tuple2<>(2, "上海"));
info2.add(new Tuple2<>(3, "深圳"));
info2.add(new Tuple2<>(5, "廣州"));
DataSource<Tuple2<Integer, String>> data1 = env.fromCollection(info1);
DataSource<Tuple2<Integer, String>> data2 = env.fromCollection(info2);
data1.join(data2).where(0).equalTo(0).with(new JoinFunction<Tuple2<Integer, String>, Tuple2<Integer, String>, Tuple3<Integer, String, String>>() {
@Override
public Tuple3<Integer, String, String> join(Tuple2<Integer, String> first, Tuple2<Integer, String> second) throws Exception {
return new Tuple3<Integer, String, String>(first.f0, first.f1,second.f1);
}
}).print();
}Tuple2<Integer, String>, Tuple2<Integer, String>表示兩個輸入的集合，Tuple3<Integer, String, String>>表示輸出的Tuple3
上面講的join是內(nèi)連接，這個OuterJoin是外連接，包括左外連接，右外連接，全連接在兩個數(shù)據(jù)集上。
def outJoinFunction(env: ExecutionEnvironment): Unit = {
val info1 = ListBuffer[(Int, String)]() //編號 名字
info1.append((1, "hadoop"))
info1.append((2, "spark"))
info1.append((3, "flink"))
info1.append((4, "java"))
val info2 = ListBuffer[(Int, String)]() //編號 城市
info2.append((1, "北京"))
info2.append((2, "上海"))
info2.append((3, "深圳"))
info2.append((5, "廣州"))
val data1 = env.fromCollection(info1)
val data2 = env.fromCollection(info2)
data1.leftOuterJoin(data2).where(0).equalTo(0).apply((first, second) => {
if (second == null) {
(first._1, first._2, "-")
}else {
(first._1, first._2, second._2)
}
}).print() //左外連接 把左邊的所有數(shù)據(jù)展示出來
}左外連接，當左邊的數(shù)據(jù)在右邊沒有對應的數(shù)據(jù)時，需要進行處理，否則會出現(xiàn)空指針異常。輸出如下：
(3,flink,深圳) (1,hadoop,北京) (2,spark,上海) (4,java,-)
右外連接：
data1.rightOuterJoin(data2).where(0).equalTo(0).apply((first, second) => {
if (first == null) {
(second._1, "-", second._2)
}else {
(first._1, first._2, second._2)
}
}).print()右外連接，輸出：
(3,flink,深圳) (1,hadoop,北京) (5,-,廣州) (2,spark,上海)
全連接：
data1.fullOuterJoin(data2).where(0).equalTo(0).apply((first, second) => {
if (first == null) {
(second._1, "-", second._2)
}else if (second == null){
(second._1, "-", second._2)
} else {
(first._1, first._2, second._2)
}
}).print()(3,flink,深圳) (1,hadoop,北京) (5,-,廣州) (2,spark,上海) (4,java,-)
左外連接：
public static void outjoinFunction(ExecutionEnvironment env) throws Exception {
List<Tuple2<Integer, String>> info1 = new ArrayList<>(); //編號 名字
info1.add(new Tuple2<>(1, "hadoop"));
info1.add(new Tuple2<>(2, "spark"));
info1.add(new Tuple2<>(3, "flink"));
info1.add(new Tuple2<>(4, "java"));
List<Tuple2<Integer, String>> info2 = new ArrayList<>(); //編號 城市
info2.add(new Tuple2<>(1, "北京"));
info2.add(new Tuple2<>(2, "上海"));
info2.add(new Tuple2<>(3, "深圳"));
info2.add(new Tuple2<>(5, "廣州"));
DataSource<Tuple2<Integer, String>> data1 = env.fromCollection(info1);
DataSource<Tuple2<Integer, String>> data2 = env.fromCollection(info2);
data1.leftOuterJoin(data2).where(0).equalTo(0).with(new JoinFunction<Tuple2<Integer, String>, Tuple2<Integer, String>, Tuple3<Integer, String, String>>() {
@Override
public Tuple3<Integer, String, String> join(Tuple2<Integer, String> first, Tuple2<Integer, String> second) throws Exception {
if(second == null) {
return new Tuple3<Integer, String, String>(first.f0, first.f1, "-");
}
return new Tuple3<Integer, String, String>(first.f0, first.f1,second.f1);
}
}).print();
}右外連接：
data1.rightOuterJoin(data2).where(0).equalTo(0).with(new JoinFunction<Tuple2<Integer, String>, Tuple2<Integer, String>, Tuple3<Integer, String, String>>() {
@Override
public Tuple3<Integer, String, String> join(Tuple2<Integer, String> first, Tuple2<Integer, String> second) throws Exception {
if (first == null) {
return new Tuple3<Integer, String, String>(second.f0, "-", second.f1);
}
return new Tuple3<Integer, String, String>(first.f0, first.f1, second.f1);
}
}).print();全連接：
data1.fullOuterJoin(data2).where(0).equalTo(0).with(new JoinFunction<Tuple2<Integer, String>, Tuple2<Integer, String>, Tuple3<Integer, String, String>>() {
@Override
public Tuple3<Integer, String, String> join(Tuple2<Integer, String> first, Tuple2<Integer, String> second) throws Exception {
if (first == null) {
return new Tuple3<Integer, String, String>(second.f0, "-", second.f1);
} else if (second == null) {
return new Tuple3<Integer, String, String>(first.f0, first.f1, "-");
}
return new Tuple3<Integer, String, String>(first.f0, first.f1, second.f1);
}
}).print();笛卡爾積，左邊與右邊交叉處理
def crossFunction(env: ExecutionEnvironment): Unit = {
val info1 = List("喬峰", "慕容復")
val info2 = List(3,1,0)
val data1 = env.fromCollection(info1)
val data2 = env.fromCollection(info2)
data1.cross(data2).print()
}輸出：
(喬峰,3) (喬峰,1) (喬峰,0) (慕容復,3) (慕容復,1) (慕容復,0)
public static void crossFunction(ExecutionEnvironment env) throws Exception {
List<String> info1 = new ArrayList<>();
info1.add("喬峰");
info1.add("慕容復");
List<String> info2 = new ArrayList<>();
info2.add("3");
info2.add("1");
info2.add("0");
DataSource<String> data1 = env.fromCollection(info1);
DataSource<String> data2 = env.fromCollection(info2);
data1.cross(data2).print();
}到此，關(guān)于“Apache下Flink transformation的用法”的學習就結(jié)束了，希望能夠解決大家的疑惑。理論與實踐的搭配能更好的幫助大家學習，快去試試吧！若想繼續(xù)學習更多相關(guān)知識，請繼續(xù)關(guān)注億速云網(wǎng)站，小編會繼續(xù)努力為大家?guī)砀鄬嵱玫奈恼拢?/p>
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