如何将多个文本文件读入单个RDD？

您可以指定整个目录,使用通配符甚至CSV目录和通配符.例如:

sc.textFile("/my/dir1,/my/paths/part-00[0-5]*,/another/dir,/a/specific/file")

正如Nick Chammas所指出的,这是Hadoop的曝光,FileInputFormat因此这也适用于Hadoop(和Scalding).

您可以使用单个textFile调用来读取多个文件.斯卡拉:

sc.textFile(','.join(files)) 

你可以用它

首先,您可以获得S3路径的缓冲区/列表:

import scala.collection.JavaConverters._
import java.util.ArrayList
import com.amazonaws.services.s3.AmazonS3Client
import com.amazonaws.services.s3.model.ObjectListing
import com.amazonaws.services.s3.model.S3ObjectSummary
import com.amazonaws.services.s3.model.ListObjectsRequest

def listFiles(s3_bucket:String, base_prefix : String) = {
    var files = new ArrayList[String]

    //S3 Client and List Object Request
    var s3Client = new AmazonS3Client();
    var objectListing: ObjectListing = null;
    var listObjectsRequest = new ListObjectsRequest();

    //Your S3 Bucket
    listObjectsRequest.setBucketName(s3_bucket)

    //Your Folder path or Prefix
    listObjectsRequest.setPrefix(base_prefix)

    //Adding s3:// to the paths and adding to a list
    do {
      objectListing = s3Client.listObjects(listObjectsRequest);
      for (objectSummary <- objectListing.getObjectSummaries().asScala) {
        files.add("s3://" + s3_bucket + "/" + objectSummary.getKey());
      }
      listObjectsRequest.setMarker(objectListing.getNextMarker());
    } while (objectListing.isTruncated());

    //Removing Base Directory Name
    files.remove(0)

    //Creating a Scala List for same
    files.asScala
  }

现在将此List对象传递给下面的代码段,注意:sc是SQLContext的对象

var df: DataFrame = null;
  for (file <- files) {
    val fileDf= sc.textFile(file)
    if (df!= null) {
      df= df.unionAll(fileDf)
    } else {
      df= fileDf
    }
  }

现在你有一个最终的统一RDD即df

可选,您也可以在一个BigRDD中重新分配它

val files = sc.textFile(filename, 1).repartition(1)

重新分区始终有效:D

使用union方法如下:

val sc = new SparkContext(...)
val r1 = sc.textFile("xxx1")
val r2 = sc.textFile("xxx2")
...
val rdds = Seq(r1, r2, ...)
val bigRdd = sc.union(rdds)

然后bigRdd是包含所有文件的RDD.