下面有任何指针吗?
输入df:此处col1
的类型为string
+----------------------------------+
| col1|
+----------------------------------+
|[{a:1,g:2},{b:3,h:4},{c:5,i:6}] |
|[{d:7,j:8},{e:9,k:10},{f:11,l:12}]|
+----------------------------------+
预期的输出:(同样col1
的类型为string
)
+-------------+
| col1 |
+-------------+
| {a:1,g:2} |
| {b:3,h:4} |
| {c:5,i:6} |
| {d:7,j:8} |
| {e:9,k:10} |
| {f:11,l:12}|
+-----+
谢谢!
答案 0 :(得分:2)
您可以将Spark SQL爆炸功能与UDF一起使用:
import spark.implicits._
val df = spark.createDataset(Seq("[{a},{b},{c}]","[{d},{e},{f}]")).toDF("col1")
df.show()
+-------------+
| col1|
+-------------+
|[{a},{b},{c}]|
|[{d},{e},{f}]|
+-------------+
import org.apache.spark.sql.functions._
val stringToSeq = udf{s: String => s.drop(1).dropRight(1).split(",")}
df.withColumn("col1", explode(stringToSeq($"col1"))).show()
+----+
|col1|
+----+
| {a}|
| {b}|
| {c}|
| {d}|
| {e}|
| {f}|
+----+
编辑:对于您新的输入数据,自定义UDF可以按以下方式演变:
val stringToSeq = udf{s: String =>
val extractor = "[^{]*:[^}]*".r
extractor.findAllIn(s).map(m => s"{$m}").toSeq
}
新输出:
+-----------+
| col1|
+-----------+
| {a:1,g:2}|
| {b:3,h:4}|
| {c:5,i:6}|
| {d:7,j:8}|
| {e:9,k:10}|
|{f:11,l:12}|
+-----------+
答案 1 :(得分:2)
Spark提供了非常丰富的trim函数,该函数可用于删除前导字符和尾随字符[]
。正如@LeoC已经提到的那样,可以通过内置函数来实现所需的功能,这些函数的性能会更好:
import org.apache.spark.sql.functions.{trim, explode, split}
val df = Seq(
("[{a},{b},{c}]"),
("[{d},{e},{f}]")
).toDF("col1")
df.select(
explode(
split(
trim($"col1", "[]"), ","))).show
// +---+
// |col|
// +---+
// |{a}|
// |{b}|
// |{c}|
// |{d}|
// |{e}|
// |{f}|
// +---+
编辑:
对于新数据集,逻辑保持不变,不同之处在于您需要使用,
以外的其他字符进行拆分。您可以使用regexp_replace
将},
替换为}|
,以便以后可以使用|
而不是,
进行拆分:
import org.apache.spark.sql.functions.{trim, explode, split, regexp_replace}
val df = Seq(
("[{a:1,g:2},{b:3,h:4},{c:5,i:6}]"),
("[{d:7,j:8},{e:9,k:10},{f:11,l:12}]")
).toDF("col1")
df.select(
explode(
split(
regexp_replace(trim($"col1", "[]"), "},", "}|"), // gives: {a:1,g:2}|{b:3,h:4}|{c:5,i:6}
"\\|")
)
).show(false)
// +-----------+
// |col |
// +-----------+
// |{a:1,g:2} |
// |{b:3,h:4} |
// |{c:5,i:6} |
// |{d:7,j:8} |
// |{e:9,k:10} |
// |{f:11,l:12}|
// +-----------+
请注意:使用split(..., "\\|")
可以转义|
,这是一个特殊的正则表达式字符。
答案 2 :(得分:1)
您可以这样做:
val newDF = df.as[String].flatMap(line=>line.replaceAll("\\[", "").replaceAll("\\]", "").split(","))
newDF.show()
输出:
+-----+
|value|
+-----+
| {a}|
| {b}|
| {c}|
| {d}|
| {e}|
| {f}|
+-----+
仅需注意,此过程会将输出列命名为value
,您可以使用select
,withColumn
等轻松地将其重命名(如果需要)。
答案 3 :(得分:1)
最后有效的方法:
import spark.implicits._
val df = spark.createDataset(Seq("[{a:1,g:2},{b:3,h:4},{c:5,i:6}]","[{d:7,j:8},{e:9,k:10},{f:11,l:12}]")).toDF("col1")
df.show()
val toStr = udf((value : String) => value.split("},\\{").map(_.toString))
val addParanthesis = udf((value : String) => ("{" + value + "}"))
val removeParanthesis = udf((value : String) => (value.slice(2,value.length()-2)))
import org.apache.spark.sql.functions._
df
.withColumn("col0", removeParanthesis(col("col1")))
.withColumn("col2", toStr(col("col0")))
.withColumn("col3", explode(col("col2")))
.withColumn("col4", addParanthesis(col("col3")))
.show()
输出:
+--------------------+--------------------+--------------------+---------+-----------+
| col1| col0| col2| col3| col4|
+--------------------+--------------------+--------------------+---------+-----------+
|[{a:1,g:2},{b:3,h...|a:1,g:2},{b:3,h:4...|[a:1,g:2, b:3,h:4...| a:1,g:2| {a:1,g:2}|
|[{a:1,g:2},{b:3,h...|a:1,g:2},{b:3,h:4...|[a:1,g:2, b:3,h:4...| b:3,h:4| {b:3,h:4}|
|[{a:1,g:2},{b:3,h...|a:1,g:2},{b:3,h:4...|[a:1,g:2, b:3,h:4...| c:5,i:6| {c:5,i:6}|
|[{d:7,j:8},{e:9,k...|d:7,j:8},{e:9,k:1...|[d:7,j:8, e:9,k:1...| d:7,j:8| {d:7,j:8}|
|[{d:7,j:8},{e:9,k...|d:7,j:8},{e:9,k:1...|[d:7,j:8, e:9,k:1...| e:9,k:10| {e:9,k:10}|
|[{d:7,j:8},{e:9,k...|d:7,j:8},{e:9,k:1...|[d:7,j:8, e:9,k:1...|f:11,l:12|{f:11,l:12}|
+--------------------+--------------------+--------------------+---------+-----------+