Removing nulls while mapping a JavaRDD
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I am developing a Java + Apache Spark application. One of its funcionalities is to trim Strings. For example:
// Having JavaSparkContext jsc previously defined and a function like
// the following already defined:
private String trim(String s, int charsToTrim);
// that returns a new String with length = s.length() - charsToTrim or
// null if its new length is <= 0
JavaRDD<String> original = jsc.parallelize(Arrays.asList("aaa", "bbbb", "ccccc");
int num = 4;
JavaRDD<String> mapped = original.map(s -> trim(s, num));
The new JavaRDD mapped will contain (null, null, "c"), and I will have to clean null values with a filter.
The question I have is: Is there any way to remove those values while doing the map or to don't return any value at while mapping all if the result is null?
Thank you.
java apache-spark null
asked Nov 19 at 22:26
Roi G.
33
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up vote
0
down vote
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I am developing a Java + Apache Spark application. One of its funcionalities is to trim Strings. For example:
// Having JavaSparkContext jsc previously defined and a function like
// the following already defined:
private String trim(String s, int charsToTrim);
// that returns a new String with length = s.length() - charsToTrim or
// null if its new length is <= 0
JavaRDD<String> original = jsc.parallelize(Arrays.asList("aaa", "bbbb", "ccccc");
int num = 4;
JavaRDD<String> mapped = original.map(s -> trim(s, num));
The new JavaRDD mapped will contain (null, null, "c"), and I will have to clean null values with a filter.
The question I have is: Is there any way to remove those values while doing the map or to don't return any value at while mapping all if the result is null?
Thank you.
java apache-spark null
asked Nov 19 at 22:26
Roi G.
33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am developing a Java + Apache Spark application. One of its funcionalities is to trim Strings. For example:
// Having JavaSparkContext jsc previously defined and a function like
// the following already defined:
private String trim(String s, int charsToTrim);
// that returns a new String with length = s.length() - charsToTrim or
// null if its new length is <= 0
JavaRDD<String> original = jsc.parallelize(Arrays.asList("aaa", "bbbb", "ccccc");
int num = 4;
JavaRDD<String> mapped = original.map(s -> trim(s, num));
The new JavaRDD mapped will contain (null, null, "c"), and I will have to clean null values with a filter.
The question I have is: Is there any way to remove those values while doing the map or to don't return any value at while mapping all if the result is null?
Thank you.
java apache-spark null
asked Nov 19 at 22:26
Roi G.
33
I am developing a Java + Apache Spark application. One of its funcionalities is to trim Strings. For example:
// Having JavaSparkContext jsc previously defined and a function like
// the following already defined:
private String trim(String s, int charsToTrim);
// that returns a new String with length = s.length() - charsToTrim or
// null if its new length is <= 0
JavaRDD<String> original = jsc.parallelize(Arrays.asList("aaa", "bbbb", "ccccc");
int num = 4;
JavaRDD<String> mapped = original.map(s -> trim(s, num));
The new JavaRDD mapped will contain (null, null, "c"), and I will have to clean null values with a filter.
The question I have is: Is there any way to remove those values while doing the map or to don't return any value at while mapping all if the result is null?
Thank you.
java apache-spark null
java apache-spark null
asked Nov 19 at 22:26
Roi G.
33
asked Nov 19 at 22:26
Roi G.
33
asked Nov 19 at 22:26
Roi G.
33
asked Nov 19 at 22:26
Roi G.
33
asked Nov 19 at 22:26
Roi G.
33
33
add a comment |
add a comment |
1 Answer
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To remove or filter some values you can always use .filter over an RDD.
But you can not use .map for skipping the elements or filtering them. However, you can use flatMap method here which supports the intended functionality. Try the following line in your code:
mapped.flatMap(x -> Arrays.asList((x == null)? (new String[0]) : (new String{x})).iterator());
answered Nov 19 at 23:43
anuj saxena
23917
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
To remove or filter some values you can always use .filter over an RDD.
But you can not use .map for skipping the elements or filtering them. However, you can use flatMap method here which supports the intended functionality. Try the following line in your code:
mapped.flatMap(x -> Arrays.asList((x == null)? (new String[0]) : (new String{x})).iterator());
answered Nov 19 at 23:43
anuj saxena
23917
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
add a comment |
up vote
0
down vote
accepted
To remove or filter some values you can always use .filter over an RDD.
But you can not use .map for skipping the elements or filtering them. However, you can use flatMap method here which supports the intended functionality. Try the following line in your code:
mapped.flatMap(x -> Arrays.asList((x == null)? (new String[0]) : (new String{x})).iterator());
answered Nov 19 at 23:43
anuj saxena
23917
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
To remove or filter some values you can always use .filter over an RDD.
But you can not use .map for skipping the elements or filtering them. However, you can use flatMap method here which supports the intended functionality. Try the following line in your code:
mapped.flatMap(x -> Arrays.asList((x == null)? (new String[0]) : (new String{x})).iterator());
answered Nov 19 at 23:43
anuj saxena
23917
To remove or filter some values you can always use .filter over an RDD.
But you can not use .map for skipping the elements or filtering them. However, you can use flatMap method here which supports the intended functionality. Try the following line in your code:
mapped.flatMap(x -> Arrays.asList((x == null)? (new String[0]) : (new String{x})).iterator());
answered Nov 19 at 23:43
anuj saxena
23917
answered Nov 19 at 23:43
anuj saxena
23917
answered Nov 19 at 23:43
anuj saxena
23917
answered Nov 19 at 23:43
anuj saxena
23917
23917
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
add a comment |
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
Thank you so much, that gave me the hint I needed!
– Roi G.
Nov 20 at 21:40
add a comment |
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