List.indexByRange
return a sublist of the list
Calling Sequence
Parameters
Description
Examples
void indexByRange( int a, int b ) throws MapleException
a
-
index of the beginning of the range
b
index of the end of the range
The indexByRange function returns a List object corresponding to the sublist from positions a to b inclusive.
Valid values of a and b are 1..n where n is the length of the list. The length can be obtained by calling numElements.
The input indices are one-based; that is, the index of the first element is 1 and the index of the last element equals the result of numElements. See subList for an equivalent which is zero-based.
import com.maplesoft.openmaple.*;
import com.maplesoft.externalcall.MapleException;
class Example
{
public static void main( String notused[] ) throws MapleException
String[] mapleArgs = { "java" };
Engine engine = new Engine( mapleArgs, new EngineCallBacksDefault(), null, null );
List l = (List)engine.evaluate( "[2,4,6,8,10]:" );
List l2 = l.indexByRange( 2, 4 );
System.out.prinln( "Sublist: " + l2 );
}
Executing this code produces the following output.
[4, 6, 8]
See Also
ExternalCalling/Java/MapleException
OpenMaple
OpenMaple/Java/Algebraic
OpenMaple/Java/API
OpenMaple/Java/List
OpenMaple/Java/List/contains
OpenMaple/Java/List/containsAll
OpenMaple/Java/List/select
OpenMaple/Java/List/subList
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