I\'m trying to use the below code to calculate the average of a set of values that a user enters and display it in a jTextArea
but it does not work properly. Sa
sum += i;
You're adding the index; you should be adding the actual item in the ArrayList
:
sum += marks.get(i);
Also, to ensure the return value isn't truncated, force one operand to double
and change your method signature to double
:
return (double)sum / marks.size();
If using Java8 you can get the average of the values from a List as follows:
List<Integer> intList = Arrays.asList(1,2,2,3,1,5);
Double average = intList.stream().mapToInt(val -> val).average().orElse(0.0);
This has the advantage of having no moving parts. It can be easily adapted to work with a List of other types of object by changing the map method call.
For example with Doubles:
List<Double> dblList = Arrays.asList(1.1,2.1,2.2,3.1,1.5,5.3);
Double average = dblList.stream().mapToDouble(val -> val).average().orElse(0.0);
NB. mapToDouble is required because it returns a DoubleStream which has an average
method, while using map
does not.
or BigDecimals:
@Test
public void bigDecimalListAveragedCorrectly() {
List<BigDecimal> bdList = Arrays.asList(valueOf(1.1),valueOf(2.1),valueOf(2.2),valueOf(3.1),valueOf(1.5),valueOf(5.3));
Double average = bdList.stream().mapToDouble(BigDecimal::doubleValue).average().orElse(0.0);
assertEquals(2.55, average, 0.000001);
}
using orElse(0.0)
removes problems with the Optional object returned from the average
being 'not present'.
List.stream().mapToDouble(a->a).average()
You can use standard looping constructs or iterator/listiterator for the same :
List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8);
double sum = 0;
Iterator<Integer> iter1 = list.iterator();
while (iter1.hasNext()) {
sum += iter1.next();
}
double average = sum / list.size();
System.out.println("Average = " + average);
If using Java 8, you could use Stream or IntSream operations for the same :
OptionalDouble avg = list.stream().mapToInt(Integer::intValue).average();
System.out.println("Average = " + avg.getAsDouble());
Reference : Calculating average of arraylist
When the number is not big, everything seems just right. But if it isn't, great caution is required to achieve correctness.
Take double as an example:
If it is not big, as others mentioned you can just try this simply:
doubles.stream().mapToDouble(d -> d).average().orElse(0.0);
However, if it's out of your control and quite big, you have to turn to BigDecimal as follows (methods in the old answers using BigDecimal actually are wrong).
doubles.stream().map(BigDecimal::valueOf).reduce(BigDecimal.ZERO, BigDecimal::add)
.divide(BigDecimal.valueOf(doubles.size())).doubleValue();
Enclose the tests I carried out to demonstrate my point:
@Test
public void testAvgDouble() {
assertEquals(5.0, getAvgBasic(Stream.of(2.0, 4.0, 6.0, 8.0)), 1E-5);
List<Double> doubleList = new ArrayList<>(Arrays.asList(Math.pow(10, 308), Math.pow(10, 308), Math.pow(10, 308), Math.pow(10, 308)));
// Double.MAX_VALUE = 1.7976931348623157e+308
BigDecimal doubleSum = BigDecimal.ZERO;
for (Double d : doubleList) {
doubleSum = doubleSum.add(new BigDecimal(d.toString()));
}
out.println(doubleSum.divide(valueOf(doubleList.size())).doubleValue());
out.println(getAvgUsingRealBigDecimal(doubleList.stream()));
out.println(getAvgBasic(doubleList.stream()));
out.println(getAvgUsingFakeBigDecimal(doubleList.stream()));
}
private double getAvgBasic(Stream<Double> doubleStream) {
return doubleStream.mapToDouble(d -> d).average().orElse(0.0);
}
private double getAvgUsingFakeBigDecimal(Stream<Double> doubleStream) {
return doubleStream.map(BigDecimal::valueOf)
.collect(Collectors.averagingDouble(BigDecimal::doubleValue));
}
private double getAvgUsingRealBigDecimal(Stream<Double> doubleStream) {
List<Double> doubles = doubleStream.collect(Collectors.toList());
return doubles.stream().map(BigDecimal::valueOf).reduce(BigDecimal.ZERO, BigDecimal::add)
.divide(valueOf(doubles.size()), BigDecimal.ROUND_DOWN).doubleValue();
}
As for Integer
or Long
, correspondingly you can use BigInteger
similarly.