假设我们有 2 个测试 String[],类似于:
String[] test = {"","","","",""};
String[] test2 = {"Test","Name", "5.00", "NY", "Single"};
然后我们将它们附加到这样的列表中:
List<String[]> testList = new ArrayList<>();
testList.add(test);
testList.add(test2);
我想做的目标是使用 Java 8 流在每个 String[] 的索引 2 中找到支付的美元总和。但我似乎无法过滤掉每个索引中包含空值的 String[] 。这是我的尝试:
public static void main(String[] args) {
String[] test = {"","","","",""};
String[] test2 = {"Test","Name", "5.00", "NY", "Single"};
List<String[]> testList = new ArrayList<>();
testList.add(test);
testList.add(test2);
System.out.println(testList);
testList.stream().map(Arrays::asList).filter(i -> !i.isEmpty())
.mapToDouble(columnsPerRow ->
Double.parseDouble(columnsPerRow.get(2)))
.sum();
System.out.println(testList);
}
我尝试使用 filter(i -> !i.isEmpty()) 来过滤它
它抛出的错误仍然是:
Exception in thread "main" java.lang.NumberFormatException: empty String
at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:1842)
at sun.misc.FloatingDecimal.parseDouble(FloatingDecimal.java:110)
at java.lang.Double.parseDouble(Double.java:538)
at Main.lambda$main$0(Main.java:23)
at java.util.stream.ReferencePipeline$6$1.accept(ReferencePipeline.java:244)
at java.util.stream.ReferencePipeline$3$1.accept(ReferencePipeline.java:193)
at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1382)
at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:482)
at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:472)
at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
at java.util.stream.DoublePipeline.collect(DoublePipeline.java:500)
at java.util.stream.DoublePipeline.sum(DoublePipeline.java:411)
at Main.main(Main.java:23)
在
i -> !i.isEmpty()
中,i
是ArrayList
,它包含("","","","","")
,如果它不为空,你可以检查索引2处的值,你可以这样做
double r = testList.stream().map(Arrays::asList)
.filter(i -> !i.isEmpty()) // ensure list not empty
.map(i -> i.get(2)) // keep only 3rd element
.filter(i -> !i.isEmpty()) // ensure string isn't empty
.mapToDouble(Double::parseDouble) // map to double
.sum();
第一个
.filter(i -> !i.isEmpty())
可以替换为.filter(i -> i.size() >= 3)
您过滤了错误的实体:
public static void main(String[] args)
{
String[] test = {"","","","",""};
String[] test2 = {"Test","Name", "5.00", "NY", "Single"};
String[] test3 = {"Test","Name", "13.00", "NY", "Single"};
List<String[]> testList = new ArrayList<>();
testList.add(test);
testList.add(test2);
testList.add(test3);
double sum = testList.stream()
.map(Arrays::asList)
.filter(i -> i.size() > 2 && !i.get(2).isEmpty())
.mapToDouble(columnsPerRow -> Double.parseDouble(columnsPerRow.get(2)))
.sum();
System.out.println(sum);
}
产生预期的
18.0
。
过滤阶段删除少于 2 个元素或第二个元素为空的数组。
相信你的话,过滤掉每个索引中包含空值的 String[]。
String[] test = {"","","","",""};
String[] test2 = {"Test","Name", "5.00", "NY", "Single"};
double total = Stream.of(test, test2)
.filter(arr -> !Arrays.stream(arr).allMatch(String::isEmpty))
.mapToDouble(arr -> Double.parseDouble(arr[2]))
.sum();
System.out.println(total);
输出是:
5.0
我的代码仅过滤掉字符串数组,其中all的值都是空字符串。因此,如果美元值为
""
并且数组包含其他非空字符串,则会出现异常。这就是您想要的没有美元值的数组只需要保存空字符串的情况。为了验证。您可能想要捕获 NumerFormatException
并报告验证错误。或者在进行计算之前验证流。