我有两个数组列表,声明为:
ArrayList<JRadioButton> category = new ArrayList<JRadioButton>();
ArrayList<Integer> cat_ids = new ArrayList<Integer>();
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这两个字段都包含完全相同的值,它们在自然界中具有实际对应性.
我知道我可以像这样迭代其中一个循环:
for(JRadioButton button: category)
{
if(button.isSelected())
{
buttonName = button.getName();
System.out.println(buttonName);
}
}
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但是,我想同时迭代两个LISTS.我知道他们的尺寸完全相同.我怎么做?
Mar*_*oun 94
你可以使用Collection#iterator
:
Iterator<JRadioButton> it1 = category.iterator();
Iterator<Integer> it2 = cats_ids.iterator();
while (it1.hasNext() && it2.hasNext()) {
...
}
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Zol*_*asi 13
如果您经常这样做,您可以考虑使用帮助函数将两个列表压缩成一对列表:
public static <A, B> List<Pair<A, B>> zip(List<A> listA, List<B> listB) {
if (listA.size() != listB.size()) {
throw new IllegalArgumentException("Lists must have same size");
}
List<Pair<A, B>> pairList = new LinkedList<>();
for (int index = 0; index < listA.size(); index++) {
pairList.add(Pair.of(listA.get(index), listB.get(index)));
}
return pairList;
}
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您还需要一个Pair实现.Apache commons lang包有一个合适的.
有了这些,你现在可以优雅地迭代pairlist:
ArrayList<JRadioButton> category = new ArrayList<JRadioButton>();
ArrayList<Integer> cat_ids = new ArrayList<Integer>();
for (Pair<JRadioButton, Integer> item : zip(category , cat_ids)) {
// do something with JRadioButton
item.getLeft()...
// do something with Integer
item.getRight()...
}
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eug*_*e82 13
java8风格:
private static <T1, T2> void iterateSimultaneously(Iterable<T1> c1, Iterable<T2> c2, BiConsumer<T1, T2> consumer) {
Iterator<T1> i1 = c1.iterator();
Iterator<T2> i2 = c2.iterator();
while (i1.hasNext() && i2.hasNext()) {
consumer.accept(i1.next(), i2.next());
}
}
//
iterateSimultaneously(category, cay_id, (JRadioButton b, Integer i) -> {
// do stuff...
});
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JRR*_*JRR 11
试试这个
ArrayList<JRadioButton> category = new ArrayList<JRadioButton>();
ArrayList<Integer> cat_ids = new ArrayList<Integer>();
for (int i = 0; i < category.size(); i++) {
JRadioButton cat = category.get(i);
Integer id= cat_ids.get(i);
..
}
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