按值满足某些条件删除元素 [英] Remove elements by value satisfying certain condition
问题描述
从这些数据结构中,我想按值删除元素,满足某些条件
From these data structures, I want to remove elements by value, that satisfies certain condition
<Data Structures>
- RowSortedTable<String, String, Double> a; (Guava Table)
- HashMap<String, Double> b;
来自上一个问题,我找到了优雅的答案使用 Collections.Singleton
然而,似乎需要完全匹配。
From the previous question, I found the elegant answer using Collections.Singleton
however, it seems like exact matching is required.
hmap.values().removeAll(Collections.singleton("Two"));
在这里,我想从表格或地图中删除其值小于特定阈值的元素。你可以用什么方式编写代码?
Here, I want to remove elements from a table or map where their values are smaller than certain threshold. What would be your way to write the code?
我刚检查了两个答案,那些是关于地图的答案,怎么样表格案例?我的解决方案如下。
I just checked two answers and those are answers about map, how about the table case? My solution is as follows.
for (Iterator<String> it1 = proptypeconf.columnKeySet().iterator(); it1.hasNext();) {
String type = it1.next();
System.out.println(type);
for (Iterator<Map.Entry<String, Double>> it2 = proptypeconf.column(type).entrySet().iterator(); it2.hasNext();){
Map.Entry<String, Double> e = it2.next();
if (e.getValue() < conflist.get(index-1)) {
it2.remove();
}
}
}
推荐答案
Iterator<Integer> iterator = hmap.values().iterator();
while (iterator.hasNext()) {
if (iterator.next() < threshold) {
iterator.remove();
}
}
当然,如果你使用的是Java 8,它更容易:
Of course, if you're on Java 8, it's much easier:
hmap.values().removeIf(value -> value < threshold);
表的工作方式完全相同;只需使用 table.values()
而不是 hmap.values()
。
Tables work exactly the same; just use table.values()
instead of hmap.values()
.
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