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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