如何有效地小写集合的每个元素？ [英] How to lowercase every element of a collection efficiently?

问题描述

降低List或Set的每个元素的最有效方法是什么？

What's the most efficient way to lower case every element of a List or Set?

我对列表的想法：

final List<String> strings = new ArrayList<String>();
strings.add("HELLO");
strings.add("WORLD");

for(int i=0,l=strings.size();i<l;++i)
{
  strings.add(strings.remove(0).toLowerCase());
}

有更好，更快的方式吗？这个例子对于一个Set怎么样？由于目前没有方法将一个操作应用于Set（或List）的每个元素，是否可以在不创建额外临时Set的情况下完成？

Is there a better, faster way? How would this example look like for a Set? As there is currently no method for applying an operation to each element of a Set (or List) can it be done without creating an additional temporary Set?

这样的事情会很好：

Set<String> strings = new HashSet<String>();
strings.apply(
  function (element)
  { this.replace(element, element.toLowerCase();) } 
);

谢谢，

推荐答案

这似乎是一个相当干净的列表解决方案。它应该允许特定的List实现用于提供一个实现，该实现对于列表的遍历（线性时间）和字符串的替换都是最佳的。

This seems like a fairly clean solution for lists. It should allow for the particular List implementation being used to provide an implementation that is optimal for both the traversal of the list--in linear time--and the replacing of the string--in constant time.

public static void replace(List<String> strings)
{
    ListIterator<String> iterator = strings.listIterator();
    while (iterator.hasNext())
    {
        iterator.set(iterator.next().toLowerCase());
    }
}

---

这是我能想到的最好的套装。正如其他人所说，由于多种原因，该操作不能在该组中就地执行。小写字符串可能需要放在集合中的不同位置，而不是它要替换的字符串。此外，如果小写字符串与已经添加的另一个小写字符串相同（例如，HELLO和Hello都将产生hello），则根本不会将小写字符串添加到集合中，这将是只被添加到集合中一次）。

---

This is the best that I can come up with for sets. As others have said, the operation cannot be performed in-place in the set for a number of reasons. The lower-case string may need to be placed in a different location in the set than the string it is replacing. Moreover, the lower-case string may not be added to the set at all if it is identical to another lower-case string that has already been added (e.g., "HELLO" and "Hello" will both yield "hello", which will only be added to the set once).

public static void replace(Set<String> strings)
{
    String[] stringsArray = strings.toArray(new String[0]);
    for (int i=0; i<stringsArray.length; ++i)
    {
        stringsArray[i] = stringsArray[i].toLowerCase();
    }
    strings.clear();
    strings.addAll(Arrays.asList(stringsArray));
}

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