blocks|key|1861378|text|据我所知，iterate和instantiate是唯一的方法。例如(对于其他潜在的帮助，因为我相信你知道如何做到这一点)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1861379|List<Integer>+oldList+=+...
/*+Specify+the+size+of+the+list+up+front+to+prevent+resizing.+*/
List<String>+newList+=+new+ArrayList<>(oldList.size());
for+(Integer+myInt+:+oldList)+{+
++newList.add(String.valueOf(myInt));+
}|code-block|syntax|javascript|1861380|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

As far as I know, iterate and instantiate is the only way to do this. Something like (for others potential help, since I'm sure you know how to do this):

<pre><code>List&lt;Integer&gt; oldList = ...
/* Specify the size of the list up front to prevent resizing. */
List&lt;String&gt; newList = new ArrayList&lt;&gt;(oldList.size());
for (Integer myInt : oldList) { 
 newList.add(String.valueOf(myInt)); 
}
</code></pre>

blocks|key|2075936|text|通过使用Google+Collections+from+Guava-Project，您可以在Lists类中使用transform方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2075937|import+com.google.common.collect.Lists;
import+com.google.common.base.Functions

List<Integer>+integers+=+Arrays.asList(1,+2,+3,+4);

List<String>+strings+=+Lists.transform(integers,+Functions.toStringFunction());|code-block|syntax|javascript|2075938|transform返回的List是支持列表上的一个视图-转换将应用于对转换后的列表的每次访问。|2075939|请注意，当应用于null时，Functions.toStringFunction()将抛出NullPointerException，因此只有在您确定列表中不包含null时才使用它。|2075940|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/google/guava/|1|https://google.github.io/guava/releases/23.0/api/docs/com/google/common/collect/Lists.html^0|1J|9|4|11|0|1A|5|1|0|0|0|9|C|4|0|E|S|19|K|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

Using <a href="https://github.com/google/guava/" rel="noreferrer">Google Collections from Guava-Project</a>, you could use the <code>transform</code> method in the <a href="https://google.github.io/guava/releases/23.0/api/docs/com/google/common/collect/Lists.html" rel="noreferrer">Lists</a> class

<pre><code>import com.google.common.collect.Lists;
import com.google.common.base.Functions

List&lt;Integer&gt; integers = Arrays.asList(1, 2, 3, 4);

List&lt;String&gt; strings = Lists.transform(integers, Functions.toStringFunction());
</code></pre>

The <code>List</code> returned by <code>transform</code> is a view on the backing list - the transformation will be applied on each access to the transformed list.

Be aware that <code>Functions.toStringFunction()</code> will throw a <code>NullPointerException</code> when applied to null, so only use it if you are sure your list will not contain null.

blocks|key|2076200|text|Java+8的解决方案。比Guava要长一点，但至少你不需要安装库。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2076201|import+java.util.Arrays;
import+java.util.List;
import+java.util.stream.Collectors;

//...

List<Integer>+integers+=+Arrays.asList(1,+2,+3,+4);
List<String>+strings+=+integers.stream().map(Object::toString)
++++++++++++++++++++++++++++++++++++++++.collect(Collectors.toList());|code-block|syntax|javascript|2076202|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Solution for Java 8. A bit longer than the Guava one, but at least you don't have to install a library.

<pre><code>import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

//...

List&lt;Integer&gt; integers = Arrays.asList(1, 2, 3, 4);
List&lt;String&gt; strings = integers.stream().map(Object::toString)
 .collect(Collectors.toList());
</code></pre>

blocks|key|1861790|text|您所做的一切都很好，但是如果您觉得需要'Java-it-up‘，您可以使用来自Apache+Commons的Transformer和collect+method，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1861791|public+class+IntegerToStringTransformer+implements+Transformer<Integer,+String>+{
+++public+String+transform(final+Integer+i)+{
++++++return+(i+==+null+?+null+:+i.toString());
+++}
}|code-block|syntax|javascript|1861792|然后..and+..|1861793|CollectionUtils.collect(
+++collectionOfIntegers,+
+++new+IntegerToStringTransformer(),+
+++newCollectionOfStrings);|1861794|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/|1|http://commons.apache.org/collections/api/org/apache/commons/collections/Transformer.html|2|http://commons.apache.org/collections/api/org/apache/commons/collections/CollectionUtils.html^0|13|E|0|1I|B|1|1U|E|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]|$A|12|B|13|1|14]|$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|19|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|1B|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]|W|$5|P|Q|R|C|$S|X]]]]

What you're doing is fine, but if you feel the need to 'Java-it-up' you could use a <a href="http://commons.apache.org/collections/api/org/apache/commons/collections/Transformer.html" rel="nofollow noreferrer">Transformer</a> and the <a href="http://commons.apache.org/collections/api/org/apache/commons/collections/CollectionUtils.html" rel="nofollow noreferrer">collect method</a> from <a href="http://commons.apache.org/" rel="nofollow noreferrer">Apache Commons</a>, e.g.:

<pre><code>public class IntegerToStringTransformer implements Transformer&lt;Integer, String&gt; {
 public String transform(final Integer i) {
 return (i == null ? null : i.toString());
 }
}
</code></pre>

..and then..

<pre><code>CollectionUtils.collect(
 collectionOfIntegers, 
 new IntegerToStringTransformer(), 
 newCollectionOfStrings);
</code></pre>

blocks|key|2075350|text|我不使用String.valueOf，而是使用.toString()；它避免了@johnathan.holland描述的一些自动装箱|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075351|javadoc说明valueOf返回的内容与Integer.toString()相同。|2075352|List<Integer>+oldList+=+...
List<String>+newList+=+new+ArrayList<String>(oldList.size());

for+(Integer+myInt+:+oldList)+{+
++newList.add(myInt.toString());+
}|code-block|syntax|javascript|2075353|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Instead of using String.valueOf I'd use .toString(); it avoids some of the auto boxing described by @johnathan.holland

The javadoc says that valueOf returns the same thing as Integer.toString().

<pre><code>List&lt;Integer&gt; oldList = ...
List&lt;String&gt; newList = new ArrayList&lt;String&gt;(oldList.size());

for (Integer myInt : oldList) { 
 newList.add(myInt.toString()); 
}
</code></pre>

blocks|key|2075449|text|String.valueOf的源代码如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075450|public+static+String+valueOf(Object+obj)+{
++++return+(obj+==+null)+?+"null"+:+obj.toString();
}|code-block|syntax|javascript|2075451|这并不重要，但我会使用toString。|2075452|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The source for String.valueOf shows this:

<pre><code>public static String valueOf(Object obj) {
 return (obj == null) ? "null" : obj.toString();
}
</code></pre>

Not that it matters much, but I would use toString.

blocks|key|2076105|text|这里有一个没有欺骗非JDK库的一行程序解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2076106|List<String>+strings+=+Arrays.asList(list.toString().replaceAll("\\[(.*)\\]",+"$1").split(",+"));|code-block|syntax|javascript|2076107|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a one-liner solution without cheating with a non-JDK library.

<pre><code>List&lt;String&gt; strings = Arrays.asList(list.toString().replaceAll("\\[(.*)\\]", "$1").split(", "));
</code></pre>

blocks|key|2076212|text|另一个使用Guava和Java+8的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2076213|List<Integer>+numbers+=+Arrays.asList(1,+2,+3,+4,+5);
List<String>+strings+=+Lists.transform(numbers,+number+->+String.valueOf(number));|code-block|syntax|javascript|2076214|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another Solution using Guava and Java 8

<pre><code>List&lt;Integer&gt; numbers = Arrays.asList(1, 2, 3, 4, 5);
List&lt;String&gt; strings = Lists.transform(numbers, number -&gt; String.valueOf(number));
</code></pre>

blocks|key|2075557|text|不是核心Java，也不是泛型，但是流行的Jakarta+commons集合库对这类任务有一些有用的抽象。具体地说，请看一下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075558|CollectionUtils|offset|length|2075559|如果您已经在项目中使用commons集合，则需要考虑的事项。|2075560|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/collections/api-release/org/apache/commons/collections/CollectionUtils.html^0|0|0|F|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@$D|R|E|S|1|T]]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

Not core Java, and not generic-ified, but the popular Jakarta commons collections library has some useful abstractions for this sort of task. Specifically, have a look at the collect methods on

<a href="http://commons.apache.org/collections/api-release/org/apache/commons/collections/CollectionUtils.html" rel="nofollow noreferrer">CollectionUtils</a>

Something to consider if you are already using commons collections in your project.

blocks|key|1861883|text|对于jsight中关心“拳击”的人来说，答案是:没有。这里使用的是String.valueOf(Object)，并且不会对int执行拆箱操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1861884|使用Integer.toString()还是String.valueOf(Object)取决于您希望如何处理可能的空值。你想抛出一个异常(可能)，或者在你的列表中有"null“字符串(可能)。如果是前者，您想抛出NullPointerException还是其他类型？|1861885|另外，在jsight的响应中有一个小缺陷：List是一个接口，你不能在它上面使用新操作符。在这种情况下，我可能会使用java.util.ArrayList，特别是因为我们预先知道列表可能有多长。|1861886|entityMap^0|X|M|1P|3|0|2|I|M|M|2Y|K|0|L|4|1M|J|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|K|$]]

To the people concerned about "boxing" in jsight's answer: there is none. <code>String.valueOf(Object)</code> is used here, and no unboxing to <code>int</code> is ever performed.

Whether you use <code>Integer.toString()</code> or <code>String.valueOf(Object)</code> depends on how you want to handle possible nulls. Do you want to throw an exception (probably), or have "null" Strings in your list (maybe). If the former, do you want to throw a <code>NullPointerException</code> or some other type?

Also, one small flaw in jsight's response: <code>List</code> is an interface, you can't use the new operator on it. I would probably use a <code>java.util.ArrayList</code> in this case, especially since we know up front how long the list is likely to be.

blocks|key|2076216|text|List<String>+stringList+=+integerList.stream().map((Object+s)->String.valueOf(s)).collect(Collectors.toList())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2076217|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;String&gt; stringList = integerList.stream().map((Object s)-&gt;String.valueOf(s)).collect(Collectors.toList())
</code></pre>

blocks|key|1862528|text|在原始列表上使用forEach方法的稍微更简洁的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1862529|++++List<Integer>+oldList+=+Arrays.asList(1,+2,+3,+4,+5);
++++List<String>+newList+=+new+ArrayList<>(oldList.size());
++++oldList.forEach(e+->+newList.add(String.valueOf(e)));|code-block|syntax|javascript|1862530|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A slightly more concise solution using the forEach method on the original list:

<pre><code> List&lt;Integer&gt; oldList = Arrays.asList(1, 2, 3, 4, 5);
 List&lt;String&gt; newList = new ArrayList&lt;&gt;(oldList.size());
 oldList.forEach(e -&gt; newList.add(String.valueOf(e)));
</code></pre>

blocks|key|2075274|text|@Jonathan:我可能弄错了，但我相信本例中的String.valueOf()将调用String.valueOf(Object)函数，而不是被装箱为String.valueOf(int)。如果String.valueOf(Object)为null，则只返回"null“；如果为非null，则调用Object.toString()，这不应该涉及装箱(尽管显然需要实例化新的字符串对象)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075275|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

@Jonathan: I could be mistaken, but I believe that String.valueOf() in this case will call the String.valueOf(Object) function rather than getting boxed to String.valueOf(int). String.valueOf(Object) just returns "null" if it is null or calls Object.toString() if non-null, which shouldn't involve boxing (although obviously instantiating new string objects is involved).

blocks|key|2075433|text|我认为将Object.toString()用于除调试之外的任何目的可能是一个非常糟糕的想法，即使在这种情况下两者在功能上是等价的(假设列表中没有空值)。开发人员可以随意更改任何toString()方法的行为，而无需任何警告，包括标准库中任何类的toString()方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075434|甚至不用担心装箱/拆箱过程导致的性能问题。如果性能很重要，只需使用数组。如果它真的很关键，就不要使用Java。试图智胜JVM只会导致心痛。|2075435|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

I think using Object.toString() for any purpose other than debugging is probably a really bad idea, even though in this case the two are functionally equivalent (assuming the list has no nulls). Developers are free to change the behavior of any toString() method without any warning, including the toString() methods of any classes in the standard library.

Don't even worry about the performance problems caused by the boxing/unboxing process. If performance is critical, just use an array. If it's really critical, don't use Java. Trying to outsmart the JVM will only lead to heartache.

blocks|key|2075732|text|仅供专家回答：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075733|++++List<Integer>+ints+=+...;
++++String+all+=+new+ArrayList<Integer>(ints).toString();
++++String[]+split+=+all.substring(1,+all.length()-1).split(",+");
++++List<String>+strs+=+Arrays.asList(split);|code-block|syntax|javascript|2075734|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

An answer for experts only:

<pre><code> List&lt;Integer&gt; ints = ...;
 String all = new ArrayList&lt;Integer&gt;(ints).toString();
 String[] split = all.substring(1, all.length()-1).split(", ");
 List&lt;String&gt; strs = Arrays.asList(split);
</code></pre>

blocks|key|1862296|text|Lambdaj允许以一种非常简单和可读的方式来实现这一点。例如，假设你有一个Integer列表，你想把它们转换成相应的字符串表示，你可以这样写；|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1862297|List<Integer>+ints+=+asList(1,+2,+3,+4);
Iterator<String>+stringIterator+=+convertIterator(ints,+new+Converter<Integer,+String>+{
++++public+String+convert(Integer+i)+{+return+Integer.toString(i);+}
}|code-block|syntax|javascript|1862298|Lambdaj仅在迭代结果时应用转换函数。|1862299|entityMap|0|LINK|mutability|MUTABLE|url|http://code.google.com/p/lambdaj/^0|0|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<a href="http://code.google.com/p/lambdaj/" rel="nofollow noreferrer">Lambdaj</a> allows to do that in a very simple and readable way. For example, supposing you have a list of Integer and you want to convert them in the corresponding String representation you could write something like that;

<pre><code>List&lt;Integer&gt; ints = asList(1, 2, 3, 4);
Iterator&lt;String&gt; stringIterator = convertIterator(ints, new Converter&lt;Integer, String&gt; {
 public String convert(Integer i) { return Integer.toString(i); }
}
</code></pre>

Lambdaj applies the conversion function only while you're iterating on the result.

blocks|key|1861705|text|您无法避免“装箱开销”；Java的人工泛型容器只能存储对象，因此您的int必须被装箱为整数。原则上，它可以避免从Object到Integer的向下转换(因为它没有意义，因为Object对于String.valueOf和Object.toString都足够好)，但我不知道编译器是否足够聪明来做到这一点。从字符串到对象的转换应该或多或少是无操作的，所以我不愿意担心这个问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1861706|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can't avoid the "boxing overhead"; Java's faux generic containers can only store Objects, so your ints must be boxed into Integers. In principle it could avoid the downcast from Object to Integer (since it's pointless, because Object is good enough for both String.valueOf and Object.toString) but I don't know if the compiler is smart enough to do that. The conversion from String to Object should be more or less a no-op, so I would be disinclined to worry about that one.

blocks|key|2076259|text|使用流:如果假设结果是整数列表(List<Integer>+result)，则：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2076260|List<String>+ids+=+(List<String>)+result.stream().map(intNumber+->+Integer.toString(intNumber)).collect(Collectors.toList());|code-block|syntax|javascript|2076261|解决这个问题的方法之一。希望这能有所帮助。|2076262|entityMap^0|G|K|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

Using Streams :
If let say result is list of integers (<code>List&lt;Integer&gt; result</code>) then :

<pre><code>List&lt;String&gt; ids = (List&lt;String&gt;) result.stream().map(intNumber -&gt; Integer.toString(intNumber)).collect(Collectors.toList());
</code></pre>

One of the ways to solve it. Hope this helps.

blocks|key|1861985|text|为了好玩，我给出了一个使用应该在JDK7中实现的jsr166y分叉-连接框架的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1861986|import+java.util.concurrent.forkjoin.*;

private+final+ForkJoinExecutor+executor+=+new+ForkJoinPool();
...
List<Integer>+ints+=+...;
List<String>+strs+=
++++ParallelArray.create(ints.size(),+Integer.class,+executor)
++++.withMapping(new+Ops.Op<Integer,String>()+{+public+String+op(Integer+i)+{
++++++++return+String.valueOf(i);
++++}})
++++.all()
++++.asList();|code-block|syntax|javascript|1861987|(免责声明:未编译。规范尚未最终确定。等)|1861988|在JDK7中不太可能存在一些类型推断和语法糖，以使withMapping调用不那么冗长：|1861989|++++.withMapping(#(Integer+i)+String.valueOf(i))|1861990|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Just for fun, a solution using the jsr166y fork-join framework that should in JDK7.

<pre><code>import java.util.concurrent.forkjoin.*;

private final ForkJoinExecutor executor = new ForkJoinPool();
...
List&lt;Integer&gt; ints = ...;
List&lt;String&gt; strs =
 ParallelArray.create(ints.size(), Integer.class, executor)
 .withMapping(new Ops.Op&lt;Integer,String&gt;() { public String op(Integer i) {
 return String.valueOf(i);
 }})
 .all()
 .asList();
</code></pre>

(Disclaimer: Not compiled. Spec is not finalised. Etc.)

Unlikely to be in JDK7 is a bit of type inference and syntactical sugar to make that withMapping call less verbose:

<pre><code> .withMapping(#(Integer i) String.valueOf(i))
</code></pre>

blocks|key|2075866|text|这是一件非常基本的事情，我不会使用外部库(它会在你的项目中产生一个你可能不需要的依赖)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2075867|我们有一类专门为完成这类工作而设计的静态方法。因为这方面的代码非常简单，我们让Hotspot为我们做优化。这似乎是我最近代码中的一个主题:编写非常简单(直接)的代码，让Hotspot发挥它的魔力。我们很少有像这样的代码性能问题--当一个新的VM版本出现时，你会得到所有额外的速度优势，等等。|2075868|尽管我非常喜欢雅加达的收藏品，但它们不支持泛型，并且使用1.4作为LCD。我对Google+Collections保持警惕，因为它们被列为Alpha支持级别！|2075869|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

This is such a basic thing to do I wouldn't use an external library (it will cause a dependency in your project that you probably don't need).

We have a class of static methods specifically crafted to do these sort of jobs. Because the code for this is so simple we let Hotspot do the optimization for us. This seems to be a theme in my code recently: write very simple (straightforward) code and let Hotspot do its magic. We rarely have performance issues around code like this - when a new VM version comes along you get all the extra speed benefits etc.

As much as I love Jakarta collections, they don't support Generics and use 1.4 as the LCD. I am wary of Google Collections because they are listed as Alpha support level!

blocks|key|2076187|text|我只是想为这个问题提供一个面向对象的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2076188|如果您建模域对象，则解决方案在域对象中。这里的域是我们想要的字符串值的整数列表。|2076189|最简单的方法是根本不转换列表。|2076190|也就是说，为了在不进行转换的情况下进行转换，将原始的Integer列表更改为值列表，其中的值如下所示...|2076191|class+Value+{
++++Integer+value;
++++public+Integer+getInt()
++++{
+++++++return+value;
++++}
++++public+String+getString()
++++{
+++++++return+String.valueOf(value);
++++}
}|code-block|syntax|javascript|2076192|这将比复制列表更快，占用更少的内存。|2076193|祝好运!|2076194|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I just wanted to chime in with an object oriented solution to the problem.

If you model domain objects, then the solution is in the domain objects. The domain here is a List of integers for which we want string values.

The easiest way would be to not convert the list at all.

That being said, in order to convert without converting, change the original list of Integer to List of Value, where Value looks something like this...

<pre><code>class Value {
 Integer value;
 public Integer getInt()
 {
 return value;
 }
 public String getString()
 {
 return String.valueOf(value);
 }
}
</code></pre>

This will be faster and take up less memory than copying the List.

Good Luck!

I have a list of integers, <code>List&lt;Integer&gt;</code> and I'd like to convert all the integer objects into Strings, thus finishing up with a new <code>List&lt;String&gt;</code>.

Naturally, I could create a new <code>List&lt;String&gt;</code> and loop through the list calling <code>String.valueOf()</code> for each integer, but I was wondering if there was a better (read: more automatic) way of doing it?

Converting List<Integer> to List<String>

我有一个整数列表，List<Integer>，我想把所有的整数对象转换成List<String>，这样就完成了一个新的String。当然，我可以创建一个新的List<String>并遍历列表，为每个整数调用String.valueOf()，但我想知道是否有更好的(阅读:更自动化)方法来做这件事？

问将List<Integer>转换为List<String>
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