blocks|key|4472179|text|你可以这样写：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4472180|filePaths.stream()
+++++++++.filter(p+->+acceptedIds.stream().anyMatch(id+->+p.toString().contains(id)))
+++++++++.collect(toList());|code-block|syntax|javascript|4472181|这会过滤每个路径，以便路径的字符串表示中至少包含一个acceptedIds。您可能希望在这里实现比contains更好的东西，这取决于您的用例(例如，匹配文件名的开头)。|offset|length|style|CODE|4472182|anyMatch是一个操作，它确定是否至少有一个元素与给定的谓词匹配。|4472183|请注意，此答案并未对过滤出元素的路径做出任何假设。如果您可以放心地说，在每个路径中，父目录都是用id命名的，那么出于性能原因，您绝对应该使用@Bohemian答案。|4472184|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html#anyMatch-java.util.function.Predicate-^0|0|0|Q|B|1D|8|0|0|8|0|8|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@$I|16|J|17|K|L]]|9|@$I|18|J|19|1|1A]]|A|$]]|$1|O|3|P|5|6|7|1B|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

You could write:

<pre><code>filePaths.stream()
 .filter(p -&gt; acceptedIds.stream().anyMatch(id -&gt; p.toString().contains(id)))
 .collect(toList());
</code></pre>

This filters each path such that at least one of the <code>acceptedIds</code> is contained in the string representation of the path. You might want to implement something better than <code>contains</code> here, depending on your use-case (matching the beginning of the filename for example).

<a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html#anyMatch-java.util.function.Predicate-" rel="nofollow"><code>anyMatch</code></a> is an operation that determines if at least one element matches the given predicate.

Note that this answer does not make any assumption about the path to filter out elements. If you can safely say that in each path, the parent directory is named with the id, you should definitely go with @Bohemian answer, for performance reason.

blocks|key|4471702|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4471703|List+removeMissing(List+l1,+List+l2)+{
++++List+ret+=+l1.stream()
++++++++.filter(o+->+l2.contains(o))+//Keep+if+object+o+satisfies+the+condition+"l2+contains+a+reference+to+this+object"
++++++++.collect(Collectors.toList());
++++return+ret;
}|code-block|syntax|javascript|4471704|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|$]]

Like so:

<pre><code>List removeMissing(List l1, List l2) {
 List ret = l1.stream()
 .filter(o -&gt; l2.contains(o)) //Keep if object o satisfies the condition "l2 contains a reference to this object"
 .collect(Collectors.toList());
 return ret;
}
</code></pre>

blocks|key|4472231|text|最有效的方法是从路径中提取ID，然后尝试在集合中找到它，使每个过滤器在恒定的时间内执行，即O(1)给出总体O(n)，其中n是路径的数量：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4472232|filePaths.stream()
++.filter(p+->+acceptedIds.contains(p.getParent().getFileName().toString()))
++.collect(Collectors.toList());|code-block|syntax|javascript|4472233|如果执行了相反的方法，即在路径中搜索每个acceptedIds+(就像在其他答案中一样)，则每个过滤器都是O(m*k)，其中m是acceptedIds的数量，k是平均路径长度，这给出了总体O(n+*+m+*+k)，即使是中等大小的集合，其性能也会非常差。|4472234|entityMap^0|19|4|1H|4|1O|1|0|0|K|B|1H|6|1Q|1|1S|B|27|1|2M|C|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|N|$]]

The most efficient way is to extract the ID from the path, then attempt to find it in the Set, making each filter execute in constant time, ie <code>O(1)</code> giving an overall <code>O(n)</code>, where <code>n</code> is the number of paths:

<pre><code>filePaths.stream()
 .filter(p -&gt; acceptedIds.contains(p.getParent().getFileName().toString()))
 .collect(Collectors.toList());
</code></pre>

If the reverse approach is done, where each <code>acceptedIds</code> is searched for in the path (as in other answers), each filter is <code>O(m*k)</code>, where <code>m</code> is the number of <code>acceptedIds</code> and <code>k</code> is the average Path length, giving an overall <code>O(n * m * k)</code>, which will perform very poorly for even moderate sizes of collections.

blocks|key|4471769|text|如果您的文件名结构是常量，我将首先使用正则表达式来提取数字，然后检查它是否在所需的ids中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4471770|final+Set<String>+acceptedIds+=+...
//+Matches+the+number+of+the+file,+concluded+with+the+underscore
final+Pattern+extractor+=+Pattern.compile("\.*(?<number>\d%2B)_")
filePaths.stream().filter(+path+->+{
++++final+Matcher+m+=+extractor
++++++++.matcher(path.getFileName().toString());
++++m.find();
++++return+acceptedIds.contains(m.group("number"));
})
...|code-block|syntax|javascript|4471771|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|$]]

If your file name structure is constant, I'd use a regex first to extract the number, and then will check if it is among the desired ids.

<pre class="lang-java prettyprint-override"><code>final Set&lt;String&gt; acceptedIds = ...
// Matches the number of the file, concluded with the underscore
final Pattern extractor = Pattern.compile("\.*(?&lt;number&gt;\d+)_")
filePaths.stream().filter( path -&gt; {
 final Matcher m = extractor
 .matcher(path.getFileName().toString());
 m.find();
 return acceptedIds.contains(m.group("number"));
})
...
</code></pre>

I have List of file Paths: . 

<pre><code>List&lt;Path&gt; filePaths; //e.g. [src\test\resources\file\15\54\54_exampleFile.pdf]
</code></pre>

<code>54</code> above refers to file ID

I then obtain a <code>Set</code> of <code>String</code> Ids which my application can handle as follows:

<pre><code>Set&lt;String&gt; acceptedIds = connection.getAcceptedIDs(); //e.g. elements [64, 101, 33]
</code></pre>

How can I use Java 8 lambdas to <code>filter</code> out all elements in <code>filePaths</code> that do not contain any of the acceptable Ids that are contained in <code>acceptedIds</code> collection Set.

In other words, I would like to retain in <code>filePaths</code> only the paths that have ids which are in <code>acceptedIds</code> set. For example, 54 is not in the above list so is removed.

<pre><code>filePaths.stream().filter(...).collect(Collectors.toList());
</code></pre>

How to remove elements from a list with lambda based on another list

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个文件路径列表：。List<Path> filePaths; //e.g. [src\test\resources\file\15\54\54_exampleFile.pdf]上面的54指的是文件ID然后，我获得一个String Ids的Set，我的应用程序可以按如下方式处理它：Set<String> accep...

问如何使用lambda基于另一个列表从列表中删除元素
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