Java 获取匹配布尔值的第一个元素的索引的流方式
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Stream Way to get index of first element matching boolean
提问by Edward Peters
I have a List<Users>. I want to get the index of the (first) user in the stream with a particular username. I don't want to actually require the Userto be .equals()to some described User, just to have the same username.
我有一个List<Users>. 我想使用特定用户名获取流中（第一个）用户的索引。我不想实际要求User成为.equals()某些描述的User，只是为了具有相同的用户名。
I can think of ugly ways to do this (iterate and count), but it feels like there should be a nice way to do this, probably by using Streams. So far the best I have is:
我可以想到丑陋的方法来做到这一点（迭代和计数），但感觉应该有一个很好的方法来做到这一点，可能是使用 Streams。到目前为止，我拥有的最好的是：
int index = users.stream()
.map(user -> user.getName())
.collect(Collectors.toList())
.indexOf(username);
Which isn't the worst code I've ever written, but it's not great. It's also not that flexible, as it relies on there being a mapping function to a type with a .equals()function that describes the property you're looking for; I'd much rather have something that could work for arbitrary Function<T, Boolean>
这不是我写过的最糟糕的代码，但不是很好。它也不是那么灵活，因为它依赖于一个映射函数到一个类型，该.equals()函数描述了你正在寻找的属性；我宁愿有一些可以为任意工作的东西Function<T, Boolean>
Anyone know how?
有谁知道怎么做？
采纳答案by vsminkov
Occasionally there is no pythonic zipWithIndexin java. So I came across something like that:
有时zipWithIndexjava中没有pythonic 。所以我遇到了这样的事情：
OptionalInt indexOpt = IntStream.range(0, users.size())
.filter(i -> searchName.equals(users.get(i)))
.findFirst();
Alternatively you can use zipWithIndexfrom protonpacklibrary
或者您可以使用zipWithIndex从protonpack库
Note
笔记
That solution may be time-consuming if users.get is not constant time operation.
如果 users.get 不是恒定时间操作，该解决方案可能会很耗时。
回答by AmanSinghal
Try This:
尝试这个：
IntStream.range(0, users.size())
.filter(userInd-> users.get(userInd).getName().equals(username))
.findFirst()
.getAsInt();
回答by SerCe
You can try StreamExlibrary made by Tagir Valeev. That library has a convenient #indexOf method.
您可以尝试StreamEx所作库Tagir Valeev。该库有一个方便的 #indexOf 方法。
This is a simple example:
这是一个简单的例子：
List<User> users = asList(new User("Vas"), new User("Innokenty"), new User("WAT"));
long index = StreamEx.of(users)
.indexOf(user -> user.name.equals("Innokenty"))
.getAsLong();
System.out.println(index);
回答by hibour
A solution without any external library
没有任何外部库的解决方案
AtomicInteger i = new AtomicInteger(); // any mutable integer wrapper
int index = users.stream()
.peek(v -> i.incrementAndGet())
.anyMatch(user -> user.getName().equals(username)) ? // your predicate
i.get() - 1 : -1;
peek increment index i while predicate is false hence when predicate is true i is 1 more than matched predicate => i.get() -1
在谓词为假时查看增量索引 i 因此当谓词为真时 i 比匹配的谓词多 1 => i.get() -1
回答by karmakaze
Using Guava library: int index =
Iterables.indexOf(users, u -> searchName.equals(u.getName()))
使用番石榴库： int index =
Iterables.indexOf(users, u -> searchName.equals(u.getName()))
回答by Donald Raab
There is the detectIndexmethod in the Eclipse Collectionslibrary which takes a Predicate.
还有就是detectIndex在方法Eclipse的收藏库，这需要Predicate。
int index = ListIterate.detectIndex(users, user -> username.equals(user.getName()));
If you have a method on Userclass which returns booleanif usernamematches you can use the following:
如果您在User类上有一个boolean如果username匹配则返回的方法，您可以使用以下方法：
int index = ListIterate.detectIndexWith(users, User::named, username);
Note: I a committer for Eclipse Collections
注意：我是 Eclipse Collections 的提交者
