blocks|key|1307462|text|假人型：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1307463|struct+CustomObject+{
++++let+foo:+Int
++++init(_+foo:+Int)+{+self.foo+=+foo+}
}|code-block|syntax|javascript|1307464|如果您想根据给定的一组键，以“相交”的方式修改原始字典(而不是创建新字典)：|1307465|let+keySet+=+Set(["foo",+"baz"])
var+dict+=+["foo":+CustomObject(1),+"bar":+CustomObject(2),
++++++++++++"baz":+CustomObject(3),+"bax":+CustomObject(4)]

Set(dict.keys).subtracting(keySet).forEach+{+dict.removeValue(forKey:+$0)+}

print(dict)+//+["foo":+CustomObject(foo:+1),+"baz":+CustomObject(foo:+3)]|1307466|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Dummy type:

<pre><code>struct CustomObject {
 let foo: Int
 init(_ foo: Int) { self.foo = foo }
}
</code></pre>

In case you'd like to mutate the original dictionary (instead of creating a new one) in an "intersect" manner, based on a given set of keys:

<pre><code>let keySet = Set(["foo", "baz"])
var dict = ["foo": CustomObject(1), "bar": CustomObject(2),
 "baz": CustomObject(3), "bax": CustomObject(4)]

Set(dict.keys).subtracting(keySet).forEach { dict.removeValue(forKey: $0) }

print(dict) // ["foo": CustomObject(foo: 1), "baz": CustomObject(foo: 3)]
</code></pre>

blocks|key|1505458|text|你的假设是正确的，有一个更简洁/快捷的方式来完成你所需要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505459|例如，您可以通过reduce+(Swift中提供的一种函数式编程概念)来实现它：|offset|length|style|CODE|1505460|let+subDict+=+originalDict.reduce([String:+CustomObject]())+{
++++guard+mySet.contains($1.key)+else+{+return+$0+}
++++var+d+=+$0
++++d[$1.key]+=+$1.value
++++return+d
}|code-block|syntax|javascript|1505461|或者，在两个步骤中，首先过滤有效的元素，然后用过滤的元素构建字典：|1505462|let+filteredDict+=+originalDict.filter+{+mySet.contains($0.key)+}
++++.reduce([CustomObject]()){+var+d+=+$0;+d[$1.key]=$1.value;+return+d+}|1505463|forEach还可用于构造过滤字典：|1505464|var+filteredDict+=+[CustomObject]()
mySet.forEach+{+filteredDict[$0]+=+originalDict[$0]+}|1505465|但其结果将是好的，它将是不可改变的：|1505466|let+filteredDict:+[String:CustomObject]+=+{+
++++var+result+=+[String:CustomObject]()
++++mySet.forEach+{+filteredDict2[$0]+=+originalDict[$0]+}
++++return+result
}()|1505467|entityMap^0|0|8|6|0|0|0|0|0|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|17|8|@$D|18|E|19|F|G]]|9|@]|A|$]]|$1|S|3|T|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|X|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|Y|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|Z|$]]

Your assumption is correct, there is a more concise/swift-ish way to accomplish what you need.

For example you can do it via <code>reduce</code>, a functional programming concept available in Swift:

<pre><code>let subDict = originalDict.reduce([String: CustomObject]()) {
 guard mySet.contains($1.key) else { return $0 }
 var d = $0
 d[$1.key] = $1.value
 return d
}
</code></pre>

Or, in two steps, first filtering the valid elements, and then constructing back the dictionary with the filtered elements:

<pre><code>let filteredDict = originalDict.filter { mySet.contains($0.key) }
 .reduce([CustomObject]()){ var d = $0; d[$1.key]=$1.value; return d }
</code></pre>

<code>forEach</code> can also be used to construct the filtered dictionary:

<pre><code>var filteredDict = [CustomObject]()
mySet.forEach { filteredDict[$0] = originalDict[$0] }
</code></pre>

, however the result would be good it it would be immutable:

<pre><code>let filteredDict: [String:CustomObject] = { 
 var result = [String:CustomObject]()
 mySet.forEach { filteredDict2[$0] = originalDict[$0] }
 return result
}()
</code></pre>

blocks|key|415928|text|Swift+5+-您可以非常简单地这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|415929|let+subsetDict+=+originalDict.filter({+mySet.contains($0.key)})|code-block|syntax|javascript|415930|结果是一个新字典，其类型与原始字典相同，但只包含与mySet中的键对应的键值对。|CODE|415931|entityMap^0|0|7|0|0|P|5|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|S|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|T|8|@$9|U|A|V|B|M]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|O|$]]

Swift 5 - You can do this very simply:

<pre><code>let subsetDict = originalDict.filter({ mySet.contains($0.key)})
</code></pre>

The result is a new dictionary with the same type as the original but which only contains the key-value pairs corresponding to the keys in <code>mySet</code>.

I've looked through the <a href="https://developer.apple.com/reference/swift/dictionary" rel="nofollow noreferrer">methods here</a> but I can't quite find what I'm looking for. I'm new-ish to Swift. I would like to extract a subset from a Dictionary based on a Set of key values, preferably without a loop.

For example, if my key Set is of type <code>Set&lt;String&gt;</code> and I have a Dictionary of type <code>Dictionary&lt;String, CustomObject&gt;</code>, I would like to create a new Dictionary of type <code>Dictionary&lt;String, CustomObject&gt;</code> that contains only the key-value pairs associated with the keys in the Set of Strings. 

I can see that I could do this with <code>for</code> loop, by initializing a new <code>Dictionary&lt;String, CustomObj&gt;()</code>, checking if the original Dictionary contains a value at each String in the set, and adding key-value pairs to the new Dictionary. I am wondering if there is a more efficient/elegant way to do this however.

I'd be open to finding the subset with an Array of Strings instead of a Set if there is a better way to do it with an Array of keys.

Many thanks!

How to extract a subset of a swift 3 Dictionary

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我已经翻阅过，但是我找不到我要找的东西。我是斯威夫特的新手。我想从字典中提取一个子集，它基于一组键值，最好没有循环。例如，如果我的键集是Set<String>类型，并且我有一个Dictionary<String, CustomObject>类型的字典，我想创建一个新的Dictionary<String, CustomO...

问如何提取“快速3词典”的子集
EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何提取“快速3词典”的子集EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何提取“快速3词典”的子集
EN