blocks|key|1416223|text|OpenStruct|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1416224|简要示例：|1416225|require+'ostruct'

person+=+OpenStruct.new
person.name++++=+"John+Smith"
person.age+++++=+70
person.pension+=+300

puts+person.name+++++#+->+"John+Smith"
puts+person.age++++++#+->+70
puts+person.address++#+->+nil|code-block|syntax|javascript|1416226|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ruby-doc.org/stdlib/libdoc/ostruct/rdoc/classes/OpenStruct.html^0|0|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<a href="http://www.ruby-doc.org/stdlib/libdoc/ostruct/rdoc/classes/OpenStruct.html" rel="noreferrer">OpenStruct</a>?

Brief example:

<pre><code>require 'ostruct'

person = OpenStruct.new
person.name = "John Smith"
person.age = 70
person.pension = 300

puts person.name # -&gt; "John Smith"
puts person.age # -&gt; 70
puts person.address # -&gt; nil
</code></pre>

blocks|key|1416246|text|基于你所说的哈希和。我假设你指的是一种不同于(1.+"a")排序的元组。您可能正在寻找Struct类。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1416247|Person+=+Struct.new(:name,+:age)
me+=+Person.new
me.name+=+"Guy"
me.age+=++30|code-block|syntax|javascript|1416248|entityMap^0|M|8|17|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Based on the fact that you talk about hashes and . notation I'm going to assume you mean a different kind of tuple than the <code>(1. "a")</code> sort. You're probably looking for the <code>Struct</code> class. eg:

<pre><code>Person = Struct.new(:name, :age)
me = Person.new
me.name = "Guy"
me.age = 30
</code></pre>

blocks|key|1416292|text|虽然这不是严格意义上的元组(不能做成员的点表示法)，但您可以从列表中分配一个变量列表，这通常可以解决当您在寻找一个返回值列表时ruby是按值传递的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1416293|例如。|1416294|:linenum+>+(a,b,c)+=+[1,2,3]
:linenum+>+a
++=>+1
:linenum+>+b
++=>+2
:linenum+>+c
++=>+3|code-block|syntax|javascript|1416295|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|$]]

While this isn't strictly a tuple (can't do dot notation of members), you can assign a list of variables from a list, which often will solve issues with ruby being pass-by-value when you are after a list of return values.

E.g. 

<pre><code>:linenum &gt; (a,b,c) = [1,2,3]
:linenum &gt; a
 =&gt; 1
:linenum &gt; b
 =&gt; 2
:linenum &gt; c
 =&gt; 3
</code></pre>

blocks|key|1416324|text|你可以做一些类似于解构的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1416325|def+something((a,+b))
++a+%2B+b
end

p+something([1,+2])|code-block|syntax|javascript|1416326|这会打印出预期的3。|offset|length|style|CODE|1416327|entityMap^0|0|0|8|1|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can do something similiar with destructuring:

<pre><code>def something((a, b))
 a + b
end

p something([1, 2])
</code></pre>

This prints out <code>3</code> as expected.

blocks|key|1416370|text|您可以使用此技巧模拟Scala元组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1416371|Tuple+=+Struct.new(:_1,+:_2)

2.2.5+:003+>+t+=+Tuple.new("a",+"b")
+=>+#<struct+Tuple+_1="a",+_2="b">
2.2.5+:004+>+t._1
+=>+"a"
2.2.5+:005+>+t._2
+=>+"b"|code-block|syntax|javascript|1416372|但在这里，你不能进行解构：|1416373|2.2.5+:012+>+a,+b+=+t
+=>+{:_1=>"a",+:_2=>"b"}
2.2.5+:013+>+a
+=>+{:_1=>"a",+:_2=>"b"}
2.2.5+:014+>+b
+=>+nil|1416374|但多亏了这个技巧：https://gist.github.com/stevecj/9ace6a70370f6d1a1511解构将会起作用：|offset|length|1416375|2.2.5+:001+>+Tuple+=+Struct.new(:_1,+:_2)
+=>+Tuple
2.2.5+:002+>+t+=+Tuple.new("a",+"b")
+=>+#<struct+Tuple+_1="a",+_2="b">
2.2.5+:003+>+t._1
+=>+"a"
2.2.5+:004+>+class+Tuple+;+def+to_ary+;+to_a+;+end+;+end
+=>+:to_ary
2.2.5+:005+>+a,+b+=+t
+=>+#<struct+Tuple+_1="a",+_2="b">
2.2.5+:006+>+a
+=>+"a"
2.2.5+:007+>+b
+=>+"b"|1416376|entityMap|0|LINK|mutability|MUTABLE|url|https://gist.github.com/stevecj/9ace6a70370f6d1a1511^0|0|0|0|0|9|1G|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]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@$M|13|N|14|1|15]]|A|$]]|$1|O|3|P|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

You can mock the Scala tuples with this trick :

<pre><code>Tuple = Struct.new(:_1, :_2)

2.2.5 :003 &gt; t = Tuple.new("a", "b")
 =&gt; #&lt;struct Tuple _1="a", _2="b"&gt;
2.2.5 :004 &gt; t._1
 =&gt; "a"
2.2.5 :005 &gt; t._2
 =&gt; "b"
</code></pre>

but here you can't have destructuring:

<pre><code>2.2.5 :012 &gt; a, b = t
 =&gt; {:_1=&gt;"a", :_2=&gt;"b"}
2.2.5 :013 &gt; a
 =&gt; {:_1=&gt;"a", :_2=&gt;"b"}
2.2.5 :014 &gt; b
 =&gt; nil
</code></pre>

But thanks to this trick : <a href="https://gist.github.com/stevecj/9ace6a70370f6d1a1511" rel="noreferrer">https://gist.github.com/stevecj/9ace6a70370f6d1a1511</a>
destructuring will work:

<pre><code>2.2.5 :001 &gt; Tuple = Struct.new(:_1, :_2)
 =&gt; Tuple
2.2.5 :002 &gt; t = Tuple.new("a", "b")
 =&gt; #&lt;struct Tuple _1="a", _2="b"&gt;
2.2.5 :003 &gt; t._1
 =&gt; "a"
2.2.5 :004 &gt; class Tuple ; def to_ary ; to_a ; end ; end
 =&gt; :to_ary
2.2.5 :005 &gt; a, b = t
 =&gt; #&lt;struct Tuple _1="a", _2="b"&gt;
2.2.5 :006 &gt; a
 =&gt; "a"
2.2.5 :007 &gt; b
 =&gt; "b"
</code></pre>

blocks|key|1631171|text|由于解构，将数组用作元组很酷|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1631172|a+=+[[1,2],+[2,3],+[3,4]]
a.map+{%7Ca,b%7C+a%2Bb+}|code-block|syntax|javascript|1631173|结构为您提供了方便的.访问器|offset|length|style|CODE|1631174|Person+=+Struct.new(:first_name,+:last_name)
ppl+=+Person.new('John',+'Connor')
ppl.first_name+
ppl.last_name|1631175|通过使用to_ary，您可以同时享受这两个方面的便利|1631176|Person+=+Struct.new(:first_name,+:last_name)+do
++def+to_ary
++++[first_name,+last_name]
++end
end
#+=>
[
++Person.new('John',+'Connor'),+
++Person.new('John',+'Conway')
].map+{+%7Ca,+b%7C+a+%2B+'+'+%2B+b++}
#+=>+["John+Connor",+"John+Conway"]|1631177|entityMap^0|0|0|A|1|0|0|4|6|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

Arrays are cool to use as tuples because of destructuring

<pre class="lang-rb prettyprint-override"><code>a = [[1,2], [2,3], [3,4]]
a.map {|a,b| a+b }
</code></pre>

Struct give you convenient <code>.</code> accessors

<pre class="lang-rb prettyprint-override"><code>Person = Struct.new(:first_name, :last_name)
ppl = Person.new('John', 'Connor')
ppl.first_name 
ppl.last_name
</code></pre>

You can get the convenience of both worlds with <code>to_ary</code>

<pre class="lang-rb prettyprint-override"><code>Person = Struct.new(:first_name, :last_name) do
 def to_ary
 [first_name, last_name]
 end
end
# =&gt;
[
 Person.new('John', 'Connor'), 
 Person.new('John', 'Conway')
].map { |a, b| a + ' ' + b }
# =&gt; ["John Connor", "John Conway"]
</code></pre>

Does anyone use tuples in Ruby? If so, how may one implement a tuple? Ruby hashes are nice and work almost as well, but I'd really like to see something like the Tuple class in Python, where you can use <code>.</code> notation to find the value for which you are looking. I'm wanting this so that I can create an implementation of <a href="https://rads.stackoverflow.com/amzn/click/com/0321399420" rel="noreferrer" rel="nofollow noreferrer">D</a>, similar to <a href="http://www.quicksort.co.uk/" rel="noreferrer">Dee</a> for Python.

Using Tuples in Ruby?

Python

有人在Ruby中使用元组吗？如果是这样，那么如何实现元组呢？Ruby散列很好，而且工作得几乎一样好，但我真的希望看到类似Python中的元组类的东西，您可以使用.表示法来查找您正在查找的值。我想要这个，这样我就可以创建一个类似于Python的的实现。

问在Ruby中使用元组？
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