blocks|key|1594110|text|您可以使用slice()来执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1594111|>>+foo+=+[1,2,3,4,5,6]
=>+[1,+2,+3,+4,+5,+6]
>>+bar+=+[10,20,30,40,50,60]
=>+[10,+20,+30,+40,+50,+60]
>>+half+=+foo.length+/+2
=>+3
>>+foobar+=+foo.slice(0,+half)+%2B+bar.slice(half,+foo.length)
=>+[1,+2,+3,+40,+50,+60]|code-block|syntax|javascript|1594112|顺便说一句，据我所知，Python“列表”只是高效地实现了动态增长的数组。开始插入为O(n)，结束插入为O(1)，随机访问为O(1)。|1594113|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ruby-doc.org/core/classes/Array.html#M002228^0|5|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]]]]

You can use <a href="http://www.ruby-doc.org/core/classes/Array.html#M002228" rel="noreferrer">slice()</a> for this:

<pre><code>&gt;&gt; foo = [1,2,3,4,5,6]
=&gt; [1, 2, 3, 4, 5, 6]
&gt;&gt; bar = [10,20,30,40,50,60]
=&gt; [10, 20, 30, 40, 50, 60]
&gt;&gt; half = foo.length / 2
=&gt; 3
&gt;&gt; foobar = foo.slice(0, half) + bar.slice(half, foo.length)
=&gt; [1, 2, 3, 40, 50, 60]
</code></pre>

By the way, to the best of my knowledge, Python "lists" are just efficiently implemented dynamically growing arrays. Insertion at the beginning is in O(n), insertion at the end is amortized O(1), random access is O(1).

blocks|key|1594159|text|是的，Ruby的数组切片语法与Python非常相似。下面是数组索引方法的ri文档：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1594160|---------------------------------------------------------------+Array#[]
+++++array[index]++++++++++++++++->+obj++++++or+nil
+++++array[start,+length]++++++++->+an_array+or+nil
+++++array[range]++++++++++++++++->+an_array+or+nil
+++++array.slice(index)++++++++++->+obj++++++or+nil
+++++array.slice(start,+length)++->+an_array+or+nil
+++++array.slice(range)++++++++++->+an_array+or+nil
------------------------------------------------------------------------
+++++Element+Reference---Returns+the+element+at+index,+or+returns+a+
+++++subarray+starting+at+start+and+continuing+for+length+elements,+or+
+++++returns+a+subarray+specified+by+range.+Negative+indices+count+
+++++backward+from+the+end+of+the+array+(-1+is+the+last+element).+
+++++Returns+nil+if+the+index+(or+starting+index)+are+out+of+range.

++++++++a+=+[+"a",+"b",+"c",+"d",+"e"+]
++++++++a[2]+%2B++a[0]+%2B+a[1]++++#=>+"cab"
++++++++a[6]+++++++++++++++++++#=>+nil
++++++++a[1,+2]++++++++++++++++#=>+[+"b",+"c"+]
++++++++a[1..3]++++++++++++++++#=>+[+"b",+"c",+"d"+]
++++++++a[4..7]++++++++++++++++#=>+[+"e"+]
++++++++a[6..10]+++++++++++++++#=>+nil
++++++++a[-3,+3]+++++++++++++++#=>+[+"c",+"d",+"e"+]
++++++++#+special+cases
++++++++a[5]+++++++++++++++++++#=>+nil
++++++++a[6,+1]++++++++++++++++#=>+nil
++++++++a[5,+1]++++++++++++++++#=>+[]
++++++++a[5..10]+++++++++++++++#=>+[]|code-block|syntax|javascript|1594161|entityMap^0|10|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Yes, Ruby has very similar array-slicing syntax to Python. Here is the <code>ri</code> documentation for the array index method:

<pre><code>--------------------------------------------------------------- Array#[]
 array[index] -&gt; obj or nil
 array[start, length] -&gt; an_array or nil
 array[range] -&gt; an_array or nil
 array.slice(index) -&gt; obj or nil
 array.slice(start, length) -&gt; an_array or nil
 array.slice(range) -&gt; an_array or nil
------------------------------------------------------------------------
 Element Reference---Returns the element at index, or returns a 
 subarray starting at start and continuing for length elements, or 
 returns a subarray specified by range. Negative indices count 
 backward from the end of the array (-1 is the last element). 
 Returns nil if the index (or starting index) are out of range.

 a = [ "a", "b", "c", "d", "e" ]
 a[2] + a[0] + a[1] #=&gt; "cab"
 a[6] #=&gt; nil
 a[1, 2] #=&gt; [ "b", "c" ]
 a[1..3] #=&gt; [ "b", "c", "d" ]
 a[4..7] #=&gt; [ "e" ]
 a[6..10] #=&gt; nil
 a[-3, 3] #=&gt; [ "c", "d", "e" ]
 # special cases
 a[5] #=&gt; nil
 a[6, 1] #=&gt; nil
 a[5, 1] #=&gt; []
 a[5..10] #=&gt; []
</code></pre>

blocks|key|1379022|text|如果您想拆分/剪切索引i上的数组，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1379023|arr+=+arr.drop(i)

>+arr+=+[1,2,3,4,5]
+=>+[1,+2,+3,+4,+5]+
>+arr.drop(2)
+=>+[3,+4,+5]+|code-block|syntax|javascript|1379024|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 you want to split/cut the array on an index i,

<pre><code>arr = arr.drop(i)

&gt; arr = [1,2,3,4,5]
 =&gt; [1, 2, 3, 4, 5] 
&gt; arr.drop(2)
 =&gt; [3, 4, 5] 
</code></pre>

blocks|key|1379045|text|另一种方法是使用range方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1379046|foo+=+[1,2,3,4,5,6]
bar+=+[10,20,30,40,50,60]
a+=+foo[0...3]
b+=+bar[3...6]

print+a+%2B+b+
=>+[1,+2,+3,+40,+50+,+60]|code-block|syntax|javascript|1379047|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 way is to use the range method

<pre><code>foo = [1,2,3,4,5,6]
bar = [10,20,30,40,50,60]
a = foo[0...3]
b = bar[3...6]

print a + b 
=&gt; [1, 2, 3, 40, 50 , 60]
</code></pre>

blocks|key|1594266|text|我喜欢这样的范围：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1594267|def+first_half(list)
++list[0...(list.length+/+2)]
end

def+last_half(list)
++list[(list.length+/+2)..list.length]
end|code-block|syntax|javascript|1594268|但是，要非常小心终结点是否包括在您的范围内。这在奇数长度的列表中变得至关重要，因为您需要选择在哪里打破中间位置。否则，您最终将重复计算中间元素。|1594269|上面的示例将始终将中间元素放在后半部分。|1594270|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I like ranges for this:

<pre><code>def first_half(list)
 list[0...(list.length / 2)]
end

def last_half(list)
 list[(list.length / 2)..list.length]
end
</code></pre>

However, be very careful about whether the endpoint is included in your range. This becomes critical on an odd-length list where you need to choose where you're going to break the middle. Otherwise you'll end up double-counting the middle element.

The above example will consistently put the middle element in the last half.

blocks|key|1379127|text|Ruby+2.6无边界/无止境范围|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1379128|(..1)
#+or
(...1)|code-block|syntax|javascript|1379129|​|1379130|(1..)
#+or
(1...)|1379131|1379132|[1,2,3,4,5,6][..3]
=>+[1,+2,+3,+4]

[1,2,3,4,5,6][...3]
+=>+[1,+2,+3]|1379133|1379134|ROLES+=+%25w[superadmin+manager+admin+contact+user]
ROLES[ROLES.index('admin')..]
=>+["admin",+"contact",+"user"]|1379135|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|L|3|M|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|N|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|R|$]]

Ruby 2.6 Beginless/Endless Ranges

<pre><code>(..1)
# or
(...1)
</code></pre>

 

<pre><code>(1..)
# or
(1...)
</code></pre>

 

<pre><code>[1,2,3,4,5,6][..3]
=&gt; [1, 2, 3, 4]

[1,2,3,4,5,6][...3]
 =&gt; [1, 2, 3]
</code></pre>

 

<pre><code>ROLES = %w[superadmin manager admin contact user]
ROLES[ROLES.index('admin')..]
=&gt; ["admin", "contact", "user"]
</code></pre>

With a list in Python I can return a part of it using the following code:

<pre><code>foo = [1,2,3,4,5,6]
bar = [10,20,30,40,50,60]
half = len(foo) / 2
foobar = foo[:half] + bar[half:]
</code></pre>

Since Ruby does everything in arrays I wonder if there is something similar to that.

How to return a part of an array in Ruby?

Python

使用Python中的列表，我可以使用以下代码返回其中的一部分：foo = [1,2,3,4,5,6]bar = [10,20,30,40,50,60]half = len(foo) / 2foobar = foo[:half] + bar[half:]由于Ruby在数组中做所有的事情，我想知道是否有类似的东西。

问如何在Ruby中返回数组的一部分？
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