blocks|key|1409311|text|您可以使用%2B运算符来组合它们：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1409312|listone+=+[1,+2,+3]
listtwo+=+[4,+5,+6]

joinedlist+=+listone+%2B+listtwo|code-block|syntax|javascript|1409313|输出：|1409314|>>>+joinedlist
[1,+2,+3,+4,+5,+6]|1409315|entityMap^0|5|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You can use the <code>+</code> operator to combine them:
<pre><code>listone = [1, 2, 3]
listtwo = [4, 5, 6]

joinedlist = listone + listtwo
</code></pre>
Output:
<pre><code>&gt;&gt;&gt; joinedlist
[1, 2, 3, 4, 5, 6]
</code></pre>

blocks|key|1409537|text|也可以创建一个生成器，使用itertools.chain()简单地遍历两个列表中的项。这允许您将列表(或任何可迭代的)链接在一起进行处理，而无需将项复制到新的列表中：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1409538|import+itertools
for+item+in+itertools.chain(listone,+listtwo):
++++#+Do+something+with+each+list+item|code-block|syntax|javascript|1409539|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/library/itertools.html#itertools.chain^0|D|H|D|H|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

It's also possible to create a generator that simply iterates over the items in both lists using <a href="https://docs.python.org/library/itertools.html#itertools.chain" rel="noreferrer"><code>itertools.chain()</code></a>. This allows you to chain lists (or any iterable) together for processing without copying the items to a new list:

<pre class="lang-python prettyprint-override"><code>import itertools
for item in itertools.chain(listone, listtwo):
 # Do something with each list item
</code></pre>

blocks|key|1194280|text|您还可以使用list.extend()方法将一个list添加到另一个use的末尾：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1194281|listone+=+[1,2,3]
listtwo+=+[4,5,6]

listone.extend(listtwo)|code-block|syntax|javascript|1194282|如果希望保持原始列表不变，可以创建一个新的list对象，并将两个列表都extend到该对象：|1194283|mergedlist+=+[]
mergedlist.extend(listone)
mergedlist.extend(listtwo)|1194284|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2.7/tutorial/datastructures.html#more-on-lists^0|6|D|O|4|6|D|0|0|0|L|4|Z|6|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

You could also use the <a href="https://docs.python.org/2.7/tutorial/datastructures.html#more-on-lists" rel="noreferrer"><code>list.extend()</code></a> method in order to add a <code>list</code> to the end of another one:

<pre><code>listone = [1,2,3]
listtwo = [4,5,6]

listone.extend(listtwo)
</code></pre>

If you want to keep the original list intact, you can create a new <code>list</code> object, and <code>extend</code> both lists to it:

<pre><code>mergedlist = []
mergedlist.extend(listone)
mergedlist.extend(listtwo)
</code></pre>

blocks|key|1409578|text|您可以使用集合来获取唯一值的合并列表|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1409579|mergedlist+=+list(set(listone+%2B+listtwo))|code-block|syntax|javascript|1409580|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|$]]

You can use sets to obtain merged list of unique values

<pre><code>mergedlist = list(set(listone + listtwo))
</code></pre>

blocks|key|1194055|text|这很简单，我想它甚至在the+tutorial中显示过|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1194056|>>>+listone+=+[1,2,3]
>>>+listtwo+=+[4,5,6]
>>>
>>>+listone+%2B+listtwo
[1,+2,+3,+4,+5,+6]|code-block|syntax|javascript|1194057|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/tutorial/^0|B|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

This is quite simple, and I think it was even shown in <a href="http://docs.python.org/tutorial/" rel="noreferrer">the tutorial</a>:

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

blocks|key|1194393|text|您可以简单地使用%2B或%2B=操作符，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1194394|a+=+[1,+2,+3]
b+=+[4,+5,+6]

c+=+a+%2B+b|code-block|syntax|javascript|1194395|或者：|1194396|c+=+[]
a+=+[1,+2,+3]
b+=+[4,+5,+6]

c+%2B=+(a+%2B+b)|1194397|此外，如果您希望合并列表中的值是唯一的，您可以这样做：|1194398|c+=+list(set(a+%2B+b))|1194399|entityMap^0|8|1|A|2|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

You could simply use the <code>+</code> or <code>+=</code> operator as follows:

<pre><code>a = [1, 2, 3]
b = [4, 5, 6]

c = a + b
</code></pre>

Or:

<pre><code>c = []
a = [1, 2, 3]
b = [4, 5, 6]

c += (a + b)
</code></pre>

Also, if you want the values in the merged list to be unique you can do:

<pre><code>c = list(set(a + b))
</code></pre>

blocks|key|1194314|text|值得注意的是，itertools.chain函数接受数量可变的参数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1194315|>>>+l1+=+['a'];+l2+=+['b',+'c'];+l3+=+['d',+'e',+'f']
>>>+[i+for+i+in+itertools.chain(l1,+l2)]
['a',+'b',+'c']
>>>+[i+for+i+in+itertools.chain(l1,+l2,+l3)]
['a',+'b',+'c',+'d',+'e',+'f']|code-block|syntax|javascript|1194316|如果可迭代(元组、列表、生成器等)是输入，则可以使用from_iterable类方法：|1194317|>>>+il+=+[['a'],+['b',+'c'],+['d',+'e',+'f']]
>>>+[i+for+i+in+itertools.chain.from_iterable(il)]
['a',+'b',+'c',+'d',+'e',+'f']|1194318|entityMap^0|7|F|0|0|Q|D|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

It's worth noting that the <code>itertools.chain</code> function accepts variable number of arguments:

<pre><code>&gt;&gt;&gt; l1 = ['a']; l2 = ['b', 'c']; l3 = ['d', 'e', 'f']
&gt;&gt;&gt; [i for i in itertools.chain(l1, l2)]
['a', 'b', 'c']
&gt;&gt;&gt; [i for i in itertools.chain(l1, l2, l3)]
['a', 'b', 'c', 'd', 'e', 'f']
</code></pre>

If an iterable (tuple, list, generator, etc.) is the input, the <code>from_iterable</code> class method may be used:

<pre><code>&gt;&gt;&gt; il = [['a'], ['b', 'c'], ['d', 'e', 'f']]
&gt;&gt;&gt; [i for i in itertools.chain.from_iterable(il)]
['a', 'b', 'c', 'd', 'e', 'f']
</code></pre>

blocks|key|1409935|text|对于Python+3.3%2B，您可以使用yield+from|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1409936|listone+=+[1,2,3]
listtwo+=+[4,5,6]

def+merge(l1,+l2):
++++yield+from+l1
++++yield+from+l2

>>>+list(merge(listone,+listtwo))
[1,+2,+3,+4,+5,+6]|code-block|syntax|javascript|1409937|或者，如果您希望支持任意数量的迭代器：|1409938|def+merge(*iters):
++++for+it+in+iters:
++++++++yield+from+it

>>>+list(merge(listone,+listtwo,+'abcd',+[20,+21,+22]))
[1,+2,+3,+4,+5,+6,+'a',+'b',+'c',+'d',+20,+21,+22]|1409939|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/whatsnew/3.3.html#pep-380^0|J|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

With Python 3.3+ you can use <a href="https://docs.python.org/3/whatsnew/3.3.html#pep-380" rel="noreferrer">yield from</a>:

<pre><code>listone = [1,2,3]
listtwo = [4,5,6]

def merge(l1, l2):
 yield from l1
 yield from l2

&gt;&gt;&gt; list(merge(listone, listtwo))
[1, 2, 3, 4, 5, 6]
</code></pre>

Or, if you want to support an arbitrary number of iterators:

<pre><code>def merge(*iters):
 for it in iters:
 yield from it

&gt;&gt;&gt; list(merge(listone, listtwo, 'abcd', [20, 21, 22]))
[1, 2, 3, 4, 5, 6, 'a', 'b', 'c', 'd', 20, 21, 22]
</code></pre>

blocks|key|1194636|text|如果希望以排序的形式合并这两个列表，可以使用heapq库中的merge函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1194637|from+heapq+import+merge

a+=+[1,+2,+4]
b+=+[2,+4,+6,+7]

print+list(merge(a,+b))|code-block|syntax|javascript|1194638|entityMap^0|M|5|U|5|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|$]]

If you want to merge the two lists in sorted form, you can use the <code>merge</code> function from the <code>heapq</code> library.

<pre><code>from heapq import merge

a = [1, 2, 4]
b = [2, 4, 6, 7]

print list(merge(a, b))
</code></pre>

blocks|key|1409989|text|如果不能使用加号运算符(%2B)，可以使用operator导入：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1409990|import+operator

listone+=+[1,2,3]
listtwo+=+[4,5,6]

result+=+operator.add(listone,+listtwo)
print(result)

>>>+[1,+2,+3,+4,+5,+6]|code-block|syntax|javascript|1409991|或者，您也可以使用__add__+dunder函数：|1409992|listone+=+[1,2,3]
listtwo+=+[4,5,6]

result+=+list.__add__(listone,+listtwo)
print(result)

>>>+[1,+2,+3,+4,+5,+6]|1409993|entityMap|0|LINK|mutability|MUTABLE|url|https://wiki.python.org/moin/DunderAlias^0|C|1|J|8|0|0|9|7|H|6|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

If you can't use the plus operator (<code>+</code>), you can use the <code>operator</code> import:

<pre><code>import operator

listone = [1,2,3]
listtwo = [4,5,6]

result = operator.add(listone, listtwo)
print(result)

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

Alternatively, you could also use the <code>__add__</code> <a href="https://wiki.python.org/moin/DunderAlias" rel="noreferrer">dunder</a> function:

<pre><code>listone = [1,2,3]
listtwo = [4,5,6]

result = list.__add__(listone, listtwo)
print(result)

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

blocks|key|1409889|text|如果您需要合并两个具有复杂排序规则的有序列表，则可能需要像以下代码中那样自己滚动(使用简单的排序规则以提高可读性:-)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1409890|list1+=+[1,2,5]
list2+=+[2,3,4]
newlist+=+[]

while+list1+and+list2:
++++if+list1[0]+==+list2[0]:
++++++++newlist.append(list1.pop(0))
++++++++list2.pop(0)
++++elif+list1[0]+<+list2[0]:
++++++++newlist.append(list1.pop(0))
++++else:
++++++++newlist.append(list2.pop(0))

if+list1:
++++newlist.extend(list1)
if+list2:
++++newlist.extend(list2)

assert(newlist+==+[1,+2,+3,+4,+5])|code-block|syntax|javascript|1409891|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 need to merge two ordered lists with complicated sorting rules, you might have to roll it yourself like in the following code (using a simple sorting rule for readability :-) ).

<pre><code>list1 = [1,2,5]
list2 = [2,3,4]
newlist = []

while list1 and list2:
 if list1[0] == list2[0]:
 newlist.append(list1.pop(0))
 list2.pop(0)
 elif list1[0] &lt; list2[0]:
 newlist.append(list1.pop(0))
 else:
 newlist.append(list2.pop(0))

if list1:
 newlist.extend(list1)
if list2:
 newlist.extend(list2)

assert(newlist == [1, 2, 3, 4, 5])
</code></pre>

blocks|key|1410215|text|如果您使用的是NumPy，则可以使用以下命令连接两个兼容维数的数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1410216|numpy.concatenate([a,b])|code-block|syntax|javascript|1410217|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 are using NumPy, you can concatenate two arrays of compatible dimensions with this command:
<pre><code>numpy.concatenate([a,b])
</code></pre>

blocks|key|1194796|text|使用简单的列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1194797|joined_list+=+[item+for+list_+in+[list_one,+list_two]+for+item+in+list_]|code-block|syntax|javascript|1194798|它具有使用Additional+Unpacking+Generalizations的最新方法的所有优点-即您可以以这种方式连接任意数量的不同迭代器(例如，列表、元组、范围和生成器)-而且它并不限于Python3.5或更高版本。|offset|length|1194799|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0448/^0|0|0|5|10|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Use a simple list comprehension:

<pre><code>joined_list = [item for list_ in [list_one, list_two] for item in list_]
</code></pre>

It has all the advantages of the newest approach of using <a href="https://www.python.org/dev/peps/pep-0448/" rel="noreferrer">Additional Unpacking Generalizations</a> - i.e. you can concatenate an arbitrary number of different iterables (for example, lists, tuples, ranges, and generators) that way - and it's not limited to Python 3.5 or later.

blocks|key|1194452|text|您可以使用在list对象上定义的append()方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1194453|mergedlist+=[]
for+elem+in+listone:
++++mergedlist.append(elem)
for+elem+in+listtwo:
++++mergedlist.append(elem)|code-block|syntax|javascript|1194454|entityMap^0|6|4|G|8|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|$]]

You could use the <code>append()</code> method defined on <code>list</code> objects: 

<pre><code>mergedlist =[]
for elem in listone:
 mergedlist.append(elem)
for elem in listtwo:
 mergedlist.append(elem)
</code></pre>

blocks|key|1410077|text|list(set(listone)+%7C+set(listtwo))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1410078|上面的代码不保留顺序，从每个列表(但不是从连接的列表)中删除重复项|unstyled|1410079|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>list(set(listone) | set(listtwo))
</code></pre>

The above code, does not preserve order, removes duplicate from each list (but not from the concatenated list)

blocks|key|1410115|text|一种非常简洁的组合列表列表的方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1410116|list_of_lists+=+[[1,2,3],+[4,5,6],+[7,8,9]]
reduce(list.__add__,+list_of_lists)|code-block|syntax|javascript|1410117|这给了我们|1410118|[1,+2,+3,+4,+5,+6,+7,+8,+9]|1410119|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|$]]

A really concise way to combine a list of lists is

<pre><code>list_of_lists = [[1,2,3], [4,5,6], [7,8,9]]
reduce(list.__add__, list_of_lists)
</code></pre>

which gives us

<pre><code>[1, 2, 3, 4, 5, 6, 7, 8, 9]
</code></pre>

blocks|key|1410504|text|我推荐三种方法来连接列表，但第一种方法是最推荐的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1410505|#+easiest+and+least+complexity+method+<=+recommended

listone+=+[1,+2,+3]
listtwo+=+[4,+5,+6]

newlist+=+listone+%2B+listtwo
print(newlist)

#+2nd+easiest+method
newlist+=+listone.copy()
newlist.extend(listtwo)
print(newlist)|code-block|syntax|javascript|1410506|在第二种方法中，我将newlist赋值给listone的一个副本，因为我不想更改listone。|offset|length|style|CODE|1410507|#+3rd+method
newlist+=+listone.copy()
for+j+in+listtwo:
++++newlist.append(j)

print(newlist)|1410508|这不是连接列表的好方法，因为我们使用for循环来连接列表。因此，时间复杂度比其他两种方法要高得多。|1410509|entityMap^0|0|0|A|7|K|7|14|7|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]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

I recommend three methods to concatenate the list but 1st method is most recommended,
<pre><code># easiest and least complexity method &lt;= recommended

listone = [1, 2, 3]
listtwo = [4, 5, 6]

newlist = listone + listtwo
print(newlist)

# 2nd easiest method
newlist = listone.copy()
newlist.extend(listtwo)
print(newlist)
</code></pre>
In the 2nd method, I assign <code>newlist</code> to a copy of the <code>listone</code> because I don't want to change <code>listone</code>.
<pre><code># 3rd method
newlist = listone.copy()
for j in listtwo:
 newlist.append(j)

print(newlist)
</code></pre>
This is not a good way to concatenate lists because we using for loop to concatenate the lists. So time complexity is much higher than with the other two methods.

blocks|key|1195002|text|您可以使用.extend方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1195003|listone+=+[1,+2,+3]
listtwo+=+[4,+5,+6]
listone.extend(lsittwo)|code-block|syntax|javascript|1195004|现在当你这么做的时候，|1195005|print(listone)|1195006|你会有这份名单，|1195007|[1,2,3,4,5,6]|1195008|entityMap^0|5|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

You can use <code>.extend</code> method
<pre><code>listone = [1, 2, 3]
listtwo = [4, 5, 6]
listone.extend(lsittwo)
</code></pre>
Now when you do,
<pre><code>print(listone)
</code></pre>
You'll have this list,
<pre><code>[1,2,3,4,5,6]
</code></pre>

blocks|key|1410443|text|你也可以直接使用sum。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1410444|>>>+a+=+[1,+2,+3]
>>>+b+=+[4,+5,+6]
>>>+sum([a,+b],+[])
[1,+2,+3,+4,+5,+6]
>>>+|code-block|syntax|javascript|1410445|这适用于任何长度和任何元素类型的列表：|1410446|>>>+a+=+['a',+'b',+'c',+'d']
>>>+b+=+[1,+2,+3,+4]
>>>+c+=+[1,+2]
>>>+sum([a,+b,+c],+[])
['a',+'b',+'c',+'d',+1,+2,+3,+4,+1,+2]
>>>+|1410447|我添加[]的原因是因为start参数默认设置为0，因此它循环遍历列表并添加到start中，但0+%2B+[1,+2,+3]会给出一个错误，所以如果我们将start设置为[]，它将添加到[]中，[]+%2B+[1,+2,+3]将按预期工作。|1410448|entityMap^0|8|3|0|0|0|0|3|2|B|5|N|1|12|5|1A|D|22|5|2A|2|2I|2|2M|E|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$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]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|R|$]]

You could also just use <code>sum</code>.
<pre><code>&gt;&gt;&gt; a = [1, 2, 3]
&gt;&gt;&gt; b = [4, 5, 6]
&gt;&gt;&gt; sum([a, b], [])
[1, 2, 3, 4, 5, 6]
&gt;&gt;&gt; 
</code></pre>
This works for any length and any element type of list:
<pre><code>&gt;&gt;&gt; a = ['a', 'b', 'c', 'd']
&gt;&gt;&gt; b = [1, 2, 3, 4]
&gt;&gt;&gt; c = [1, 2]
&gt;&gt;&gt; sum([a, b, c], [])
['a', 'b', 'c', 'd', 1, 2, 3, 4, 1, 2]
&gt;&gt;&gt; 
</code></pre>
The reason I add <code>[]</code>, is because the <code>start</code> argument is set to <code>0</code> by default, so it loops through the list and adds to <code>start</code>, but <code>0 + [1, 2, 3]</code> would give an error, so if we set the <code>start</code> to <code>[]</code>, it would add to <code>[]</code>, <code>[] + [1, 2, 3]</code> would work as expected.

blocks|key|1195284|text|只需使用extend()函数，您就会过上幸福的生活|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1195285|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

JUST USE THE extend() function, and you will live a bliss life

blocks|key|1410310|text|只需添加以下两个列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1410311|final_list+=+list1+%2B+list2|code-block|syntax|javascript|1410312|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|$]]

Just add these two lists:
<pre><code>final_list = list1 + list2
</code></pre>

blocks|key|1410582|text|您可以使用提供的方法'extend‘。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1410583|LIST_1+=+[1,2,3]
LIST_2+=+[4,5,6]
LIST_1.extend(LIST_2)
print(LIST_1)|code-block|syntax|javascript|1410584|输出：|1410585|[1,2,3,4,5,6]|1410586|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|$]]

You can use the provided method 'extend'.
<pre><code>LIST_1 = [1,2,3]
LIST_2 = [4,5,6]
LIST_1.extend(LIST_2)
print(LIST_1)
</code></pre>
Output:
<pre><code>[1,2,3,4,5,6]
</code></pre>

blocks|key|1195179|text|提供的解决方案是针对单个列表的。在列表中存在列表并且需要合并相应列表的情况下。通过for循环的"%2B“操作完成了这项工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1195180|a=[[1,2,3],[4,5,6]]

b=[[0,1,2],[7,8,9]]

for+i+in+range(len(a)):
++++cc.append(a[i]%2Bb[i])|code-block|syntax|javascript|1195181|输出：[1，2，3，0，1，2，4，5，6，7，8，9]|1195182|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The solutions provided are for single list. In case there are lists within a list and the merging of corresponding lists is required. &quot;+&quot; operation through for loop does the work.
<pre><code>a=[[1,2,3],[4,5,6]]

b=[[0,1,2],[7,8,9]]

for i in range(len(a)):
 cc.append(a[i]+b[i])
</code></pre>
output: [[1, 2, 3, 0, 1, 2], [4, 5, 6, 7, 8, 9]]

blocks|key|1410278|text|您可以在python中使用union()函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1410279|joinedlist+=+union(listone,+listtwo)
print(joinedlist)|code-block|syntax|javascript|1410280|本质上，它所做的是删除两个列表中的每个重复项。因为您的列表没有任何副本，所以它只返回两个列表的连接版本。|1410281|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can use the union() function in python.
<pre><code>joinedlist = union(listone, listtwo)
print(joinedlist)
</code></pre>
Essentially what this is doing is its removing one of every duplicate in the two lists. Since your lists don't have any duplicates it, it just returns the concatenated version of the two lists.

How do I concatenate two lists in Python?

Example:

<pre><code>listone = [1, 2, 3]
listtwo = [4, 5, 6]
</code></pre>

Expected outcome:

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

How do I concatenate two lists in Python?

Python

如何在Python中连接两个列表？示例：listone = [1, 2, 3]listtwo = [4, 5, 6]预期结果：>>> joinedlist[1, 2, 3, 4, 5, 6]

问如何在Python中连接两个列表？
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