blocks|key|472887|text|在Python中不能通过值传递任何内容。如果要创建a的副本，可以按照official+Python+FAQ中所述显式执行此操作|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|472888|b+=+a[:]|code-block|syntax|javascript|472889|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/faq/programming.html#how-do-i-copy-an-object-in-python^0|P|1|Y|J|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]]]]

You cannot pass anything by value in Python. If you want to make a copy of <code>a</code>, you can do so explicitly, as described in the <a href="http://docs.python.org/faq/programming.html#how-do-i-copy-an-object-in-python" rel="noreferrer">official Python FAQ</a>:
<pre><code>b = a[:]
</code></pre>

blocks|key|689166|text|要创建列表的副本，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|689167|b+=+a[:]|code-block|syntax|javascript|689168|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|$]]

To create a copy of a list do this:

<pre><code>b = a[:]
</code></pre>

blocks|key|472974|text|此外，您还可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|472975|b+=+list(a)|code-block|syntax|javascript|472976|这将适用于任何序列，即使是那些不支持索引器和切片的序列……|472977|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|$]]

Also, you can do:

<pre><code>b = list(a)
</code></pre>

This will work for any sequence, even those that don't support indexers and slices...

blocks|key|689252|text|b+=+list(a)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|689253|参见http://henry.precheur.org/python/copy_list。|689254|entityMap|0|LINK|mutability|MUTABLE|url|http://henry.precheur.org/python/copy_list^0|0|B|0|2|16|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

<code>b = list(a)</code>

See <a href="http://henry.precheur.org/python/copy_list" rel="nofollow">http://henry.precheur.org/python/copy_list</a>.

blocks|key|473087|text|要复制列表，可以使用list(a)或a[:]。在这两种情况下，都会创建一个新对象。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|473088|然而，这两种方法对可变对象的集合有限制，因为内部对象保持它们的引用不变：|473089|>>>+a+=+[[1,2],[3],[4]]

>>>+b+=+a[:]
>>>+c+=+list(a)

>>>+c[0].append(9)

>>>+a
[[1,+2,+9],+[3],+[4]]
>>>+c
[[1,+2,+9],+[3],+[4]]
>>>+b
[[1,+2,+9],+[3],+[4]]
>>>+|code-block|syntax|javascript|473090|如果您想要对象的完整副本，则需要copy.deepcopy|473091|>>>+from+copy+import+deepcopy
>>>+a+=+[[1,2],[3],[4]]

>>>+b+=+a[:]
>>>+c+=+deepcopy(a)

>>>+c[0].append(9)

>>>+a
[[1,+2],+[3],+[4]]
>>>+b
[[1,+2],+[3],+[4]]
>>>+c
[[1,+2,+9],+[3],+[4]]
>>>+|473092|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/copy.html^0|A|7|I|4|0|0|0|G|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|O|3|P|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

To copy a list you can use <code>list(a)</code> or <code>a[:]</code>. In both cases a new object is created. 
These two methods, however, have limitations with collections of mutable objects as inner objects keep their references intact:

<pre><code>&gt;&gt;&gt; a = [[1,2],[3],[4]]

&gt;&gt;&gt; b = a[:]
&gt;&gt;&gt; c = list(a)

&gt;&gt;&gt; c[0].append(9)

&gt;&gt;&gt; a
[[1, 2, 9], [3], [4]]
&gt;&gt;&gt; c
[[1, 2, 9], [3], [4]]
&gt;&gt;&gt; b
[[1, 2, 9], [3], [4]]
&gt;&gt;&gt; 
</code></pre>

If you want a full copy of your objects you need <a href="http://docs.python.org/library/copy.html" rel="noreferrer">copy.deepcopy</a>

<pre><code>&gt;&gt;&gt; from copy import deepcopy
&gt;&gt;&gt; a = [[1,2],[3],[4]]

&gt;&gt;&gt; b = a[:]
&gt;&gt;&gt; c = deepcopy(a)

&gt;&gt;&gt; c[0].append(9)

&gt;&gt;&gt; a
[[1, 2], [3], [4]]
&gt;&gt;&gt; b
[[1, 2], [3], [4]]
&gt;&gt;&gt; c
[[1, 2, 9], [3], [4]]
&gt;&gt;&gt; 
</code></pre>

blocks|key|473109|text|我发现我们可以使用extend()来实现copy()的函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|473110|a=['help',+'copyright',+'credits',+'license']
b+=+[]
b.extend(a)
b.append("XYZ")+|code-block|syntax|javascript|473111|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|$]]

I found that we can use extend() to implement the function of copy()

<pre><code>a=['help', 'copyright', 'credits', 'license']
b = []
b.extend(a)
b.append("XYZ") 
</code></pre>

blocks|key|473153|text|在性能方面，我最喜欢的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|473154|b.extend(a)|code-block|syntax|javascript|473155|检查相关替代方案在性能方面的比较：|473156|In+[1]:+import+timeit

In+[2]:+timeit.timeit('b.extend(a)',+setup='b=[];a=range(0,10)',+number=100000000)
Out[2]:+9.623248100280762

In+[3]:+timeit.timeit('b+=+a[:]',+setup='b=[];a=range(0,10)',+number=100000000)
Out[3]:+10.84756088256836

In+[4]:+timeit.timeit('b+=+list(a)',+setup='b=[];a=range(0,10)',+number=100000000)
Out[4]:+21.46313500404358

In+[5]:+timeit.timeit('b+=+[elem+for+elem+in+a]',+setup='b=[];a=range(0,10)',+number=100000000)
Out[5]:+66.99795293807983

In+[6]:+timeit.timeit('for+elem+in+a:+b.append(elem)',+setup='b=[];a=range(0,10)',+number=100000000)
Out[6]:+67.9775960445404

In+[7]:+timeit.timeit('b+=+deepcopy(a)',+setup='from+copy+import+deepcopy;+b=[];a=range(0,10)',+number=100000000)
Out[7]:+1216.1108016967773|473157|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|$]]

In terms of performance my favorite answer would be:

<pre><code>b.extend(a)
</code></pre>

Check how the related alternatives compare with each other in terms of performance:

<pre><code>In [1]: import timeit

In [2]: timeit.timeit('b.extend(a)', setup='b=[];a=range(0,10)', number=100000000)
Out[2]: 9.623248100280762

In [3]: timeit.timeit('b = a[:]', setup='b=[];a=range(0,10)', number=100000000)
Out[3]: 10.84756088256836

In [4]: timeit.timeit('b = list(a)', setup='b=[];a=range(0,10)', number=100000000)
Out[4]: 21.46313500404358

In [5]: timeit.timeit('b = [elem for elem in a]', setup='b=[];a=range(0,10)', number=100000000)
Out[5]: 66.99795293807983

In [6]: timeit.timeit('for elem in a: b.append(elem)', setup='b=[];a=range(0,10)', number=100000000)
Out[6]: 67.9775960445404

In [7]: timeit.timeit('b = deepcopy(a)', setup='from copy import deepcopy; b=[];a=range(0,10)', number=100000000)
Out[7]: 1216.1108016967773
</code></pre>

blocks|key|689411|text|如果要复制一维列表，请使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|689412|b+=+a[:]|code-block|syntax|javascript|689413|但是，如果a是一个二维列表，那么这对您就不起作用了。也就是说，a中的任何更改也将反映在b中。在这种情况下，使用|offset|length|style|CODE|689414|b+=+[[a[x][y]+for+y+in+range(len(a[0]))]+for+x+in+range(len(a))]|689415|entityMap^0|0|0|5|1|V|1|17|1|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

If you want to copy a one-dimensional list, use

<pre><code>b = a[:]
</code></pre>

However, if <code>a</code> is a 2-dimensional list, this is not going to work for you. That is, any changes in <code>a</code> will also be reflected in <code>b</code>. In that case, use

<pre><code>b = [[a[x][y] for y in range(len(a[0]))] for x in range(len(a))]
</code></pre>

blocks|key|473274|text|正如phihag在他的回答中提到的，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|473275|b+=+a[:]|code-block|syntax|javascript|473276|将适用于您的情况，因为切片列表将创建列表的新内存id+(这意味着您不再在内存中引用相同的对象，并且您对一个对象所做的更改不会反映在另一个对象中)。|473277|然而，有一个小问题。如果您的列表是多维的，就像列表中的列表一样，简单的切片不能解决这个问题。在更高维度中所做的更改，即原始列表中的列表，将在两者之间共享。|473278|不要着急，有一个解决方案。模块复制有一个很好的复制技术来解决这个问题。|473279|from+copy+import+deepcopy

b+=+deepcopy(a)|473280|将复制一个具有新内存id的列表，而不管它包含多少级别的列表！|473281|entityMap^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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

As mentioned by phihag in his answer,

<pre><code>b = a[:]
</code></pre>

will work for your case since slicing a list creates a new memory id of the list (meaning you are no longer referencing the same object in your memory and the changes you make to one will not be reflected in the other.)

However, there is a slight problem. If your list is multidimensional, as in lists within lists, simply slicing will not solve this problem. Changes made in the higher dimensions, i.e. the lists within the original list, will be shared between the two.

Do not fret, there is a solution. The module copy has a nifty copying technique that takes care of this issue.

<pre><code>from copy import deepcopy

b = deepcopy(a)
</code></pre>

will copy a list with a new memory id no matter how many levels of lists it contains!

blocks|key|473348|text|我推荐以下解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|473349|b+=+[]
b[:]+=+a|code-block|syntax|javascript|473350|这将把所有元素从a复制到b。复制将是值复制，而不是引用复制。|473351|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|$]]

I would recommend the following solution:

<pre><code>b = []
b[:] = a
</code></pre>

This will copy all the elements from a to b. The copy will be value copy, not reference copy.

Let's take an example

<pre><code>a=['help', 'copyright', 'credits', 'license']
b=a
b.append('XYZ')
b
['help', 'copyright', 'credits', 'license', 'XYZ']
a
['help', 'copyright', 'credits', 'license', 'XYZ']
</code></pre>

I wanted to append value in list 'b' but the value of list 'a' have also changed.
 
I think I have little idea why its like this (python passes lists by reference).
 
My question is "how can I pass it by value so that appending 'b' does't change values in 'a' ?"

python list by value not by reference

Python

让我们举个例子a=['help', 'copyright', 'credits', 'license']b=ab.append('XYZ')b['help', 'copyright', 'credits', 'license', 'XYZ']a['help', 'copyright', 'credits', 'lice...

问python按值而不是按引用列出
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问python按值而不是按引用列出EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问python按值而不是按引用列出
EN