blocks|key|170990|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|170991|B=A创建reference|ordered-list-item|offset|length|style|CODE|170992|B[:]=A+makes+a+copy|170993|numpy.copy(B,A)+makes+a+copy|170994|170995|后两个需要额外的内存。|170996|要制作深度副本，您需要使用B+=+copy.deepcopy(A)|170997|entityMap^0|0|0|3|0|0|6|0|0|F|0|0|0|D|K|0^^$0|@$1|2|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|T|7|@$D|U|E|V|F|G]]|8|@]|9|$]]|$1|H|3|I|4|C|6|W|7|@$D|X|E|Y|F|G]]|8|@]|9|$]]|$1|J|3|K|4|C|6|Z|7|@$D|10|E|11|F|G]]|8|@]|9|$]]|$1|L|3|-4|4|5|6|12|7|@]|8|@]|9|$]]|$1|M|3|N|4|5|6|13|7|@]|8|@]|9|$]]|$1|O|3|P|4|5|6|14|7|@$D|15|E|16|F|G]]|8|@]|9|$]]|$1|Q|3|-4|4|5|6|17|7|@]|8|@]|9|$]]]|R|$]]

<ol>
<li><code>B=A</code> creates a reference</li>
<li><code>B[:]=A</code> makes a copy</li>
<li><code>numpy.copy(B,A)</code> makes a copy</li>
</ol>

the last two need additional memory.

To make a deep copy you need to use <code>B = copy.deepcopy(A)</code>

blocks|key|171827|text|这三个版本做了不同的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|171828|171829|B+=+A|ordered-list-item|offset|length|style|CODE|171830|171831|这会将一个新名称B绑定到已命名为A的现有对象。之后，它们引用相同的对象，所以如果您就地修改其中一个对象，您将通过另一个too.|171832|B[:]+=+A+(与B[:]=A[:]相同？)看到更改。|unordered-list-item|171833|171834|这会将值从A复制到现有数组B中。这两个数组必须具有相同的形状才能工作。B[:]+=+A[:]做同样的事情(但是B+=+A[:]会做更像1).|171835|numpy.copy(B,+A)的事情|171836|171837|这不是合法的语法。你可能指的是B+=+numpy.copy(A)。这几乎与2相同，但它创建了一个新数组，而不是重用B数组。如果没有对先前B值的其他引用，则最终结果将与2相同，但它将在复制过程中临时使用更多内存。|171838|或者你指的是numpy.copyto(B,+A)，这是合法的，相当于2?。|171839|entityMap^0|0|0|0|5|0|0|8|1|G|1|0|0|8|B|9|0|0|5|1|D|1|Z|B|1J|8|0|0|G|0|0|F|H|1L|1|1W|1|0|6|I|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|13|8|@$F|14|G|15|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|17|8|@$F|18|G|19|H|I]|$F|1A|G|1B|H|I]]|9|@]|A|$]]|$1|M|3|N|5|O|7|1C|8|@$F|1D|G|1E|H|I]|$F|1F|G|1G|H|I]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1I|8|@$F|1J|G|1K|H|I]|$F|1L|G|1M|H|I]|$F|1N|G|1O|H|I]|$F|1P|G|1Q|H|I]]|9|@]|A|$]]|$1|S|3|T|5|O|7|1R|8|@$F|1S|G|1T|H|I]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1V|8|@$F|1W|G|1X|H|I]|$F|1Y|G|1Z|H|I]|$F|20|G|21|H|I]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|22|8|@$F|23|G|24|H|I]]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|25|8|@]|9|@]|A|$]]]|10|$]]

All three versions do different things:

<ol>
<li><code>B = A</code>

This binds a new name <code>B</code> to the existing object already named <code>A</code>. Afterwards they refer to the same object, so if you modify one in place, you'll see the change through the other one too.</li>
<li><code>B[:] = A</code> (same as <code>B[:]=A[:]</code>?)

This copies the values from <code>A</code> into an existing array <code>B</code>. The two arrays must have the same shape for this to work. <code>B[:] = A[:]</code> does the same thing (but <code>B = A[:]</code> would do something more like 1).</li>
<li><code>numpy.copy(B, A)</code>

This is not legal syntax. You probably meant <code>B = numpy.copy(A)</code>. This is almost the same as 2, but it creates a new array, rather than reusing the <code>B</code> array. If there were no other references to the previous <code>B</code> value, the end result would be the same as 2, but it will use more memory temporarily during the copy.

Or maybe you meant <code>numpy.copyto(B, A)</code>, which is legal, and is equivalent to 2?</li>
</ol>

blocks|key|169837|text|对我来说，这是唯一有效的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|169838|B=numpy.array(A)|code-block|syntax|javascript|169839|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|$]]

This is the only working answer for me:

<pre><code>B=numpy.array(A)
</code></pre>

For example, if we have a <code>numpy</code> array <code>A</code>, and we want a <code>numpy</code> array <code>B</code> with the same elements.

What is the difference between the following (see below) methods? When is additional memory allocated, and when is it not?

<ol>
<li><code>B = A</code></li>
<li><code>B[:] = A</code> (same as <code>B[:]=A[:]</code>?)</li>
<li><code>numpy.copy(B, A)</code></li>
</ol>

Numpy array assignment with copy

例如，如果我们有一个numpy数组A，并且我们想要一个具有相同元素的numpy数组B。以下两种方法(见下文)有什么不同？什么时候会分配额外的内存，什么时候不会？B = AB[:] = A (与B[:]=A[:]?)numpy.copy(B, A)相同

问使用copy进行Numpy数组赋值
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