blocks|key|911539|text|你能提供一个代码示例，让我看看你是如何使用copy()的，在什么上下文中使用？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|911540|你可以使用|911541|new+=+dict(old)|code-block|syntax|javascript|911542|但我不认为它会更快。|911543|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Can you provide a code sample so I can see how you are using copy() and in what context?

You could use 

<pre><code>new = dict(old)
</code></pre>

But I dont think it will be faster.

blocks|key|911606|text|正如你所说，看起来dict.copy更快。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|911607|[utdmr@utdmr-arch+~]$+python+-m+timeit+-s+"d={1:1,+2:2,+3:3}"+"new+=+d.copy()"
1000000+loops,+best+of+3:+0.238+usec+per+loop
[utdmr@utdmr-arch+~]$+python+-m+timeit+-s+"d={1:1,+2:2,+3:3}"+"new+=+dict(d)"
1000000+loops,+best+of+3:+0.621+usec+per+loop
[utdmr@utdmr-arch+~]$+python+-m+timeit+-s+"from+copy+import+copy;+d={1:1,+2:2,+3:3}"+"new+=+copy(d)"
1000000+loops,+best+of+3:+1.58+usec+per+loop|code-block|syntax|javascript|911608|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|$]]

Appearantly dict.copy is faster, as you say.

<pre><code>[utdmr@utdmr-arch ~]$ python -m timeit -s "d={1:1, 2:2, 3:3}" "new = d.copy()"
1000000 loops, best of 3: 0.238 usec per loop
[utdmr@utdmr-arch ~]$ python -m timeit -s "d={1:1, 2:2, 3:3}" "new = dict(d)"
1000000 loops, best of 3: 0.621 usec per loop
[utdmr@utdmr-arch ~]$ python -m timeit -s "from copy import copy; d={1:1, 2:2, 3:3}" "new = copy(d)"
1000000 loops, best of 3: 1.58 usec per loop
</code></pre>

blocks|key|695575|text|查看Python+dict操作的C+source，您可以看到它们执行了非常简单(但高效)的复制。它本质上归结为对PyDict_Merge的调用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|695576|PyDict_Merge(PyObject+*a,+PyObject+*b,+int+override)|code-block|syntax|javascript|695577|这会快速检查它们是否是相同的对象，以及其中是否有对象。在此之后，它对目标dict执行一次慷慨的调整大小/分配，然后逐个复制元素。我不认为你会比内置的copy()更快。|695578|entityMap|0|LINK|mutability|MUTABLE|url|http://hg.python.org/cpython/file/tip/Objects/dictobject.c^0|9|4|1K|C|G|8|0|0|0|22|6|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Looking at the <a href="http://hg.python.org/cpython/file/tip/Objects/dictobject.c">C source</a> for the Python <code>dict</code> operations, you can see that they do a pretty naive (but efficient) copy. It essentially boils down to a call to <code>PyDict_Merge</code>:

<pre><code>PyDict_Merge(PyObject *a, PyObject *b, int override)
</code></pre>

This does the quick checks for things like if they're the same object and if they've got objects in them. After that it does a generous one-time resize/alloc to the target dict and then copies the elements one by one. I don't see you getting much faster than the built-in <code>copy()</code>.

blocks|key|695635|text|但是，度量依赖于字典的大小。对于10000个条目，copy(d)和d.copy()几乎相同。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|695636|a+=+{b:+b+for+b+in+range(10000)}+
In+[5]:+%25timeit+copy(a)
10000+loops,+best+of+3:+186+µs+per+loop
In+[6]:+%25timeit+deepcopy(a)
100+loops,+best+of+3:+14.1+ms+per+loop
In+[7]:+%25timeit+a.copy()
1000+loops,+best+of+3:+180+µs+per+loop|code-block|syntax|javascript|695637|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|$]]

The measurments are dependent on the dictionary size though. For 10000 entries copy(d) and d.copy() are almost the same.

<pre><code>a = {b: b for b in range(10000)} 
In [5]: %timeit copy(a)
10000 loops, best of 3: 186 µs per loop
In [6]: %timeit deepcopy(a)
100 loops, best of 3: 14.1 ms per loop
In [7]: %timeit a.copy()
1000 loops, best of 3: 180 µs per loop
</code></pre>

blocks|key|695699|text|我意识到这是一个老帖子，但这是搜索引擎中"dict+copy+python“的高结果，也是"dict+copy+performance”的最高结果，我相信这是相关的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|695700|从Python3.7开始，newDict+=+oldDict.copy()的速度比以前快了5.5倍。值得注意的是，目前newDict+=+dict(oldDict)似乎没有这样的性能提升。|offset|length|style|CODE|695701|这里有更多的信息，here。|695702|entityMap|0|LINK|mutability|MUTABLE|url|https://bugs.python.org/issue31179^0|0|D|O|1N|N|0|9|4|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]|$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@$D|Y|E|Z|1|10]]|A|$]]|$1|J|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

I realise this is an old thread, but this is a high result in search engines for "dict copy python", and the top result for "dict copy performance", and I believe this is relevant.

From Python 3.7, <code>newDict = oldDict.copy()</code> is up to 5.5x faster than it was previously. Notably, right now, <code>newDict = dict(oldDict)</code> does not seem to have this performance increase.

There is a little more information <a href="https://bugs.python.org/issue31179" rel="noreferrer">here</a>.

I have a Python program that works with dictionaries a lot. I have to make copies of dictionaries thousands of times. I need a copy of both the keys and the associated contents. The copy will be edited and must not be linked to the original (e.g. changes in the copy must not affect the original.)

Keys are Strings, Values are Integers (0/1).

I currently use a simple way:

<pre><code>newDict = oldDict.copy()
</code></pre>

Profiling my Code shows that the copy operation takes most of the time.

Are there faster alternatives to the <code>dict.copy()</code> method? What would be fastest?

Fast way to copy dictionary in Python

Python

我有一个Python程序，它可以很好地处理字典。我得把字典复印几千次。我需要一份钥匙和相关内容的副本。副本将被编辑，并且不得链接到原始副本(例如，副本中的更改不得影响原始副本)。键是字符串，值是整数(0/1)。我目前使用一种简单的方法：newDict = oldDict.copy()分析我的代码显示，复制操作花费了大部...

问在Python中复制字典的快速方法
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