blocks|key|2682483|text|使用np.unique：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2682484|a=np.random.randint(0,10**6,3*10**6)
uniques,counts=np.unique(a,return_counts=True)
In+[293]:+uniques[counts==14]
Out[293]:+array([++4541,+250510,+622471,+665409,+881697,+920586])|code-block|syntax|javascript|2682485|这只需要不到一秒钟。但我不明白为什么您的where语句很慢。对我来说，你的解决方案更快：|2682486|In+[313]:+%25timeit+b=np.bincount(a,minlength=a.size)
61.5+ms+±+4.82+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+10+loops+each)

In+[314]:+%25timeit+np.where(b==3)[0]
11.8+ms+±+271+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+100+loops+each)

In+[315]:+%25timeit+uniques,counts=np.unique(a,return_counts=True)
424+ms+±+6.82+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)

In+[316]:+%25timeit+Counter(a)
1.41+s+±+18+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+1+loop+each)|2682487|编辑|BOLD|2682488|@numba.jit()
def+count(a,n):
++++counters=np.zeros(10**6,np.int32)
++++for+i+in+a:
++++++++counters[i]+%2B=+1
++++res=np.empty_like(counters)
++++k+=+0++++
++++for+i,j+in+enumerate(counters):
++++++++if+j+==+n:
++++++++++++res[k]+=+i
++++++++++++k+%2B=+1
++++return+res[:k]++++++++|2682489|这个numba功能可以给你一个3X的改进。要获得更多信息，您必须找到并行解决方案，以GPU为例。|2682490|In+[26]:+%25timeit++count(a,13)
23.6+ms+±+1.56+ms+per+loop+(mean+±+std.+dev.+of+7+runs,+10+loops+each)|2682491|entityMap|0|LINK|mutability|MUTABLE|url|http://numba.pydata.org/numba-doc/0.13/CUDAJit.html^0|2|9|0|0|K|5|0|0|0|2|0|0|15|6|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1D|8|@$9|1E|A|1F|B|Q]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|T|3|U|5|6|7|1H|8|@]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|V|3|W|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|X|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]]]

Use <code>np.unique</code> :

<pre><code>a=np.random.randint(0,10**6,3*10**6)
uniques,counts=np.unique(a,return_counts=True)
In [293]: uniques[counts==14]
Out[293]: array([ 4541, 250510, 622471, 665409, 881697, 920586])
</code></pre>

This takes less than a second. but I don't understand why your <code>where</code> statement is slow. For me your solution is faster :

<pre><code>In [313]: %timeit b=np.bincount(a,minlength=a.size)
61.5 ms ± 4.82 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [314]: %timeit np.where(b==3)[0]
11.8 ms ± 271 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [315]: %timeit uniques,counts=np.unique(a,return_counts=True)
424 ms ± 6.82 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [316]: %timeit Counter(a)
1.41 s ± 18 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
</code></pre>

EDIT

<pre><code>@numba.jit()
def count(a,n):
 counters=np.zeros(10**6,np.int32)
 for i in a:
 counters[i] += 1
 res=np.empty_like(counters)
 k = 0 
 for i,j in enumerate(counters):
 if j == n:
 res[k] = i
 k += 1
 return res[:k] 
</code></pre>

This numba function can give you a 3X improvement. for more you must find parallel solutions, <a href="http://numba.pydata.org/numba-doc/0.13/CUDAJit.html" rel="nofollow noreferrer">on GPU for example</a>. 

<pre><code>In [26]: %timeit count(a,13)
23.6 ms ± 1.56 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
</code></pre>

blocks|key|526918|text|您可以使用Counter|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|526919|>>>+a+=+np.array([2000000000,3005670000,2000000000,12345667,4000789000,12345687,12345667,2000000000,12345667])
>>>+c+=+Counter(a)
>>>+np.array([i+for+i+in+c+if+c[i]+>=+3])
array([2000000000,+++12345667])|code-block|syntax|javascript|526920|entityMap^0|5|7|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|$]]

You can use a <code>Counter</code>:

<pre><code>&gt;&gt;&gt; a = np.array([2000000000,3005670000,2000000000,12345667,4000789000,12345687,12345667,2000000000,12345667])
&gt;&gt;&gt; c = Counter(a)
&gt;&gt;&gt; np.array([i for i in c if c[i] &gt;= 3])
array([2000000000, 12345667])
</code></pre>

GOAL:
I have a large 1d array (3000000+) of distances with many duplicate distances. I am trying to write the swiftest function that returns all distances that appear n times in the array. I have written a function in numpy but there is a bottleneck at one line in the code. Swift performance is an issue because the calculations are done in a for loop for 2400 different large distance arrays.



<pre><code>import numpy as np
for t in range(0, 2400):
 a=np.random.randint(1000000000, 5000000000, 3000000)
 b=np.bincount(a,minlength=np.size(a))
 c=np.where(b == 3)[0] #SLOW STATEMENT/BOTTLENECK
 return c
</code></pre>


EXPECTED RESULTS:
Given a 1d array of distances [2000000000,3005670000,2000000000,12345667,4000789000,12345687,12345667,2000000000,12345667]
I would expect back an array of [2000000000,12345667] when queried to return an array of all distances that appear 3 times in the main array.


What should I do?

Python/Numpy have I already written the swiftest code for large array?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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目标：I有一个大的一维数组(3000000+)，它包含许多重复的距离。我正在尝试编写最快速的函数，它返回数组中出现的所有距离。我用numpy编写了一个函数，但是在代码的一行中有一个瓶颈。Swift性能是一个问题，因为计算是在一个for循环中为2400个不同的大距离数组进行的。import numpy as npfor ...

问Python/Numpy我已经为大型数组编写了最快的代码了吗？
EN

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问Python/Numpy我已经为大型数组编写了最快的代码了吗？EN