blocks|key|2351397|text|请不要使用.vectorize(..)，它的效率很低，因为它仍然会在Python级别进行处理。例如，如果函数由于其“结构”太复杂而不能以“批量”计算，则.vectorize()只能作为最后的手段使用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2351398|但是您不需要在这里使用.vectorize，您可以使用以下命令来实现您的函数来处理数组：|2351399|r_xy+=+np.sqrt(xx+**+2+%2B+yy+**+2)
density+=+(r_xy+<=+R0)+&+(-hz+<=+zz)+&+(zz+<=+hz)|code-block|syntax|javascript|2351400|甚至更快：|2351401|r_xy+=+xx+*+xx+%2B+yy+*+yy
density+=+(r_xy+<=+R0+*+R0)+&+(-hz+<=+zz)+&+(zz+<=+hz)|2351402|这将构建一个2000×2000×20的布尔值数组。我们可以使用：|2351403|intdens+=+density.astype(int)|2351404|来构造一个int数组。|2351405|在这里打印数组非常麻烦，但它总共包含2'356'047个数组：|2351406|>>>+density.astype(int).sum()
2356047|2351407|基准测试：如果我在本地运行10次，我会得到：|BOLD|2351408|>>>+timeit(f,+number=10)
18.040479518999973
>>>+timeit(f2,+number=10)++#+f2+is+the+optimized+variant
13.287886952000008|2351409|因此，平均而言，我们在1.3-1.8秒内计算这个矩阵(包括将其转换为ints)。|2351410|entityMap^0|5|E|24|C|0|B|A|0|0|0|0|0|0|5|3|0|0|0|0|4|0|0|Y|3|0^^$0|@$1|2|3|4|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1H|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1I|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1J|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1K|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1L|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|1P|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|J|7|1Q|8|@]|D|@]|E|$K|L]]|$1|10|3|11|5|6|7|1R|8|@$9|1S|A|1T|B|12]]|D|@]|E|$]]|$1|13|3|14|5|J|7|1U|8|@]|D|@]|E|$K|L]]|$1|15|3|16|5|6|7|1V|8|@$9|1W|A|1X|B|C]]|D|@]|E|$]]|$1|17|3|-4|5|6|7|1Y|8|@]|D|@]|E|$]]]|18|$]]

Please do not use <code>.vectorize(..)</code>, it is not efficient since it will still do the processing at the Python level. <code>.vectorize()</code> should only be used as a last resort if for example the function can not be calculated in "bulk" because its "structure" is too complex.

But you do not need to use <code>.vectorize</code> here, you can implement your function to work over arrays with:

<pre><code>r_xy = np.sqrt(xx ** 2 + yy ** 2)
density = (r_xy &lt;= R0) &amp; (-hz &lt;= zz) &amp; (zz &lt;= hz)
</code></pre>

or even a bit faster:

<pre><code>r_xy = xx * xx + yy * yy
density = (r_xy &lt;= R0 * R0) &amp; (-hz &lt;= zz) &amp; (zz &lt;= hz)
</code></pre>

This will construct a 2000&times;2000&times;20 array of booleans. We can use:

<pre><code>intdens = density.astype(int)
</code></pre>

to construct an array of <code>int</code>s.

Printing the array here is quite combersome, but it contains a total of 2'356'047 ones:

<pre><code>&gt;&gt;&gt; density.astype(int).sum()
2356047
</code></pre>

Benchmarks: If I run this locally 10 times, I get:

<pre><code>&gt;&gt;&gt; timeit(f, number=10)
18.040479518999973
&gt;&gt;&gt; timeit(f2, number=10) # f2 is the optimized variant
13.287886952000008
</code></pre>

So on average, we calculate this matrix (including casting it to <code>int</code>s) in 1.3-1.8 seconds.

blocks|key|2562091|text|这是对Deepak+Saini的回答的冗长评论。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2562092|主要的改变是不使用由np.meshgrid生成的坐标，它包含不必要的重复。如果你能避免它(无论是在内存使用还是性能方面)，这是不可取的。|offset|length|style|CODE|2562093|代码|BOLD|2562094|import+numba+as+nb
import+numpy+as+np

@nb.jit(nopython=True,parallel=True)
def+calc_density_2(x,+y,+z,R0,hz):
++++threshold+=+R0+*+R0

++++density+=+0
++++for+i+in+nb.prange(y.shape[0]):
++++++++for+j+in+range(x.shape[0]):
++++++++++++r_xy+=+x[j]+**+2+%2B+y[i]+**+2
++++++++++++for+k+in+range(z.shape[0]):
++++++++++++++++if(r_xy+<=+threshold+and+-hz+<=+z[k]+<=+hz):
++++++++++++++++++++density%2B=1

++++return+density|code-block|syntax|javascript|2562095|计时|2562096|R0+=+500
hz+=+1

x+=+np.arange(-1000,+1000,+1)
y+=+np.arange(-1000,+1000,+1)
z+=+np.arange(-10,+10,+1)

xx,+yy,+zz+=+np.meshgrid(x,+y,+z)

#after+the+first+call+(compilation+overhead)
#calc_density_2++++++++++9.7+ms
#calc_density_2+parallel+3.9+ms
#@Deepak+Saini+++++++++++115+ms|2562097|entityMap^0|0|A|B|0|0|2|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@$D|10|E|11|F|J]]|9|@]|A|$]]|$1|K|3|L|5|M|7|12|8|@]|9|@]|A|$N|O]]|$1|P|3|Q|5|6|7|13|8|@$D|14|E|15|F|J]]|9|@]|A|$]]|$1|R|3|S|5|M|7|16|8|@]|9|@]|A|$N|O]]|$1|T|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|U|$]]

This is meant as a lengthy comment on the answer of Deepak Saini.

The main change is to not use the coordinates generated by <code>np.meshgrid</code> which contains unecessary repetitions. This isn't recommandable if you can avoid it (both in terms of memory usage and performance)

Code

<pre><code>import numba as nb
import numpy as np

@nb.jit(nopython=True,parallel=True)
def calc_density_2(x, y, z,R0,hz):
 threshold = R0 * R0

 density = 0
 for i in nb.prange(y.shape[0]):
 for j in range(x.shape[0]):
 r_xy = x[j] ** 2 + y[i] ** 2
 for k in range(z.shape[0]):
 if(r_xy &lt;= threshold and -hz &lt;= z[k] &lt;= hz):
 density+=1

 return density
</code></pre>

Timings

<pre><code>R0 = 500
hz = 1

x = np.arange(-1000, 1000, 1)
y = np.arange(-1000, 1000, 1)
z = np.arange(-10, 10, 1)

xx, yy, zz = np.meshgrid(x, y, z)

#after the first call (compilation overhead)
#calc_density_2 9.7 ms
#calc_density_2 parallel 3.9 ms
#@Deepak Saini 115 ms
</code></pre>

I tried to model voxels of 3D cylinder with the following code:

<pre><code>import math
import numpy as np

R0 = 500
hz = 1

x = np.arange(-1000, 1000, 1)
y = np.arange(-1000, 1000, 1)
z = np.arange(-10, 10, 1)

xx, yy, zz = np.meshgrid(x, y, z)


def density_f(x, y, z):
 r_xy = math.sqrt(x ** 2 + y ** 2)
 if r_xy &lt;= R0 and -hz &lt;= z &lt;= hz:
 return 1
 else:
 return 0


density = np.vectorize(density_f)(xx, yy, zz)
</code></pre>

and it took many minutes to compute.

Equivalent suboptimal Java code runs 10-15 seconds.

How to make Python compute voxels at the same speed? Where to optimize?

How to fill 3D array in numpy with functional values at the same speed as Java?

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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我试着用下面的代码来模拟3D圆柱体的体素：import mathimport numpy as npR0 = 500hz = 1x = np.arange(-1000, 1000, 1)y = np.arange(-1000, 1000, 1)z = np.arange(-10, 10, 1)xx, yy, zz = ...

问如何在numpy中用函数值以与Java相同的速度填充3D数组？
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