blocks|key|1831526|text|我不知道为什么IndexedBase方法不起作用(我也想知道)。但是，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1831527|import+sympy+as+sp

numSpecies+=+10
n+=+sp.symbols('n0:%25d'%25numSpecies)+++#+note+that+n+equals+the+tuple+(n0,+n1,+...,+n9)

ntot+=+sum(n)+++++++#+sum+elements+of+n+using+the+standard
++++++++++++++++++++#+Python+function+for+summing+tuple+elements
#ntot+=+sp.Add(*n)++#+same+result+using+Sympy+function

sp.diff(ntot,+n[5])|code-block|syntax|javascript|1831528|entityMap^0|7|B|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|$]]

I don't know why the <code>IndexedBase</code> approach does not work (I would be interested to know also). You can, however, do the following:

<pre><code>import sympy as sp

numSpecies = 10
n = sp.symbols('n0:%d'%numSpecies) # note that n equals the tuple (n0, n1, ..., n9)

ntot = sum(n) # sum elements of n using the standard
 # Python function for summing tuple elements
#ntot = sp.Add(*n) # same result using Sympy function

sp.diff(ntot, n[5])
</code></pre>

blocks|key|52277|text|我不清楚你想做什么。然而，这也许会有所帮助。根据收到的两条评论进行了编辑。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|52278|from+sympy+import+*

nspecies+=+10
[var('n%25s'%25_)+for+_+in+range(nspecies)]

expr+=+sympify('%2B'.join(['n%25s'%25_+for+_+in+range(nspecies)]))
expr
print+(+diff(expr,n1)+)

expr+=+sympify('n0**n1%2Bn1**n2')
expr
print+(+diff(expr,n1)+)|code-block|syntax|javascript|52279|只有第一个表达式对原始问题作出响应。这是输出。|52280|1
n0**n1*log(n0)+%2B+n1**n2*n2/n1|52281|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|$]]

I'm not clear about what you want to do. However, perhaps this will help. Edited in response to two comments received.

<pre><code>from sympy import *

nspecies = 10
[var('n%s'%_) for _ in range(nspecies)]

expr = sympify('+'.join(['n%s'%_ for _ in range(nspecies)]))
expr
print ( diff(expr,n1) )

expr = sympify('n0**n1+n1**n2')
expr
print ( diff(expr,n1) )
</code></pre>

Only the first expression responds to the original question. This is the output.

<pre><code>1
n0**n1*log(n0) + n1**n2*n2/n1
</code></pre>

blocks|key|2221118|text|使用SymPy的开发版本，您的示例可以正常工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2221119|要安装SymPy的开发版本，只需使用git将其下载|offset|length|style|CODE|2221120|git+clone+git://github.com/sympy/sympy.git
cd+sympy|code-block|syntax|javascript|2221121|然后从该路径运行python，或者在Python的默认安装之前将PYTHONPATH设置为包含该目录。|2221122|您在开发版本上的示例：|2221123|In+[3]:+numSpecies+=+10

In+[4]:+n+=+IndexedBase('n')

In+[5]:+i+=+symbols("i",cls=Idx)

In+[6]:+nges+=+summation(n[i],[i,1,numSpecies])

In+[7]:+nges
Out[7]:+n[10]+%2B+n[1]+%2B+n[2]+%2B+n[3]+%2B+n[4]+%2B+n[5]+%2B+n[6]+%2B+n[7]+%2B+n[8]+%2B+n[9]

In+[8]:+diff(nges,n[5])
Out[8]:+1|2221124|您还可以使用求和的简写形式：|2221125|In+[9]:+nges_uneval+=+Sum(n[i],+[i,1,numSpecies])

In+[10]:+nges_uneval
Out[10]:+
++10++++++
+___++++++
+╲++++++++
++╲+++n[i]
++╱+++++++
+╱++++++++
+‾‾‾++++++
i+=+1+++++

In+[11]:+diff(nges_uneval,+n[5])
Out[11]:+
++10++++++
+___++++++
+╲++++++++
++╲+++δ+++
++╱++++5,i
+╱++++++++
+‾‾‾++++++
i+=+1+++++

In+[12]:+diff(nges_uneval,+n[5]).doit()
Out[12]:+1|2221126|还要注意，在下一个SymPy版本中，您将能够使用符号索引派生符号：|2221127|In+[13]:+j+=+symbols("j")

In+[13]:+diff(n[i],+n[j])
Out[13]:+
δ+++
+j,i|2221128|你从哪里得到Kronecker+delta的。|2221129|如果您不想安装SymPy开发版本，请等待下一个完整版本(可能在今年秋天发布)，它将支持IndexedBase的派生版本。|2221130|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Kronecker_delta^0|0|I|3|0|0|W|A|0|0|0|0|0|0|0|6|F|0|0|17|B|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1D|8|@$D|1E|E|1F|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1H|8|@$D|1I|E|1J|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1K|8|@]|9|@]|A|$]]|$1|Q|3|R|5|J|7|1L|8|@]|9|@]|A|$K|L]]|$1|S|3|T|5|6|7|1M|8|@]|9|@]|A|$]]|$1|U|3|V|5|J|7|1N|8|@]|9|@]|A|$K|L]]|$1|W|3|X|5|6|7|1O|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|J|7|1P|8|@]|9|@]|A|$K|L]]|$1|10|3|11|5|6|7|1Q|8|@]|9|@$D|1R|E|1S|1|1T]]|A|$]]|$1|12|3|13|5|6|7|1U|8|@$D|1V|E|1W|F|G]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]]|15|$16|$5|17|18|19|A|$1A|1B]]]]

With SymPy's development version, your example works.

To install SymPy's development version, just pull it down with <code>git</code>:

<pre><code>git clone git://github.com/sympy/sympy.git
cd sympy
</code></pre>

Then run python from that path or set the <code>PYTHONPATH</code> to include that directory before Python's default installation.

Your example on the development version:

<pre><code>In [3]: numSpecies = 10

In [4]: n = IndexedBase('n')

In [5]: i = symbols("i",cls=Idx)

In [6]: nges = summation(n[i],[i,1,numSpecies])

In [7]: nges
Out[7]: n[10] + n[1] + n[2] + n[3] + n[4] + n[5] + n[6] + n[7] + n[8] + n[9]

In [8]: diff(nges,n[5])
Out[8]: 1
</code></pre>

You could also use the contracted form of summation:

<pre><code>In [9]: nges_uneval = Sum(n[i], [i,1,numSpecies])

In [10]: nges_uneval
Out[10]: 
 10 
 ___ 
 ╲ 
 ╲ n[i]
 ╱ 
 ╱ 
 ‾‾‾ 
i = 1 

In [11]: diff(nges_uneval, n[5])
Out[11]: 
 10 
 ___ 
 ╲ 
 ╲ δ 
 ╱ 5,i
 ╱ 
 ‾‾‾ 
i = 1 

In [12]: diff(nges_uneval, n[5]).doit()
Out[12]: 1
</code></pre>

Also notice that in the next SymPy version you will be able to derive symbols with symbolic indices:

<pre><code>In [13]: j = symbols("j")

In [13]: diff(n[i], n[j])
Out[13]: 
δ 
 j,i
</code></pre>

Where you get the <a href="https://en.wikipedia.org/wiki/Kronecker_delta" rel="nofollow">Kronecker delta</a>.

If you don't feel like installing the SymPy development version, just wait for the next full version (probably coming out this autumn), it will support derivatives of <code>IndexedBase</code>.

I am using sympy from time to time, but am not very good at it. At the moment I am stuck with defining a list of indexed variables, i.e. n1 to nmax and performing a summation on it. Then I want to be able to take the derivative:

So far I tried the following:

<pre><code>numSpecies = 10
n = IndexedBase('n')
i = symbols("i",cls=Idx)
nges = summation(n[i],[i,1,numSpecies])
</code></pre>

However, if i try to take the derivative with respect to one variable, this fails:

<pre><code>diff(nges,n[5])
</code></pre>

I also tried to avoid working with <a href="http://docs.sympy.org/latest/modules/tensor/indexed.html" rel="nofollow"><code>IndexedBase</code></a>.

<pre><code>numSpecies = 10
n = symbols('n0:%d'%numSpecies)
k = symbols('k',integer=True)
ntot = summation(n[k],[k,0,numSpecies])
</code></pre>

However, here already the summation fails because mixing python tuples and the sympy summation.

How I can perform indexedbase derivatives or some kind of workaround?

Derivative of summations

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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我会时不时地使用渐变，但我不太擅长它。目前，我一直在定义一个索引变量列表，即从n1到nmax，然后对其进行求和。然后我想可以求导数：到目前为止，我尝试了以下几种方法：numSpecies = 10n = IndexedBase('n')i = symbols("i",cls=Idx)nges = summation(n...

问求和的导数
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