blocks|key|2737746|text|使用以下任一选项|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2737747|numpy.sqrt(numpy.sum((A+-+B)**2))|code-block|syntax|javascript|2737748|或者更简单地说|2737749|numpy.linalg.norm(A+-+B)|2737750|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|$]]

<p>Use either</p>

<pre><code>numpy.sqrt(numpy.sum((A - B)**2))
</code></pre>

<p>or more simply</p>

<pre><code>numpy.linalg.norm(A - B)
</code></pre>


blocks|key|49048|text|A和B是24维空间中的2个点。您应该使用scipy.spatial.distance.euclidean。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|49049|Doc+here|49050|scipy.spatial.distance.euclidean(A,+B)|code-block|syntax|javascript|49051|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.euclidean.html^0|0|1|2|1|K|W|0|0|8|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

<p><code>A</code> and <code>B</code> are 2 points in the 24-D space. You should use <code>scipy.spatial.distance.euclidean</code>. </p>

<p><a href="http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.euclidean.html">Doc here</a></p>

<pre><code>scipy.spatial.distance.euclidean(A, B)
</code></pre>


blocks|key|2528203|text|启动Python+3.8后，您可以使用标准库的math模块及其新的dist函数，该函数返回两点之间的欧几里德距离(以坐标列表或坐标元组的形式给出)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2528204|from+math+import+dist

dist([1,+0,+0],+[0,+1,+0])+#+1.4142135623730951|code-block|syntax|javascript|2528205|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/dev/library/math.html|1|https://docs.python.org/dev/library/math.html#math.dist^0|2|A|N|4|X|4|N|4|0|X|4|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

<p>Starting <code>Python 3.8</code>, you can use standard library's <a href="https://docs.python.org/dev/library/math.html" rel="noreferrer"><code>math</code></a> module and its new <a href="https://docs.python.org/dev/library/math.html#math.dist" rel="noreferrer"><code>dist</code></a> function, which returns the euclidean distance between two points (given as lists or tuples of coordinates):</p>

<pre><code>from math import dist

dist([1, 0, 0], [0, 1, 0]) # 1.4142135623730951
</code></pre>


blocks|key|48241|text|除了已经提到的计算欧几里得距离的方法之外，这里有一种与您的原始代码接近的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|48242|scipy.spatial.distance.cdist([A],+[B],+'euclidean')|code-block|syntax|javascript|48243|或|48244|scipy.spatial.distance.cdist(np.atleast_2d(A),+np.atleast_2d(B),+'euclidean')|48245|这将返回一个保存L2距离的1×1+np.ndarray。|offset|length|style|CODE|48246|entityMap^0|0|0|0|0|H|A|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

<p>Apart from the already mentioned ways of computing the Euclidean distance, here's one that's close to your original code:</p>

<pre><code>scipy.spatial.distance.cdist([A], [B], 'euclidean')
</code></pre>

<p>or</p>

<pre><code>scipy.spatial.distance.cdist(np.atleast_2d(A), np.atleast_2d(B), 'euclidean')
</code></pre>

<p>This returns a 1×1 <code>np.ndarray</code> holding the L2 distance.</p>


blocks|key|2528183|text|由于上面的所有答案都涉及numpy和/或scipy，所以我只想指出一些非常简单的事情可以在这里使用reduce来完成。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2528184|def+n_dimensional_euclidean_distance(a,+b):
+++"""
+++Returns+the+euclidean+distance+for+n>=2+dimensions
+++:param+a:+tuple+with+integers
+++:param+b:+tuple+with+integers
+++:return:+the+euclidean+distance+as+an+integer
+++"""
+++dimension+=+len(a)+#+notice,+this+will+definitely+throw+a+IndexError+if+len(a)+!=+len(b)

+++return+sqrt(reduce(lambda+i,j:+i+%2B+((a[j]+-+b[j])+**+2),+range(dimension),+0))|code-block|syntax|javascript|2528185|这将对维数中所有j的所有(aj+-+bj)%5E2对求和(请注意，为简单起见，这不支持n<2维度距离)。|2528186|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<p>Since all of the above answers refer to numpy and or scipy, just wanted to point out that something really simple can be done with reduce here</p>

<pre><code>def n_dimensional_euclidean_distance(a, b):
   """
   Returns the euclidean distance for n&gt;=2 dimensions
   :param a: tuple with integers
   :param b: tuple with integers
   :return: the euclidean distance as an integer
   """
   dimension = len(a) # notice, this will definitely throw a IndexError if len(a) != len(b)

   return sqrt(reduce(lambda i,j: i + ((a[j] - b[j]) ** 2), range(dimension), 0))
</code></pre>

<p>This will sum all pairs of (a[j] - b[j])^2 for all j in the number of dimensions (note that for simplicity this doesn't support n&lt;2 dimensional distance).</p>


blocks|key|2737886|text|编写自己的自定义sqaure+root+sum+square并不总是安全的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2737887|您可以使用math.hypot、numpy.hypot或scipy+distance函数，而不是编写numpy.sqrt(numpy.sum((A+-+B)**2))或(i**2+%2B+j**2)**0.5。在您的情况下，它们可能会溢出|offset|length|style|CODE|2737888|refer|2737889|速度智慧|2737890|%25%25timeit
math.hypot(*(A+-+B))
#+3+µs+±+64.8+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+100000+loops+each)|code-block|syntax|javascript|2737891|%25%25timeit
numpy.sqrt(numpy.sum((A+-+B)**2))
#+5.65+µs+±+50.7+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+100000+loops+each)|2737892|安全智慧|2737893|下溢|2737894|i,+j+=+1e-200,+1e-200
np.sqrt(i**2%2Bj**2)
#+0.0|2737895|溢出|2737896|i,+j+=+1e%2B200,+1e%2B200
np.sqrt(i**2%2Bj**2)
#+inf|2737897|无下溢|2737898|i,+j+=+1e-200,+1e-200
np.hypot(i,+j)
#+1.414213562373095e-200|2737899|无溢出|2737900|i,+j+=+1e%2B200,+1e%2B200
np.hypot(i,+j)
#+1.414213562373095e%2B200|2737901|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/69233528/7035448^0|0|1E|X|2C|I|0|0|5|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1J|8|@$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1O|8|@]|9|@$D|1P|E|1Q|1|1R]]|A|$]]|$1|J|3|K|5|6|7|1S|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1T|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|N|7|1U|8|@]|9|@]|A|$O|P]]|$1|S|3|T|5|6|7|1V|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1W|8|@]|9|@]|A|$]]|$1|W|3|X|5|N|7|1X|8|@]|9|@]|A|$O|P]]|$1|Y|3|Z|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|10|3|11|5|N|7|1Z|8|@]|9|@]|A|$O|P]]|$1|12|3|13|5|6|7|20|8|@]|9|@]|A|$]]|$1|14|3|15|5|N|7|21|8|@]|9|@]|A|$O|P]]|$1|16|3|17|5|6|7|22|8|@]|9|@]|A|$]]|$1|18|3|19|5|N|7|23|8|@]|9|@]|A|$O|P]]|$1|1A|3|-4|5|6|7|24|8|@]|9|@]|A|$]]]|1B|$1C|$5|1D|1E|1F|A|$1G|1H]]]]

<p>Writing your own custom sqaure root sum square is not always safe</p>
<p>You can use math.hypot, numpy.hypot or scipy distance function rather than writing <code>numpy.sqrt(numpy.sum((A - B)**2))</code> or <code>(i**2 + j**2)**0.5</code>. In your case maybe they can overflow</p>
<p><a href="https://stackoverflow.com/a/69233528/7035448">refer</a></p>
<h2>Speed wise</h2>
<pre class="lang-py prettyprint-override"><code>%%timeit
math.hypot(*(A - B))
# 3 µs ± 64.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
</code></pre>
<pre class="lang-py prettyprint-override"><code>%%timeit
numpy.sqrt(numpy.sum((A - B)**2))
# 5.65 µs ± 50.7 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
</code></pre>
<h1>Safety wise</h1>
<h2>Underflow</h2>
<pre class="lang-py prettyprint-override"><code>i, j = 1e-200, 1e-200
np.sqrt(i**2+j**2)
# 0.0
</code></pre>
<h2>Overflow</h2>
<pre class="lang-py prettyprint-override"><code>i, j = 1e+200, 1e+200
np.sqrt(i**2+j**2)
# inf
</code></pre>
<h2>No Underflow</h2>
<pre class="lang-py prettyprint-override"><code>i, j = 1e-200, 1e-200
np.hypot(i, j)
# 1.414213562373095e-200
</code></pre>
<h2>No Overflow</h2>
<pre class="lang-py prettyprint-override"><code>i, j = 1e+200, 1e+200
np.hypot(i, j)
# 1.414213562373095e+200
</code></pre>


<p>I want to calculate the Euclidean distance in multiple dimensions (24 dimensions) between 2 arrays. I'm using numpy-Scipy. </p>

<p>Here is my code:</p>

<pre><code>import numpy,scipy;

A=numpy.array([116.629, 7192.6, 4535.66, 279714, 176404, 443608, 295522, 1.18399e+07, 7.74233e+06, 2.85839e+08, 2.30168e+08, 5.6919e+08, 168989, 7.48866e+06, 1.45261e+06, 7.49496e+07, 2.13295e+07, 3.74361e+08, 54.5, 3349.39, 262.614, 16175.8, 3693.79, 205865]);

B=numpy.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 151246, 6795630, 4566625, 2.0355328e+08, 1.4250515e+08, 3.2699482e+08, 95635, 4470961, 589043, 29729866, 6124073, 222.3]);
</code></pre>

<p>However, I used <code>scipy.spatial.distance.cdist(A[numpy.newaxis,:],B,'euclidean')</code> to calcuate the eucleidan distance.</p>

<p><strong>But it gave me an error</strong></p>

<pre><code>raise ValueError('XB must be a 2-dimensional array.');
</code></pre>

<p>I don't seem to understand it. </p>

<p>I looked up <code>scipy.spatial.distance.pdist</code> but don't understand how to use it? </p>

<p><strong>Is there any other better way to do it?</strong></p>


Multidimensional Euclidean Distance in Python

Python

我想计算两个数组在多维(24维)上的欧几里得距离。我用的是numpy-Scipy。下面是我的代码：import numpy,scipy;A=numpy.array([116.629, 7192.6, 4535.66, 279714, 176404, 443608, 295522, 1.18399e+07, 7.74233e+06, 2.85839e+08, 2.30168e+08, 5.6919e

问Python中的多维欧几里德距离
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回答 6

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问Python中的多维欧几里德距离EN

回答 6

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问Python中的多维欧几里德距离
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