blocks|key|179572|text|请参阅numpy.argmax文档(numpy.argmin文档中引用了该文档)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|179573|179574|如果最大值多次出现，则返回第一次出现的索引。|blockquote|179575|179576|179577|文档的措辞(“索引”而不是“索引”)指的是提供axis时的多维情况。|179578|所以，你不能用np.argmin做到这一点。取而代之的是，这将起作用：|179579|np.where(arr+==+arr.min())|code-block|syntax|javascript|179580|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.argmax.html#numpy.argmax^0|3|C|I|C|3|C|0|0|0|0|0|0|N|4|0|7|9|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|1B|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|1H|8|@$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1K|8|@]|D|@]|E|$S|T]]|$1|U|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

See the documentation for <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.argmax.html#numpy.argmax" rel="noreferrer"><code>numpy.argmax</code></a> (which is referred to by the docs for <code>numpy.argmin</code>):

<blockquote>
 In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned.
</blockquote>

The phrasing of the documentation ("indices" instead of "index") refers to the multidimensional case when <code>axis</code> is provided.

So, you can't do it with <code>np.argmin</code>. Instead, this will work:

<pre><code>np.where(arr == arr.min())
</code></pre>

blocks|key|2997906|text|当您考虑多维数组时，该文档更有意义。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2997907|>>>+x+=+numpy.array([[0,+1],
...++++++++++++++++++[3,+2]])
>>>+x.argmin(axis=0)
array([0,+0])
>>>+x.argmin(axis=1)
array([0,+1])|code-block|syntax|javascript|2997908|如果指定了轴，argmin将获取沿给定轴的一维子数组，并返回每个子数组的最小值的第一个索引。它不会返回单个最小值的所有索引。|offset|length|style|CODE|2997909|要获取最小值的所有索引，您可以这样做|2997910|numpy.where(x+==+x.min())|2997911|entityMap^0|0|0|7|6|0|0|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]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

That documentation makes more sense when you think about multidimensional arrays.

<pre><code>&gt;&gt;&gt; x = numpy.array([[0, 1],
... [3, 2]])
&gt;&gt;&gt; x.argmin(axis=0)
array([0, 0])
&gt;&gt;&gt; x.argmin(axis=1)
array([0, 1])
</code></pre>

With an axis specified, <code>argmin</code> takes one-dimensional subarrays along the given axis and returns the first index of each subarray's minimum value. It doesn't return all indices of a single minimum value.

To get all indices of the minimum value, you could do

<pre><code>numpy.where(x == x.min())
</code></pre>

blocks|key|177739|text|假设您需要一个列表的索引，而不是一个numpy数组，请尝试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|177740|my_list+=+[5,+3,+2,+1,+1,+1,+6,+1]
np.where(my_list+==+min(my_list))[0]|code-block|syntax|javascript|177741|索引是因为numpy返回的是答案的元组，而不返回任何内容(答案是numpy数组)。别问我为什么。|177742|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|$]]

Assuming that you want the indices of a list, not a numpy array, try 

<pre><code>my_list = [5, 3, 2, 1, 1, 1, 6, 1]
np.where(my_list == min(my_list))[0]
</code></pre>

The index [0] is because numpy returns a tuple of your answer and nothing (answer as a numpy array). Don't ask me why.

blocks|key|2785512|text|我想快速补充一下，就像用户grofte提到的，np.where返回一个元组，它声明它是nonzero的简写，它有一个对应的方法flatnonzero，它直接返回一个数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2785513|所以，最干净的版本似乎是|2785514|my_list+=+np.array([5,+3,+2,+1,+1,+1,+6,+1])
np.flatnonzero(my_list+==+my_list.min())
=>+array([3,+4,+5,+7])|code-block|syntax|javascript|2785515|entityMap^0|N|8|17|7|1R|B|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

I would like to quickly add that as user grofte mentioned, <code>np.where</code> returns a tuple and it states that it is a shorthand for <code>nonzero</code> which has a corresponding method <code>flatnonzero</code> which returns an array directly. 

So, the cleanest version seems to be

<pre><code>my_list = np.array([5, 3, 2, 1, 1, 1, 6, 1])
np.flatnonzero(my_list == my_list.min())
=&gt; array([3, 4, 5, 7])
</code></pre>

I am a little bit confused reading the documentation of <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.argmin.html" rel="noreferrer">argmin function in numpy</a>.
It looks like it should do the job: 

Reading this

<blockquote>
 Return the indices of the minimum values along an axis.
</blockquote>

I might assume that 

<pre><code>np.argmin([5, 3, 2, 1, 1, 1, 6, 1])
</code></pre>

will return an array of all indices: which will be <code>[3, 4, 5, 7]</code>

But instead of this it returns only <code>3</code>. Where is the catch, or what should I do to get my result?

How to return all the minimum indices in numpy

在阅读的文档时，我有点困惑。看起来它应该可以完成这项工作：阅读这篇文章返回轴上的最小值的索引。我可能会假设np.argmin([5, 3, 2, 1, 1, 1, 6, 1])将返回一个包含所有索引的数组:它将为[3, 4, 5, 7]但是它只返回3，而不是这个。陷阱在哪里，或者我应该怎么做才能得到我的结果？

问如何返回numpy中的所有最小索引
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何返回numpy中的所有最小索引EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何返回numpy中的所有最小索引
EN