blocks|key|643958|text|如果您使用numpy，则可以使用argsort()函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|643959|>>>+import+numpy
>>>+numpy.argsort(myList)
array([0,+1,+2,+4,+3])|code-block|syntax|javascript|643960|http://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html|offset|length|643961|这将返回对数组或列表进行排序的参数。|643962|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|1Y|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@]|9|@$I|W|J|X|1|Y]]|A|$]]|$1|K|3|L|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|H]]]]

If you are using numpy, you have the argsort() function available:

<pre><code>&gt;&gt;&gt; import numpy
&gt;&gt;&gt; numpy.argsort(myList)
array([0, 1, 2, 4, 3])
</code></pre>

<a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html" rel="noreferrer">http://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html</a>

This returns the arguments that would sort the array or list.

blocks|key|643920|text|myList+=+[1,+2,+3,+100,+5]++++
sorted(range(len(myList)),key=myList.__getitem__)

[0,+1,+2,+4,+3]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|643921|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>myList = [1, 2, 3, 100, 5] 
sorted(range(len(myList)),key=myList.__getitem__)

[0, 1, 2, 4, 3]
</code></pre>

blocks|key|861398|text|我用perfplot+(我的一个项目)对它们做了一个快速的性能检查，发现很难推荐其他东西，除了|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|861399|np.argsort(x)|code-block|syntax|javascript|861400|(请注意日志刻度)：|861401|​|861402|📷|atomic|861403|861404|861405|重现该图的代码：|861406|import+perfplot
import+numpy+as+np


def+sorted_enumerate(seq):
++++return+[i+for+(v,+i)+in+sorted((v,+i)+for+(i,+v)+in+enumerate(seq))]


def+sorted_enumerate_key(seq):
++++return+[x+for+x,+y+in+sorted(enumerate(seq),+key=lambda+x:+x[1])]


def+sorted_range(seq):
++++return+sorted(range(len(seq)),+key=seq.__getitem__)


b+=+perfplot.bench(
++++setup=np.random.rand,
++++kernels=[sorted_enumerate,+sorted_enumerate_key,+sorted_range,+np.argsort],
++++n_range=[2+**+k+for+k+in+range(15)],
++++xlabel="len(x)",
)
b.save("out.png")|861407|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/nschloe/perfplot|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/0eb09216c47f93827c39de8259599283.png|imageAlt^0|2|8|0|0|0|0|0|0|1|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@$A|1A|B|1B|1|1C]]|C|$]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1E|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|1F|8|@]|9|@]|C|$]]|$1|M|3|N|5|O|7|1G|8|@]|9|@$A|1H|B|1I|1|1J]]|C|$]]|$1|P|3|L|5|6|7|1K|8|@]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|1L|8|@]|9|@]|C|$]]|$1|R|3|S|5|6|7|1M|8|@]|9|@]|C|$]]|$1|T|3|U|5|F|7|1N|8|@]|9|@]|C|$G|H]]|$1|V|3|-4|5|6|7|1O|8|@]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|14|Z|15|C|$16|17|18|-4]]]]

I did a quick performance check on these with <a href="https://github.com/nschloe/perfplot" rel="nofollow noreferrer">perfplot</a> (a project of mine) and found that it's hard to recommend anything else but
<pre class="lang-py prettyprint-override"><code>np.argsort(x)
</code></pre>
(note the log scale):
<a href="https://i.stack.imgur.com/HuApW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HuApW.png" alt="enter image description here" /></a>
<hr />
Code to reproduce the plot:
<pre class="lang-py prettyprint-override"><code>import perfplot
import numpy as np


def sorted_enumerate(seq):
 return [i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]


def sorted_enumerate_key(seq):
 return [x for x, y in sorted(enumerate(seq), key=lambda x: x[1])]


def sorted_range(seq):
 return sorted(range(len(seq)), key=seq.__getitem__)


b = perfplot.bench(
 setup=np.random.rand,
 kernels=[sorted_enumerate, sorted_enumerate_key, sorted_range, np.argsort],
 n_range=[2 ** k for k in range(15)],
 xlabel=&quot;len(x)&quot;,
)
b.save(&quot;out.png&quot;)
</code></pre>

blocks|key|643998|text|用enumerate得到的答案很好，但我个人不喜欢用来按值排序的lambda。下面的代码只是颠倒索引和值，并对其进行排序。因此，它将首先按值排序，然后按索引排序。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|643999|sorted((e,i)+for+i,e+in+enumerate(myList))|code-block|syntax|javascript|644000|entityMap^0|1|9|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|$]]

The answers with <code>enumerate</code> are nice, but I personally don't like the lambda used to sort by the value. The following just reverses the index and the value, and sorts that. So it'll first sort by value, then by index.

<pre><code>sorted((e,i) for i,e in enumerate(myList))
</code></pre>

blocks|key|643905|text|使用enumerate和itemgetter更新答案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|643906|sorted(enumerate(a),+key=lambda+x:+x[1])
#+[(0,+1),+(1,+2),+(2,+3),+(4,+5),+(3,+100)]|code-block|syntax|javascript|643907|将列表压缩在一起:元组中的第一个元素是索引，第二个是值(然后使用元组x[1]的第二个值对其进行排序，x是元组)|643908|或者从operator模块`使用itemgetter：|643909|from+operator+import+itemgetter
sorted(enumerate(a),+key=itemgetter(1))|643910|entityMap^0|2|9|C|A|0|0|Y|4|0|3|8|G|A|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

Updated answer with <code>enumerate</code> and <code>itemgetter</code>:

<pre><code>sorted(enumerate(a), key=lambda x: x[1])
# [(0, 1), (1, 2), (2, 3), (4, 5), (3, 100)]
</code></pre>

Zip the lists together: The first element in the tuple will the index, the second is the value (then sort it using the second value of the tuple <code>x[1]</code>, x is the tuple)

Or using <code>itemgetter</code> from the <code>operator</code>module`:

<pre><code>from operator import itemgetter
sorted(enumerate(a), key=itemgetter(1))
</code></pre>

blocks|key|644038|text|如果您不想使用numpy，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|644039|sorted(range(len(seq)),+key=seq.__getitem__)|code-block|syntax|javascript|644040|是最快的，如here所示。|offset|length|644041|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/6979121/3703716^0|0|0|6|4|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

If you do not want to use numpy, 

<pre><code>sorted(range(len(seq)), key=seq.__getitem__)
</code></pre>

is fastest, as demonstrated <a href="https://stackoverflow.com/a/6979121/3703716">here</a>.

blocks|key|644209|text|我们将创建另一个从0到n-1的索引数组，然后将其压缩到原始数组中，然后根据原始值对其进行排序|type|unstyled|depth|inlineStyleRanges|entityRanges|data|644210|ar+=+[1,2,3,4,5]
new_ar+=+list(zip(ar,[i+for+i+in+range(len(ar))]))
new_ar.sort()|code-block|syntax|javascript|644211|`|644212|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|$]]

We will create another array of indexes from 0 to n-1
Then zip this to the original array and then sort it on the basis of the original values

<pre><code>ar = [1,2,3,4,5]
new_ar = list(zip(ar,[i for i in range(len(ar))]))
new_ar.sort()
</code></pre>

`

blocks|key|861651|text|s+=+[2,+3,+1,+4,+5]
print([sorted(s,+reverse=False).index(val)+for+val+in+s])+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|861652|它甚至适用于包含重复元素的列表。|unstyled|861653|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>s = [2, 3, 1, 4, 5]
print([sorted(s, reverse=False).index(val) for val in s]) 
</code></pre>
It works even for a list with duplicate elements.

blocks|key|644283|text|首先将你的列表转换为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|644284|myList+=+[1,+2,+3,+100,+5]|code-block|syntax|javascript|644285|将索引添加到列表的项目|644286|myList+=+[[0,+1],+[1,+2],+[2,+3],+[3,+100],+[4,+5]]|644287|接下来：|644288|sorted(myList,+key=lambda+k:k[1])|644289|结果：|644290|[[0,+1],+[1,+2],+[2,+3],+[4,+5],+[3,+100]]|644291|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

firstly convert your list to this:
<pre><code>myList = [1, 2, 3, 100, 5]
</code></pre>
add a index to your list's item
<pre><code>myList = [[0, 1], [1, 2], [2, 3], [3, 100], [4, 5]]
</code></pre>
next :
<pre><code>sorted(myList, key=lambda k:k[1])
</code></pre>
result:
<pre><code>[[0, 1], [1, 2], [2, 3], [4, 5], [3, 100]]
</code></pre>

I have a numerical list:

<pre><code>myList = [1, 2, 3, 100, 5]
</code></pre>

Now if I sort this list to obtain <code>[1, 2, 3, 5, 100]</code>. 
What I want is the indices of the elements from the 
original list in the sorted order i.e. <code>[0, 1, 2, 4, 3]</code> 
--- ala MATLAB's sort function that returns both 
values and indices.

How to get indices of a sorted array in Python

我有一个数字列表：myList = [1, 2, 3, 100, 5]现在，如果我对这个列表进行排序以获得[1, 2, 3, 5, 100]。我想要的是原始列表中元素按排序顺序的索引，即[0, 1, 2, 4, 3] -这是MATLAB的排序函数，它同时返回值和索引。

问如何在Python中获取排序数组的索引
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