blocks|key|4571977|text|你可以直接使用你的"+index“列表，就像一个索引数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4571978|>>>+arr+=+np.array([10,+20,+30,+40,+50])
>>>+idx+=+[1,+0,+3,+4,+2]
>>>+arr[idx]
array([20,+10,+40,+50,+30])|code-block|syntax|javascript|4571979|如果idx已经是一个ndarray而不是list，那么它往往会快得多，尽管它可以以任何一种方式工作：|offset|length|style|CODE|4571980|>>>+%25timeit+arr[idx]
100000+loops,+best+of+3:+2.11+µs+per+loop
>>>+ai+=+np.array(idx)
>>>+%25timeit+arr[ai]
1000000+loops,+best+of+3:+296+ns+per+loop|4571981|entityMap^0|0|0|2|3|A|7|K|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

You can simply use your "index" list directly, as, well, an index array:

<pre><code>&gt;&gt;&gt; arr = np.array([10, 20, 30, 40, 50])
&gt;&gt;&gt; idx = [1, 0, 3, 4, 2]
&gt;&gt;&gt; arr[idx]
array([20, 10, 40, 50, 30])
</code></pre>

It tends to be much faster if <code>idx</code> is already an <code>ndarray</code> and not a <code>list</code>, even though it'll work either way:

<pre><code>&gt;&gt;&gt; %timeit arr[idx]
100000 loops, best of 3: 2.11 µs per loop
&gt;&gt;&gt; ai = np.array(idx)
&gt;&gt;&gt; %timeit arr[ai]
1000000 loops, best of 3: 296 ns per loop
</code></pre>

blocks|key|4572001|text|对于索引为2d数组的，可以使用map函数。下面是一个示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4572002|a+=+np.random.randn(3,+3)
print(a)
print(np.argsort(a))

print(np.array(list(map(lambda+x,+y:+y[x],+np.argsort(a),+a))))|code-block|syntax|javascript|4572003|输出为|4572004|[[-1.42167035++0.62520498++2.02054623]
+[-0.17966393+-0.01561566++0.24480554]
+[+1.10568543++0.00298402+-0.71397599]]
[[0+1+2]
+[0+1+2]
+[2+1+0]]
[[-1.42167035++0.62520498++2.02054623]
+[-0.17966393+-0.01561566++0.24480554]
+[-0.71397599++0.00298402++1.10568543]]|4572005|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|$]]

for those whose index is 2d array, you can use map function.
Here is an example:

<pre><code>a = np.random.randn(3, 3)
print(a)
print(np.argsort(a))

print(np.array(list(map(lambda x, y: y[x], np.argsort(a), a))))
</code></pre>

the output is 

<pre><code>[[-1.42167035 0.62520498 2.02054623]
 [-0.17966393 -0.01561566 0.24480554]
 [ 1.10568543 0.00298402 -0.71397599]]
[[0 1 2]
 [0 1 2]
 [2 1 0]]
[[-1.42167035 0.62520498 2.02054623]
 [-0.17966393 -0.01561566 0.24480554]
 [-0.71397599 0.00298402 1.10568543]]
</code></pre>

blocks|key|65654|text|如果你想对它进行降序排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|65655|a+=+np.array([1,2,3,4,5])
np.argsort(a)
>+array([0,+1,+2,+3,+4])
np.argsort(-a)
>+array([4,+3,+2,+1,+0])|code-block|syntax|javascript|65656|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to sort it but descending:

<pre><code>a = np.array([1,2,3,4,5])
np.argsort(a)
&gt; array([0, 1, 2, 3, 4])
np.argsort(-a)
&gt; array([4, 3, 2, 1, 0])
</code></pre>

blocks|key|4572054|text|对于那些有同样困惑的人，我实际上是在寻找一个略有不同的“基于索引重新排列数组”的版本。在我的情况下，索引数组索引的是目标数组，而不是源数组。换句话说，我正在尝试根据数组在新数组中位置重新排列数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4572055|在这种情况下，只需在索引之前应用argsort即可。例如。|offset|length|style|CODE|4572056|>>>+arr+=+np.array([10,+20,+30,+40,+50])
>>>+idx+=+[1,+0,+3,+4,+2]
>>>+arr[np.argsort(idx)]
array([20,+10,+50,+30,+40])|code-block|syntax|javascript|4572057|请注意此结果与op期望的结果之间的差异。|4572058|可以来回验证|4572059|>>>+arr[np.argsort(idx)][idx]+==+arr
array([+True,++True,++True,++True,++True])
>>>+arr[idx][np.argsort(idx)]+==+arr
array([+True,++True,++True,++True,++True])|4572060|entityMap^0|0|G|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|J|7|11|8|@]|9|@]|A|$K|L]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

For those who have the same confusion, I am actually looking for a slightly different version of &quot;rearrange array based upon index&quot;. In my situation, the index array is indexing the target array instead of the source array. In other words, I am try to rearrange an array based on its position in the new array.
In this case, simply apply an <code>argsort</code> before indexing. E.g.
<pre><code>&gt;&gt;&gt; arr = np.array([10, 20, 30, 40, 50])
&gt;&gt;&gt; idx = [1, 0, 3, 4, 2]
&gt;&gt;&gt; arr[np.argsort(idx)]
array([20, 10, 50, 30, 40])
</code></pre>
Note the difference between this result and the desired result by op.
One can verify back and forth
<pre><code>&gt;&gt;&gt; arr[np.argsort(idx)][idx] == arr
array([ True, True, True, True, True])
&gt;&gt;&gt; arr[idx][np.argsort(idx)] == arr
array([ True, True, True, True, True])
</code></pre>

I'm looking for a one line solution that would help me do the following. 

Suppose I have

<pre><code>array = np.array([10, 20, 30, 40, 50])
</code></pre>

I'd like to rearrange it based upon an input ordering. If there were a numpy function called <code>arrange</code>, it would do the following:

<pre><code>newarray = np.arrange(array, [1, 0, 3, 4, 2])
print newarray

 [20, 10, 40, 50, 30]
</code></pre>

Formally, if the array to be reordered is m x n, and the "index" array is 1 x n, the ordering would be determined by the array called "index".

Does numpy have a function like this?

How to rearrange array based upon index array

我正在寻找一个单行解决方案，可以帮助我完成以下工作。假设我有array = np.array([10, 20, 30, 40, 50])我想根据输入顺序重新排列它。如果有一个名为arrange的numpy函数，它将执行以下操作：newarray = np.arrange(array, [1, 0, 3, 4, 2])p...

问如何根据索引数组重新排列数组
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