blocks|key|3022213|text|In+[1]:+import+numpy+as+np
In+[2]:+a+=+np.array([[2,0],[3,0],[3,1],[5,0],[5,1],[5,2]])
In+[3]:+b+=+np.zeros((6,3),+dtype='int32')

In+[4]:+b[a[:,0],+a[:,1]]+=+10

In+[5]:+b
Out[5]:+
array([[+0,++0,++0],
+++++++[+0,++0,++0],
+++++++[10,++0,++0],
+++++++[10,+10,++0],
+++++++[+0,++0,++0],
+++++++[10,+10,+10]])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3022214|unstyled|3022215|工作原理：|offset|length|style|BOLD|3022216|如果在赋值中使用两个numpy数组对b进行索引，|CODE|3022217|b[x,+y]+=+z|3022218|然后将NumPy看作是同时遍历x的每个元素、y的每个元素和z的每个元素(我们称它们为xval、yval和zval)，并将值zval赋给bxval，yval。当z是一个常量时，“在z上移动每次只返回相同的值。|3022219|这就是我们想要的，x是a的第一列，y是a的第二列。因此，选择x+=+a[:,+0]和y+=+a[:,+1]。|3022220|b[a[:,0],+a[:,1]]+=+10|3022221|3022222|为什么+b[a]+=+10+不起作用|3022223|当您编写b[a]时，可以将NumPy看作是通过移动a的每个元素(让我们称每个元素为idx)并将b[idx]的值放入新数组中a中idx的位置来创建一个新数组。|3022224|idx是a中的一个值。所以它是一个int32。b的形状为(6,3+)，因此b[idx]是形状(3，)的一行b。例如，当idx为|3022225|In+[37]:+a[1,1]
Out[37]:+0|3022226|b[a[1,1]]是|3022227|In+[38]:+b[a[1,1]]
Out[38]:+array([0,+0,+0])|3022228|所以|3022229|In+[33]:+b[a].shape
Out[33]:+(6,+2,+3)|3022230|因此，让我们重复一下:+NumPy通过将a的每个元素移动到新数组中，并将b[idx]的值放在a中idx的位置，从而创建了一个新的数组。当idx在a上移动时，将创建一个形状(6,2)数组。但由于b[idx]本身的形状为(3，)，因此在(6,2)-shaped数组中的每个位置都放置了一个(3，)-shaped值。结果是一个形状数组(6,2,3)。|3022231|现在，当你进行像这样的赋值|3022232|b[a]+=+10|3022233|创建一个值为b[a]的临时形状数组(6,2,3)，然后执行赋值。由于10是一个常量，此赋值将值10放在(6,2,3)-shaped数组中的每个位置。然后将临时数组中的值重新分配回b。参见reference+to+docs。因此，将(6,2,3)-shaped数组中的值复制回(6,3)-shaped+b数组。值之间会相互覆盖。但主要的一点是，你没有得到你想要的任务。|3022234|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/user/basics.indexing.html#assigning-values-to-indexed-arrays^0|0|0|0|5|0|I|1|0|0|F|1|M|1|T|1|16|4|1B|4|1G|4|1P|4|27|1|2H|1|0|9|1|B|1|H|1|J|1|U|B|16|B|0|0|0|0|3|4|9|E|4|4|9|0|4|4|P|1|15|3|1B|6|1P|1|1R|3|0|0|3|4|1|N|1|11|6|1H|1|1N|3|0|0|0|9|0|0|0|0|K|1|10|6|1A|1|1C|3|1W|3|20|1|2O|6|0|0|0|6|4|2H|1|46|1|2L|H|0|0^^$0|@$1|2|3|4|5|6|7|1T|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|1U|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|1V|8|@$H|1W|I|1X|J|K]]|9|@]|A|$]]|$1|L|3|M|5|E|7|1Y|8|@$H|1Z|I|20|J|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|21|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|E|7|22|8|@$H|23|I|24|J|N]|$H|25|I|26|J|N]|$H|27|I|28|J|N]|$H|29|I|2A|J|N]|$H|2B|I|2C|J|N]|$H|2D|I|2E|J|N]|$H|2F|I|2G|J|N]|$H|2H|I|2I|J|N]|$H|2J|I|2K|J|N]]|9|@]|A|$]]|$1|S|3|T|5|E|7|2L|8|@$H|2M|I|2N|J|N]|$H|2O|I|2P|J|N]|$H|2Q|I|2R|J|N]|$H|2S|I|2T|J|N]|$H|2U|I|2V|J|N]|$H|2W|I|2X|J|N]]|9|@]|A|$]]|$1|U|3|V|5|6|7|2Y|8|@]|9|@]|A|$B|C]]|$1|W|3|-4|5|E|7|2Z|8|@]|9|@]|A|$]]|$1|X|3|Y|5|E|7|30|8|@$H|31|I|32|J|K]|$H|33|I|34|J|K]|$H|35|I|36|J|K]|$H|37|I|38|J|N]]|9|@]|A|$]]|$1|Z|3|10|5|E|7|39|8|@$H|3A|I|3B|J|N]|$H|3C|I|3D|J|N]|$H|3E|I|3F|J|N]|$H|3G|I|3H|J|N]|$H|3I|I|3J|J|N]|$H|3K|I|3L|J|N]]|9|@]|A|$]]|$1|11|3|12|5|E|7|3M|8|@$H|3N|I|3O|J|N]|$H|3P|I|3Q|J|N]|$H|3R|I|3S|J|N]|$H|3T|I|3U|J|N]|$H|3V|I|3W|J|N]|$H|3X|I|3Y|J|N]]|9|@]|A|$]]|$1|13|3|14|5|6|7|3Z|8|@]|9|@]|A|$B|C]]|$1|15|3|16|5|E|7|40|8|@$H|41|I|42|J|N]]|9|@]|A|$]]|$1|17|3|18|5|6|7|43|8|@]|9|@]|A|$B|C]]|$1|19|3|1A|5|E|7|44|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|6|7|45|8|@]|9|@]|A|$B|C]]|$1|1D|3|1E|5|E|7|46|8|@$H|47|I|48|J|N]|$H|49|I|4A|J|N]|$H|4B|I|4C|J|N]|$H|4D|I|4E|J|N]|$H|4F|I|4G|J|N]|$H|4H|I|4I|J|N]|$H|4J|I|4K|J|N]]|9|@]|A|$]]|$1|1F|3|1G|5|E|7|4L|8|@]|9|@]|A|$]]|$1|1H|3|1I|5|6|7|4M|8|@]|9|@]|A|$B|C]]|$1|1J|3|1K|5|E|7|4N|8|@$H|4O|I|4P|J|N]|$H|4Q|I|4R|J|N]|$H|4S|I|4T|J|N]]|9|@$H|4U|I|4V|1|4W]]|A|$]]|$1|1L|3|-4|5|E|7|4X|8|@]|9|@]|A|$]]]|1M|$1N|$5|1O|1P|1Q|A|$1R|1S]]]]

<pre><code>In [1]: import numpy as np
In [2]: a = np.array([[2,0],[3,0],[3,1],[5,0],[5,1],[5,2]])
In [3]: b = np.zeros((6,3), dtype='int32')

In [4]: b[a[:,0], a[:,1]] = 10

In [5]: b
Out[5]: 
array([[ 0, 0, 0],
 [ 0, 0, 0],
 [10, 0, 0],
 [10, 10, 0],
 [ 0, 0, 0],
 [10, 10, 10]])
</code></pre>

<hr>

Why it works:

If you index <code>b</code> with two numpy arrays in an assignment, 

<pre><code>b[x, y] = z
</code></pre>

then think of NumPy as moving simultaneously over each element of <code>x</code> and each element of <code>y</code> and each element of <code>z</code> (let's call them <code>xval</code>, <code>yval</code> and <code>zval</code>), and assigning to b[xval, yval] the value <code>zval</code>. When <code>z</code> is a constant, "moving over <code>z</code> just returns the same value each time.

That's what we want, with <code>x</code> being the first column of <code>a</code> and <code>y</code> being the second column of <code>a</code>. Thus, choose <code>x = a[:, 0]</code>, and <code>y = a[:, 1]</code>.

<pre><code>b[a[:,0], a[:,1]] = 10
</code></pre>

<hr>

Why <code>b[a] = 10</code> does not work

When you write <code>b[a]</code>, think of NumPy as creating a new array by moving over each element of <code>a</code>, (let's call each one <code>idx</code>) and placing in the new array the value of <code>b[idx]</code> at the location of <code>idx</code> in <code>a</code>.

<code>idx</code> is a value in <code>a</code>. So it is an int32. <code>b</code> is of shape (6,3), so <code>b[idx]</code> is a row of <code>b</code> of shape (3,). For example, when <code>idx</code> is

<pre><code>In [37]: a[1,1]
Out[37]: 0
</code></pre>

<code>b[a[1,1]]</code> is 

<pre><code>In [38]: b[a[1,1]]
Out[38]: array([0, 0, 0])
</code></pre>

So 

<pre><code>In [33]: b[a].shape
Out[33]: (6, 2, 3)
</code></pre>

So let's repeat: NumPy is creating a new array by moving over each element of <code>a</code> and placing in the new array the value of <code>b[idx]</code> at the location of <code>idx</code> in <code>a</code>. As <code>idx</code> moves over <code>a</code>, an array of shape (6,2) would be created. But since <code>b[idx]</code> is itself of shape (3,), at each location in the (6,2)-shaped array, a (3,)-shaped value is being placed. The result is an array of shape (6,2,3).

Now, when you make an assignment like

<pre><code>b[a] = 10
</code></pre>

a temporary array of shape (6,2,3) with values <code>b[a]</code> is created, then the assignment is performed. Since 10 is a constant, this assignment places the value 10 at each location in the (6,2,3)-shaped array.
Then the values from the temporary array are reassigned back to <code>b</code>.
See <a href="http://docs.scipy.org/doc/numpy/user/basics.indexing.html#assigning-values-to-indexed-arrays" rel="noreferrer">reference to docs</a>. Thus the values in the (6,2,3)-shaped array are copied back to the (6,3)-shaped <code>b</code> array. Values overwrite each other. But the main point is you do not obtain the assignments you desire.

I am quite new to python and numpy. Can some one pls help me to understand how I can do the indexing of some arrays used as indices. I have the following six 2D arrays like this-

<pre><code>array([[2, 0],
 [3, 0],
 [3, 1],
 [5, 0],
 [5, 1],
 [5, 2]])
</code></pre>

I want to use these arrays as indices and put the value 10 in the corresponding indices of a new empty matrix. The output should look like this-

<pre><code>array([[ 0, 0, 0],
 [ 0, 0, 0],
 [10, 0, 0],
 [10, 10, 0],
 [ 0, 0, 0],
 [10, 10, 10]])
</code></pre>

So far I have tried this- 

<pre><code> from numpy import*
 a = array([[2,0],[3,0],[3,1],[5,0],[5,1],[5,2]])
 b = zeros((6,3),dtype ='int32')
 b[a] = 10
</code></pre>

But this gives me the wrong output. Any help pls.

Python numpy 2D array indexing

我对python和numpy还是个新手。有没有人可以帮助我理解如何对一些用作索引的数组进行索引。我有以下六个这样的二维数组：array([[2, 0],   [3, 0],   [3, 1],   [5, 0],   [5, 1],   [5, 2]])我想使用这些数组作为索引，并将值10放入一个新的空矩阵的相应索引中...

问Python numpy 2D数组索引
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