blocks|key|3145877|text|首先，您对“正确”数组的描述与您指定的列和行不匹配...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3145878|为了得到“正确”的数组，你可以这样做：|3145879|a[:2,+2]+=+0|code-block|syntax|javascript|3145880|要修改第一行和第三行的第二列和第三列(第0,2行和第1，2列)，您需要执行以下操作...(您对修改第0,2行和第1,2列的描述就是您得到的结果，对吧？)|3145881|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

First off, your description of the "correct" array doesn't match the columns and rows you specify...

To get your "correct" array, you'd do this:

<pre><code>a[:2, 2] = 0
</code></pre>

To modify the second and third columns of the first and third rows, (rows [0,2] and columns [1,2]) you'd do what you're doing... (Your description of modifying rows [0,2] and columns [1,2] is exactly the result you get, right?)

blocks|key|2931107|text|它应该和a0,2=0和a1,2=0一样简单。你也可以做a0:2,2=0。python中基于':‘的范围索引是一个半开区间[0，2]，实际上范围是从0到1(不包括2的终点)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2931108|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It should be as simple as a[0,2]=0 and a[1,2]=0. You could also do a[0:2,2]=0. The ':' based range indexing in python is a half-open interval [0,2) which actually range from 0 to 1 (the end point of 2 is not included).

blocks|key|254119|text|除了其他人所说的之外，您还可以使用花哨的索引修改这些元素，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|254120|In+[39]:+rows+=+[0,1]

In+[40]:+cols+=+[2,2]

In+[41]:+a+=+np.arange(1,10).reshape((3,3))

In+[42]:+a[rows,cols]+=+0

In+[43]:+a
Out[43]:+
array([[1,+2,+0],
+++++++[4,+5,+0],
+++++++[7,+8,+9]])|code-block|syntax|javascript|254121|您可能想要阅读有关索引多维数组的文档：http://docs.scipy.org/doc/numpy/user/basics.indexing.html#indexing-multi-dimensional-arrays|offset|length|254122|关键点是：|254123|254124|如果索引数组具有匹配的形状，并且被索引数组的每个维度都有一个索引数组，则结果数组具有与索引数组相同的形状，并且值对应于索引数组中每个位置的索引集。|blockquote|254125|254126|254127|重要的是，这还允许您执行以下操作：|254128|In+[60]:+a[rows,cols]+=+np.array([33,77])

In+[61]:+a
Out[61]:+
array([[+1,++2,+33],
+++++++[+4,++5,+77],
+++++++[+7,++8,++9]])|254129|其中，您可以使用相同大小的另一个数组、列表或元组单独设置每个元素。|254130|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/user/basics.indexing.html#indexing-multi-dimensional-arrays^0|0|0|J|2J|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1H|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1J|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

Adding to what others have said, you can modify these elements using fancy indexing as follows:

<pre><code>In [39]: rows = [0,1]

In [40]: cols = [2,2]

In [41]: a = np.arange(1,10).reshape((3,3))

In [42]: a[rows,cols] = 0

In [43]: a
Out[43]: 
array([[1, 2, 0],
 [4, 5, 0],
 [7, 8, 9]])
</code></pre>

You might want to read the documentation on indexing multidimensional arrays:
<a href="http://docs.scipy.org/doc/numpy/user/basics.indexing.html#indexing-multi-dimensional-arrays" rel="noreferrer">http://docs.scipy.org/doc/numpy/user/basics.indexing.html#indexing-multi-dimensional-arrays</a>

The key point is:

<blockquote>
 if the index arrays have a matching shape, and there is an index array
 for each dimension of the array being indexed, the resultant array has
 the same shape as the index arrays, and the values correspond to the
 index set for each position in the index arrays.
</blockquote>

Importantly this also allows you to do things like:

<pre><code>In [60]: a[rows,cols] = np.array([33,77])

In [61]: a
Out[61]: 
array([[ 1, 2, 33],
 [ 4, 5, 77],
 [ 7, 8, 9]])
</code></pre>

where you can set each element independently using another array, list or tuple of the same size.

blocks|key|2931277|text|一种解决方法：ndarray.flatten,+np.put(),+ndarray.reshape|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2931278|尝试使用ndarray.flatten(array)，这样您就可以处理一个可以用numpy.put(array,[indices],[values])操作的一维数组。然后使用ndarray.reshape()获取原始尺寸。|2931279|entityMap^0|7|16|0|4|M|14|Z|2F|H|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|I|$]]

one work around: <code>ndarray.flatten, np.put(), ndarray.reshape</code>

try <code>ndarray.flatten(array)</code>, that way you are dealing with a one dim array which can be manipulated with <code>numpy.put(array,[indices],[values])</code>. Then use <code>ndarray.reshape()</code> to get to the original dimensions.

I have a 2D numpy array and I have a arrays of rows and columns which should be set to a particular value. Lets consider the following example

<pre><code> a = array([[1, 2, 3],
 [4, 5, 6],
 [7, 8, 9]])
</code></pre>

I want to modify entries at rows [0,2] and columns [1,2]. This should result in the following array

<pre><code> a = array([[1, 2, 0],
 [4, 5, 0],
 [7, 8, 9]])
</code></pre>

I did following and it resulted in modifying each sequence of column in every row

<pre><code>rows = [0,1]
cols = [2,2]
b=a[numpy.ix_(rows,columns)]
</code></pre>

It resulted in the following array modifying every column of the specified array

<pre><code>array([[1, 0, 0],
 [4, 5, 6],
 [7, 0, 0]])
</code></pre>

Some one could please let me know how to do it?

Thanks a lot

EDIT: It is to be noted that rows and columns coincidently happend to be sequentia. The actual point is that these could be arbitrary and in any order. if it is rows = [a,b,c] and cols=[n x z] then I want to modify exactly three elements at locations (a,n),(b,x),(c,z).

how to modify a 2D numpy array at specific locations without a loop?

我有一个二维numpy数组，我有一个应该设置为特定值的行和列的数组。让我们考虑一下下面的例子 a = array([[1, 2, 3],            [4, 5, 6],            [7, 8, 9]])我想修改第0,2行和第1,2列的条目。 a = array([[1, 2, 0],      ...

问如何在没有循环的情况下修改特定位置的二维numpy数组？
EN

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