How to find top k elements in an array numpy
https://www.skytowner.com/explore/numpy_argpartition_method
NumPy | argpartition method with Examples
2023-08-12
None
<div><div><h2>Examples</h2><p>Suppose we have the following unsorted array:</p><p><code>
<div>
<p><a href="https://www.skytowner.com/explore/numpy_array_method">array</a></p>
</div>
</code></p><p>The sorted version of <code></code> is:</p><p><code>
<div>
<p></p>
</div>
</code></p><p>The original indices of this sorted version is:</p><p><code>
<div>
<p></p>
</div>
</code></p><p>Instead of returning this sorted indices, the <code></code> method asks you to select an index by which to partition, and ensures that all values of indices smaller than the value at this index appear before it, and all values of indices larger than this index value appear after it.</p><p>Here's an example:</p><p><code>
<div>
<div>
<p><a href="https://www.skytowner.com/explore/numpy_array_method">array</a></p><p></p>
</div>
<div><p>array([4, 5, 1, 2, 3, 0])</p></div>
</div>
</code></p><p>Here, we set <code></code>, which means that we want all values up to and including index <code></code> to be in its rightful place when sorted. In other words, the 3rd index, or the fourth (3+1) smallest value of <code></code> (7 in this case), will be in its rightful place when sorted, with all values smaller than 7 coming before it, and all those larger than 7 to come after it. Therefore, some of the valid sorted versions are as follows:</p><p><code>
<div>
<p></p><p></p><p></p><p></p>
</div>
</code></p><div><p>WARNING</p><p>The ordering of the values before after the <code></code>th index is undefined. This is why the value 4 can potentially come before the value 1. The <code></code> chooses one out of the above, and returns its indices. As such, there is no guarantee that the returned array of indices represents a perfectly sorted array.</p></div><p>In our specific example, the returned indices were:</p><p><code>
<div>
<div>
<p></p>
</div>
<div><p>array([4, 5, 1, 2, 3, 0])</p></div>
</div>
</code></p><p>Which represents the following sorted version:</p><p><code>
<div>
<div>
<p></p>
</div>
<div><p>array([6, 2, 5, 7, 9, 8])</p></div>
</div>
</code></p><p>Here, note the following:</p><ul><li><p>the values before the k-th index (index 3 in this case) are smaller than the value at the k-th index, but they are not necessarily sorted.</p></li><li><p>the values after the k-th index are larger than the value at the k-th index</p></li></ul></div></div>||||I|||| menu
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 NumPy | argpartition method

 schedule Aug 12, 2023
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 NumPy's argparition(~) method is used to partition an array based on ascending order. Explaining how it works in words is rather difficult, so let's go through a quick example.

 Examples

 Suppose we have the following unsorted array:

 x = np. array ([ 8 , 5 , 7 , 10 , 9 ])

 The sorted version of x is:

 [ 5 , 7 , 8 , 9 , 10 ]

 The original indices of this sorted version is:

 [1,2,0,4,3]

 Instead of returning this sorted indices, the argpartition() method asks you to select an index by which to partition, and ensures that all values of indices smaller than the value at this index appear before it, and all values of indices larger than this index value appear after it.

 Here's an example:

 x = np. array ([ 8 , 5 , 7 , 9 , 6 , 2 ])
 np. argpartition (x, 3 )
 array([4, 5, 1, 2, 3, 0])

 Here, we set kth=1 , which means that we want all values up to and including index 3 to be in its rightful place when sorted. In other words, the 3rd index, or the fourth (3+1) smallest value of x (7 in this case), will be in its rightful place when sorted, with all values smaller than 7 coming before it, and all those larger than 7 to come after it. Therefore, some of the valid sorted versions are as follows:

 [4, 5, 1, 2, 3, 0]
 [4, 1, 2, 5, 3, 0]
 [4, 5, 1, 2, 3, 0]
 ...
 WARNING

 The ordering of the values before after the k th index is undefined. This is why the value 4 can potentially come before the value 1. The argpartition chooses one out of the above, and returns its indices. As such, there is no guarantee that the returned array of indices represents a perfectly sorted array.

 In our specific example, the returned indices were:

 y
 array([4, 5, 1, 2, 3, 0])

 Which represents the following sorted version:

 x[y]
 array([6, 2, 5, 7, 9, 8])

 Here, note the following:

 * the values before the k-th index (index 3 in this case) are smaller than the value at the k-th index, but they are not necessarily sorted.

 * the values after the k-th index are larger than the value at the k-th index

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 Official NumPy Documentation
 https://numpy.org/doc/stable/reference/generated/numpy.argpartition.html
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Examples Suppose we have the following unsorted array:  
 array 
  The sorted version of   is:  
 
  The original indices of this sorted version is:  
 
  Instead of returning this sorted indices, the   method asks you to select an index by which to partition, and ensures that all values of indices smaller than the value at this index appear before it, and all values of indices larger than this index value appear after it. Here's an example:  
 array   
 array([4, 5, 1, 2, 3, 0]) 
  Here, we set  , which means that we want all values up to and including index   to be in its rightful place when sorted. In other words, the 3rd index, or the fourth (3+1) smallest value of   (7 in this case), will be in its rightful place when sorted, with all values smaller than 7 coming before it, and all those larger than 7 to come after it. Therefore, some of the valid sorted versions are as follows:  
       
   WARNING The ordering of the values before after the  th index is undefined. This is why the value 4 can potentially come before the value 1. The   chooses one out of the above, and returns its indices. As such, there is no guarantee that the returned array of indices represents a perfectly sorted array.  In our specific example, the returned indices were:  
 
 array([4, 5, 1, 2, 3, 0]) 
  Which represents the following sorted version:  
 
 array([6, 2, 5, 7, 9, 8]) 
  Here, note the following:   the values before the k-th index (index 3 in this case) are smaller than the value at the k-th index, but they are not necessarily sorted.   the values after the k-th index are larger than the value at the k-th index
NumPy | argpartition method with Examples
https://www.skytowner.com/explore/numpy_argpartition_method
<div><div>
<p>Similar to @Imtinan, I struggled with this. I found it useful to break up the function into the arg and the partition.</p>

<p>Take the following array:</p>

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

the corresponding indices are: [0,1,2,3,4,5,6,7,8] where 8th index = 5 and 0th = 9
</code></pre>

<p>if we do <code>np.partition(array, k=5)</code>, the code is going to take the 5th element (not index) and then place it into a new array. It is then going to put those elements &lt; 5th element before it and that &gt; 5th element after, like this:</p>

<p><code>pseudo output: [lower value elements, 5th element, higher value elements]</code></p>

<p>if we compute this we get:</p>

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

<p>This makes sense as the 5th element in the original array = 6, [1,2,3,4,5] are all lower than 6 and [7,8,9] are higher than 6. Note that the elements are not ordered. </p>

<p>The arg part of the <code>np.argpartition()</code> then goes one step further and swaps the elements out for their respective indices in the original array. So if we did:</p>

<p><code>np.argpartition(array, 5)</code> we will get:</p>

<p><code>array([5, 8, 7, 3, 1, 4, 6, 2, 0])</code></p>

<p>from above, the original array had this structure [index=value]
[0=9, 1=2, 2=7, 3=4, 4=6, 5=3, 6=8, 7=1, 8=5]</p>

<p>you can map the value of the index to the output and you with satisfy the condition:</p>

<p><code>argpartition() = partition()</code>, like this:</p>

<p>[index form] array([5, 8, 7, 3, 1, 4, 6, 2, 0]) becomes </p>

<pre><code>[3, 5, 1, 4, 2, 6, 8, 7, 9]
</code></pre>

<p>which is the same as the output of <code>np.partition(array)</code>, </p>

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

<p>Hopefully, this makes sense, it was the only way I could get my head around the arg part of the function. </p>
    </div></div>||||I||||Skip to main content
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          How does numpy's argpartition work on the documentation's example?

              Ask Question
            Asked 5 years, 9 months ago
            Modified 3 years, 10 months ago
            Viewed 12k times
                    28

                  I am trying to understand numpy's argpartition function. I have made the documentation's example as basic as possible.

                  import numpy as np
                  
                  x = np.array([3, 4, 2, 1])
                  print("x: ", x)
                  
                  a=np.argpartition(x, 3)
                  print("a: ", a)
                  
                  print("x[a]:", x[a])
                  

                  This is the output...

                  ('x: ', array([3, 4, 2, 1]))
                  ('a: ', array([2, 3, 0, 1]))
                  ('x[a]:', array([2, 1, 3, 4]))
                  

                  In the line a=np.argpartition(x, 3) isn't the kth element the last element (the number 1)? If it is number 1, when x is sorted shouldn't 1 become the first element (element 0)?

                  In x[a], why is 2 the first element "in front" of 1?

                  What fundamental thing am I missing?

                        * python
                        * numpy
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                            asked Sep 23, 2018 at 10:09
                          Mel Mel
                            6,264 11 11 gold badges 56 56 silver badges 71 71 bronze badges
              2
                        * 1
                          partition/argpartition split at the kth element by order , not by position. The latter would not require a dedicated algorithm, because you could simply do something like np.concatenate([x[x<x[3]], x[x ==x[3]], x[[x>x[3]]) . The smart bit about partition is finding the kth element without doing a full sort.
                            – Paul Panzer
                          Commented Sep 23, 2018 at 10:22
                        * @PaulPanzer: The advantage of a naive partition compared to concatenate would be that it could it in one single pass.
                            – Eric Duminil
                          Commented Sep 23, 2018 at 11:06
                  Add a comment |

                  3 Answers

                      Sorted by: Reset to default
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                      43

                    The more complete answer to what argpartition does is in the documentation of partition, and that one says:

                    Creates a copy of the array with its elements rearranged in such a way that the value of the element in k-th position is in the position it would be in a sorted array. All elements smaller than the k-th element are moved before this element and all equal or greater are moved behind it. The ordering of the elements in the two partitions is undefined.

                    So, for the input array 3, 4, 2, 1, the sorted array would be 1, 2, 3, 4.

                    The result of np.partition([3, 4, 2, 1], 3) will have the correct value (i.e. same as sorted array) in the 3rd (i.e. last) element. The correct value for the 3rd element is 4.

                    Let me show this for all values of k to make it clear:

                      * np.partition([3, 4, 2, 1], 0) - [1, 4, 2, 3]
                      * np.partition([3, 4, 2, 1], 1) - [1, 2, 4, 3]
                      * np.partition([3, 4, 2, 1], 2) - [1, 2, 3, 4]
                      * np.partition([3, 4, 2, 1], 3) - [2, 1, 3, 4]

                    In other words: the k-th element of the result is the same as the k-th element of the sorted array. All elements before k are smaller than or equal to that element. All elements after it are greater than or equal to it.

                    The same happens with argpartition, except argpartition returns indices which can then be used for form the same result.

                              Share
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                              edited Sep 23, 2018 at 11:57
                              answered Sep 23, 2018 at 10:31
                            zvone zvone
                              19.1k 5 5 gold badges 50 50 silver badges 82 82 bronze badges
                3
                          * Thank you for your detailed explanation. May I ask what could be a potential use case for this method?
                              – Anoushiravan R
                            Commented Nov 29, 2022 at 11:15
                          * I assume that potential use case could be to get top_num_of_scores from some array filled with unsorted scores ids_of_top_n_unsorted = np.argpartition(array_of_unsorted_scores, -num_top)[-num_top:]
                              – mmway
                            Commented Jan 2, 2023 at 14:23
                          * 1
                            I saw it used as selected top n largest coefficients in features selections.
                              – longlifelearner
                            Commented Jun 3, 2023 at 22:10
                    Add a comment |
                      12

                    Similar to @Imtinan, I struggled with this. I found it useful to break up the function into the arg and the partition.

                    Take the following array:

                    array = np.array([9, 2, 7, 4, 6, 3, 8, 1, 5])
                    
                    the corresponding indices are: [0,1,2,3,4,5,6,7,8] where 8th index = 5 and 0th = 9
                    

                    if we do np.partition(array, k=5), the code is going to take the 5th element (not index) and then place it into a new array. It is then going to put those elements < 5th element before it and that > 5th element after, like this:

                    pseudo output: [lower value elements, 5th element, higher value elements]

                    if we compute this we get:

                    array([3, 5, 1, 4, 2, 6, 8, 7, 9])

                    This makes sense as the 5th element in the original array = 6, [1,2,3,4,5] are all lower than 6 and [7,8,9] are higher than 6. Note that the elements are not ordered.

                    The arg part of the np.argpartition() then goes one step further and swaps the elements out for their respective indices in the original array. So if we did:

                    np.argpartition(array, 5) we will get:

                    array([5, 8, 7, 3, 1, 4, 6, 2, 0])

                    from above, the original array had this structure [index=value] [0=9, 1=2, 2=7, 3=4, 4=6, 5=3, 6=8, 7=1, 8=5]

                    you can map the value of the index to the output and you with satisfy the condition:

                    argpartition() = partition(), like this:

                    [index form] array([5, 8, 7, 3, 1, 4, 6, 2, 0]) becomes

                    [3, 5, 1, 4, 2, 6, 8, 7, 9]
                    

                    which is the same as the output of np.partition(array),

                    array([3, 5, 1, 4, 2, 6, 8, 7, 9])
                    

                    Hopefully, this makes sense, it was the only way I could get my head around the arg part of the function.

                              Share
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                              edited May 5, 2020 at 12:24
                            Nandu Raj
                              2,110 10 10 silver badges 20 20 bronze badges
                              answered May 5, 2020 at 8:05
                            DSeal6 DSeal6
                              131 1 1 silver badge 6 6 bronze badges
                    Add a comment |
                      8

                    i remember having a hard time figuring it out too, maybe the documentation is written badly but this is what it means

                    When you do a=np.argpartition(x, 3) then x is sorted in such a way that only the element at the k'th index will be sorted (in our case k=3)

                    So when you run this code basically you are asking what would the value of the 3rd index be in a sorted array. Hence the output is ('x[a]:', array([2, 1, 3, 4]))where only element 3 is sorted.

                    As the document suggests all numbers smaller than the kth element are before it (in no particular order) hence you get 2 before 1, since its no particular order.

                    i hope this clarifies it, if you are still confused then feel free to comment :)

                              Share
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                              edited Aug 16, 2020 at 7:44
                              answered Sep 23, 2018 at 10:27
                            Imtinan Azhar Imtinan Azhar
                              1,753 11 11 silver badges 26 26 bronze badges
                1
                          * 1
                            Thank you. The parts that say "x is sorted in such a way that only the element in the position 3 will be sorted" really helped. Thank you. PS. I ended up ticking the above as the correct answer because it had the nice code example and we can only pick one answer.
                              – Mel
                            Commented Sep 23, 2018 at 10:58
                    Add a comment |

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array = np.array([9, 2, 7, 4, 6, 3, 8, 1, 5])
the corresponding indices are: [0,1,2,3,4,5,6,7,8] where 8th index = 5 and 0th = 9
 
if we do np.partition(array, k=5), the code is going to take the 5th element (not index) and then place it into a new array. It is then going to put those elements &lt; 5th element before it and that &gt; 5th element after, like this:
 pseudo output: [lower value elements, 5th element, higher value elements] 
if we compute this we get:
 array([3, 5, 1, 4, 2, 6, 8, 7, 9]) 
This makes sense as the 5th element in the original array = 6, [1,2,3,4,5] are all lower than 6 and [7,8,9] are higher than 6. Note that the elements are not ordered. 
The arg part of the np.argpartition() then goes one step further and swaps the elements out for their respective indices in the original array. So if we did:
 np.argpartition(array, 5) we will get:
 array([5, 8, 7, 3, 1, 4, 6, 2, 0]) 
from above, the original array had this structure [index=value]
[0=9, 1=2, 2=7, 3=4, 4=6, 5=3, 6=8, 7=1, 8=5]
you can map the value of the index to the output and you with satisfy the condition:
 argpartition() = partition(), like this:
[index form] array([5, 8, 7, 3, 1, 4, 6, 2, 0]) becomes 
 [3, 5, 1, 4, 2, 6, 8, 7, 9]
 
which is the same as the output of np.partition(array), 
 array([3, 5, 1, 4, 2, 6, 8, 7, 9])
 
Hopefully, this makes sense, it was the only way I could get my head around the arg part of the function.
How does numpy's argpartition work on the documentation's example?
https://stackoverflow.com/a/52465229
https://numpy.org/doc/stable/reference/generated/numpy.argpartition.html
numpy.argpartition#
2022-01-01
None
<div><article>
 
 <section>
<dl>
<dt>
numpy.argpartition(<em>a</em>, <em>kth</em>, <em>axis=-1</em>, <em>kind='introselect'</em>, <em>order=None</em>)<a href="https://github.com/numpy/numpy/blob/v1.24.0/numpy/core/fromnumeric.py#L792-L871">[source]</a><a href="#numpy.argpartition">#</a></dt>
<dd><p>Perform an indirect partition along the given axis using the
algorithm specified by the <em>kind</em> keyword. It returns an array of
indices of the same shape as <em>a</em> that index data along the given
axis in partitioned order.</p>
<p>New in version 1.8.0.</p>
<dl>
<dt>Parameters:</dt>
<dd><dl>
<dt><strong>a</strong>array_like</dt><dd><p>Array to sort.</p>
</dd>
<dt><strong>kth</strong>int or sequence of ints</dt><dd><p>Element index to partition by. The k-th element will be in its
final sorted position and all smaller elements will be moved
before it and all larger elements behind it. The order of all
elements in the partitions is undefined. If provided with a
sequence of k-th it will partition all of them into their sorted
position at once.</p>
<p>Deprecated since version 1.22.0: Passing booleans as index is deprecated.</p>
</dd>
<dt><strong>axis</strong>int or None, optional</dt><dd><p>Axis along which to sort. The default is -1 (the last axis). If
None, the flattened array is used.</p>
</dd>
<dt><strong>kind</strong>{‘introselect’}, optional</dt><dd><p>Selection algorithm. Default is ‘introselect’</p>
</dd>
<dt><strong>order</strong>str or list of str, optional</dt><dd><p>When <em>a</em> is an array with fields defined, this argument
specifies which fields to compare first, second, etc. A single
field can be specified as a string, and not all fields need be
specified, but unspecified fields will still be used, in the
order in which they come up in the dtype, to break ties.</p>
</dd>
</dl>
</dd>
<dt>Returns:</dt>
<dd><dl>
<dt><strong>index_array</strong>ndarray, int</dt><dd><p>Array of indices that partition <em>a</em> along the specified axis.
If <em>a</em> is one-dimensional, <code>a[index_array]</code> yields a partitioned <em>a</em>.
More generally, <code>np.take_along_axis(a, index_array, axis=axis)</code>
always yields the partitioned <em>a</em>, irrespective of dimensionality.</p>
</dd>
</dl>
</dd>
</dl>
<p>Notes</p>
<p>See <a href="numpy.partition.html#numpy.partition"><code>partition</code></a> for notes on the different selection algorithms.</p>
<p>Examples</p>
<p>One dimensional array:</p>
<div><pre>&gt;&gt;&gt; x = np.array([3, 4, 2, 1])
&gt;&gt;&gt; x[np.argpartition(x, 3)]
array([2, 1, 3, 4])
&gt;&gt;&gt; x[np.argpartition(x, (1, 3))]
array([1, 2, 3, 4])
</pre></div>
<div><pre>&gt;&gt;&gt; x = [3, 4, 2, 1]
&gt;&gt;&gt; np.array(x)[np.argpartition(x, 3)]
array([2, 1, 3, 4])
</pre></div>
<p>Multi-dimensional array:</p>
<div><pre>&gt;&gt;&gt; x = np.array([[3, 4, 2], [1, 3, 1]])
&gt;&gt;&gt; index_array = np.argpartition(x, kth=1, axis=-1)
&gt;&gt;&gt; np.take_along_axis(x, index_array, axis=-1) # same as np.partition(x, kth=1)
array([[2, 3, 4],
 [1, 1, 3]])
</pre></div>
</dd></dl>
</section>
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 On this page
 * argpartition

 numpy.argpartition#

 numpy. argpartition ( a , kth , axis = -1 , kind = 'introselect' , order = None ) [source] #

 Perform an indirect partition along the given axis using the algorithm specified by the kind keyword. It returns an array of indices of the same shape as a that index data along the given axis in partitioned order.

 New in version 1.8.0.

 Parameters : a array_like

 Array to sort.

 kth int or sequence of ints

 Element index to partition by. The k-th element will be in its final sorted position and all smaller elements will be moved before it and all larger elements behind it. The order of all elements in the partitions is undefined. If provided with a sequence of k-th it will partition all of them into their sorted position at once.

 Deprecated since version 1.22.0: Passing booleans as index is deprecated.

 axis int or None, optional

 Axis along which to sort. The default is -1 (the last axis). If None, the flattened array is used.

 kind {‘introselect’}, optional

 Selection algorithm. Default is ‘introselect’

 order str or list of str, optional

 When a is an array with fields defined, this argument specifies which fields to compare first, second, etc. A single field can be specified as a string, and not all fields need be specified, but unspecified fields will still be used, in the order in which they come up in the dtype, to break ties.

 Returns : index_array ndarray, int

 Array of indices that partition a along the specified axis. If a is one-dimensional, a[index_array] yields a partitioned a. More generally, np.take_along_axis(a, index_array, axis=axis) always yields the partitioned a, irrespective of dimensionality.

 See also

 partition

 Describes partition algorithms used.

 ndarray.partition

 Inplace partition.

 argsort

 Full indirect sort.

 take_along_axis

 Apply index_array from argpartition to an array as if by calling partition.

 Notes

 See partition for notes on the different selection algorithms.

 Examples

 One dimensional array:

 >>> x = np.array([3, 4, 2, 1])
 >>> x[np.argpartition(x, 3)]
 array([2, 1, 3, 4])
 >>> x[np.argpartition(x, (1, 3))]
 array([1, 2, 3, 4])
 
 >>> x = [3, 4, 2, 1]
 >>> np.array(x)[np.argpartition(x, 3)]
 array([2, 1, 3, 4])
 

 Multi-dimensional array:

 >>> x = np.array([[3, 4, 2], [1, 3, 1]])
 >>> index_array = np.argpartition(x, kth=1, axis=-1)
 >>> np.take_along_axis(x, index_array, axis=-1) # same as np.partition(x, kth=1)
 array([[2, 3, 4],
 [1, 1, 3]])
 

 previous

 numpy.partition

 next

 numpy.argmax

 © Copyright 2008-2022, NumPy Developers.

 Created using Sphinx 5.3.0.

numpy.argpartition(a, kth, axis=-1, kind='introselect', order=None)[source] # 
 Perform an indirect partition along the given axis using the
algorithm specified by the kind keyword. It returns an array of
indices of the same shape as a that index data along the given
axis in partitioned order.
New in version 1.8.0.
Parameters:
 
 aarray_like  Array to sort.
 kthint or sequence of ints  Element index to partition by. The k-th element will be in its
final sorted position and all smaller elements will be moved
before it and all larger elements behind it. The order of all
elements in the partitions is undefined. If provided with a
sequence of k-th it will partition all of them into their sorted
position at once.
Deprecated since version 1.22.0: Passing booleans as index is deprecated.
 axisint or None, optional  Axis along which to sort. The default is -1 (the last axis). If
None, the flattened array is used.
 kind{‘introselect’}, optional  Selection algorithm. Default is ‘introselect’
 orderstr or list of str, optional  When a is an array with fields defined, this argument
specifies which fields to compare first, second, etc. A single
field can be specified as a string, and not all fields need be
specified, but unspecified fields will still be used, in the
order in which they come up in the dtype, to break ties.
Returns:
 
 index_arrayndarray, int  Array of indices that partition a along the specified axis.
If a is one-dimensional, a[index_array] yields a partitioned a.
More generally, np.take_along_axis(a, index_array, axis=axis)
always yields the partitioned a, irrespective of dimensionality.
Notes
See  partition  for notes on the different selection algorithms.
Examples
One dimensional array:
 &gt;&gt;&gt; x = np.array([3, 4, 2, 1])
&gt;&gt;&gt; x[np.argpartition(x, 3)]
array([2, 1, 3, 4])
&gt;&gt;&gt; x[np.argpartition(x, (1, 3))]
array([1, 2, 3, 4])
 
 &gt;&gt;&gt; x = [3, 4, 2, 1]
&gt;&gt;&gt; np.array(x)[np.argpartition(x, 3)]
array([2, 1, 3, 4])
 
Multi-dimensional array:
 &gt;&gt;&gt; x = np.array([[3, 4, 2], [1, 3, 1]])
&gt;&gt;&gt; index_array = np.argpartition(x, kth=1, axis=-1)
&gt;&gt;&gt; np.take_along_axis(x, index_array, axis=-1) # same as np.partition(x, kth=1)
array([[2, 3, 4],
 [1, 1, 3]])
numpy.argpartition#
https://numpy.org/doc/stable/reference/generated/numpy.argpartition.html
https://www.geeksforgeeks.org/numpy-argpartition-in-python/
numpy.argpartition() in Python
2018-12-26
Jana Sayantan Follow
<div><article>
<div>
<p>
</p>
</div>
<div>
<p><strong><code>numpy.argpartition() </code></strong>function is used to create a indirect partitioned copy of input array with its elements rearranged in such a way that the value of the element in k-th position is in the position it would be in a sorted array. All elements smaller than the k-th element are moved before this element and all equal or greater are moved behind it. The ordering of the elements in the two partitions is undefined.It returns an array of indices of the same shape as arr, i.e <code> arr[index_array] </code> yields a partition of arr.</p>
<blockquote><p>
<strong>Syntax : </strong> numpy.argpartition(arr, kth, axis=-1, kind=’introselect’, order=None)</p>
<p><strong>Parameters : </strong><br/>
<strong>arr : </strong>[array_like] Input array.<br/>
<strong>kth : </strong>[int or sequence of ints ] Element index to partition by.<br/>
<strong>axis : </strong>[int or None] Axis along which to sort. If None, the array is flattened before sorting. The default is -1, which sorts along the last axis.<br/>
<strong>kind : </strong>Selection algorithm. Default is ‘introselect’.<br/>
<strong>order : </strong>[str or list of str] When arr is an array with fields defined, this argument specifies which fields to compare first, second, etc.</p>
<p><strong>Return : </strong>[index_array, ndarray] Array of indices that partition arr along the specified axis.</p></blockquote>
<p><strong>Code #1 : </strong></p>
<div>
<p><code>import</code> <code>numpy as geek </code></p>
<p><code>in_arr </code><code>=</code> <code>geek.array([[ </code><code>2</code><code>, </code><code>0</code><code>,  </code><code>1</code><code>], [ </code><code>5</code><code>, </code><code>4</code><code>, </code><code>9</code><code>] ]) </code></p>
<p><code>print</code> <code>(</code><code>"Input array : \n"</code><code>, in_arr)  </code></p>
<p><code>out_arr </code><code>=</code> <code>geek.argpartition(in_arr, </code><code>1</code><code>, axis </code><code>=</code> <code>1</code><code>) </code></p>
<p><code>print</code> <code>(</code><code>"Output partitioned array indices :\n "</code><code>, out_arr) </code></p>
</div>
<div>
<p><b>Output:</b></p>
<pre>Input array : 
 [[2 0 1]
 [5 4 9]]
Output partitioned array indices :
 [[1 2 0]
 [1 0 2]]
</pre>
</div>
<p> <br/>
<strong>Code #2 : </strong></p>
<div>
<p><code>import</code> <code>numpy as geek </code></p>
<p><code>in_arr </code><code>=</code> <code>geek.array([ </code><code>2</code><code>, </code><code>0</code><code>,  </code><code>1</code><code>, </code><code>5</code><code>, </code><code>4</code><code>, </code><code>3</code><code>]) </code></p>
<p><code>print</code> <code>(</code><code>"Input array : "</code><code>, in_arr)  </code></p>
<p><code>out_arr </code><code>=</code> <code>geek.argpartition(in_arr, (</code><code>0</code><code>, </code><code>2</code><code>)) </code></p>
<p><code>print</code> <code>(</code><code>"Output partitioned array indices: "</code><code>, out_arr) </code></p>
</div>
<div>
<p><b>Output:</b></p>
<pre>Input array : [2 0 1 5 4 3]
Output partitioned array indices: [1 2 0 3 4 5]
</pre>
</div>
<br/> <br/>
<div>
<p>
</p>
</div> <div> <p> <h3>Please <u></u> to comment...</h3></p>
</div>
</div>
</article></div>||||I|||| Skip to content
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 numpy.argpartition() in Python

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 numpy.argpartition() function is used to create a indirect partitioned copy of input array with its elements rearranged in such a way that the value of the element in k-th position is in the position it would be in a sorted array. All elements smaller than the k-th element are moved before this element and all equal or greater are moved behind it. The ordering of the elements in the two partitions is undefined.It returns an array of indices of the same shape as arr, i.e arr[index_array] yields a partition of arr.

 Syntax : numpy.argpartition(arr, kth, axis=-1, kind=’introselect’, order=None)

 Parameters :
 arr : [array_like] Input array.
 kth : [int or sequence of ints ] Element index to partition by.
 axis : [int or None] Axis along which to sort. If None, the array is flattened before sorting. The default is -1, which sorts along the last axis.
 kind : Selection algorithm. Default is ‘introselect’.
 order : [str or list of str] When arr is an array with fields defined, this argument specifies which fields to compare first, second, etc.

 Return : [index_array, ndarray] Array of indices that partition arr along the specified axis.

 Code #1 :




 # Python program explaining 
 # argpartition() function 
 import numpy as geek 
 # input array 
 in_arr = geek.array([[ 2, 0, 1], [ 5, 4, 9] ]) 
 print ("Input array : \n", in_arr) 
 out_arr = geek.argpartition(in_arr, 1, axis = 1) 
 print ("Output partitioned array indices :\n ", out_arr)
 
 Output:
 
 Input array : 
 [[2 0 1]
 [5 4 9]]
 Output partitioned array indices :
 [[1 2 0]
 [1 0 2]]
 


 Code #2 :




 # Python program explaining 
 # argpartition() function 
 import numpy as geek 
 # input array 
 in_arr = geek.array([ 2, 0, 1, 5, 4, 3]) 
 print ("Input array : ", in_arr) 
 out_arr = geek.argpartition(in_arr, (0, 2)) 
 print ("Output partitioned array indices: ", out_arr)
 
 Output:
 
 Input array : [2 0 1 5 4 3]
 Output partitioned array indices: [1 2 0 3 4 5]
 


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numpy.argpartition()  function is used to create a indirect partitioned copy of input array with its elements rearranged in such a way that the value of the element in k-th position is in the position it would be in a sorted array. All elements smaller than the k-th element are moved before this element and all equal or greater are moved behind it. The ordering of the elements in the two partitions is undefined.It returns an array of indices of the same shape as arr, i.e  arr[index_array]  yields a partition of arr.
 
Syntax :  numpy.argpartition(arr, kth, axis=-1, kind=’introselect’, order=None)
 Parameters :  
arr : [array_like] Input array.
kth : [int or sequence of ints ] Element index to partition by.
axis : [int or None] Axis along which to sort. If None, the array is flattened before sorting. The default is -1, which sorts along the last axis.
kind : Selection algorithm. Default is ‘introselect’.
order : [str or list of str] When arr is an array with fields defined, this argument specifies which fields to compare first, second, etc.
 Return : [index_array, ndarray] Array of indices that partition arr along the specified axis. 
 Code #1 :  
 import numpy as geek  
 in_arr  = geek.array([[  2 ,  0 ,   1 ], [  5 ,  4 ,  9 ] ])  
 print ( "Input array : \n" , in_arr)   
 out_arr  = geek.argpartition(in_arr,  1 , axis  = 1 )  
 print ( "Output partitioned array indices :\n " , out_arr)  
 Output: 
Input array : 
 [[2 0 1]
 [5 4 9]]
Output partitioned array indices :
 [[1 2 0]
 [1 0 2]]
 
Code #2 :  
 import numpy as geek  
 in_arr  = geek.array([  2 ,  0 ,   1 ,  5 ,  4 ,  3 ])  
 print ( "Input array : " , in_arr)   
 out_arr  = geek.argpartition(in_arr, ( 0 ,  2 ))  
 print ( "Output partitioned array indices: " , out_arr)  
 Output: 
Input array : [2 0 1 5 4 3]
Output partitioned array indices: [1 2 0 3 4 5]
 
   Please   to comment...
numpy.argpartition() in Python
https://www.geeksforgeeks.org/numpy-argpartition-in-python/
https://numpy.org/doc/stable/reference/generated/numpy.argmax.html
numpy.argmax#
2022-01-01
None
<div><article>
 
 <section>
<dl>
<dt>
numpy.argmax(<em>a</em>, <em>axis=None</em>, <em>out=None</em>, <em>*</em>, <em>keepdims=&lt;no value&gt;</em>)<a href="https://github.com/numpy/numpy/blob/v1.24.0/numpy/core/fromnumeric.py#L1153-L1242">[source]</a><a href="#numpy.argmax">#</a></dt>
<dd><p>Returns the indices of the maximum values along an axis.</p>
<dl>
<dt>Parameters:</dt>
<dd><dl>
<dt><strong>a</strong>array_like</dt><dd><p>Input array.</p>
</dd>
<dt><strong>axis</strong>int, optional</dt><dd><p>By default, the index is into the flattened array, otherwise
along the specified axis.</p>
</dd>
<dt><strong>out</strong>array, optional</dt><dd><p>If provided, the result will be inserted into this array. It should
be of the appropriate shape and dtype.</p>
</dd>
<dt><strong>keepdims</strong>bool, optional</dt><dd><p>If this is set to True, the axes which are reduced are left
in the result as dimensions with size one. With this option,
the result will broadcast correctly against the array.</p>
<p>New in version 1.22.0.</p>
</dd>
</dl>
</dd>
<dt>Returns:</dt>
<dd><dl>
<dt><strong>index_array</strong>ndarray of ints</dt><dd><p>Array of indices into the array. It has the same shape as <em>a.shape</em>
with the dimension along <em>axis</em> removed. If <em>keepdims</em> is set to True,
then the size of <em>axis</em> will be 1 with the resulting array having same
shape as <em>a.shape</em>.</p>
</dd>
</dl>
</dd>
</dl>
<p>Notes</p>
<p>In case of multiple occurrences of the maximum values, the indices
corresponding to the first occurrence are returned.</p>
<p>Examples</p>
<div><pre>&gt;&gt;&gt; a = np.arange(6).reshape(2,3) + 10
&gt;&gt;&gt; a
array([[10, 11, 12],
 [13, 14, 15]])
&gt;&gt;&gt; np.argmax(a)
5
&gt;&gt;&gt; np.argmax(a, axis=0)
array([1, 1, 1])
&gt;&gt;&gt; np.argmax(a, axis=1)
array([2, 2])
</pre></div>
<p>Indexes of the maximal elements of a N-dimensional array:</p>
<div><pre>&gt;&gt;&gt; ind = np.unravel_index(np.argmax(a, axis=None), a.shape)
&gt;&gt;&gt; ind
(1, 2)
&gt;&gt;&gt; a[ind]
15
</pre></div>
<div><pre>&gt;&gt;&gt; b = np.arange(6)
&gt;&gt;&gt; b[1] = 5
&gt;&gt;&gt; b
array([0, 5, 2, 3, 4, 5])
&gt;&gt;&gt; np.argmax(b) # Only the first occurrence is returned.
1
</pre></div>
<div><pre>&gt;&gt;&gt; x = np.array([[4,2,3], [1,0,3]])
&gt;&gt;&gt; index_array = np.argmax(x, axis=-1)
&gt;&gt;&gt; # Same as np.amax(x, axis=-1, keepdims=True)
&gt;&gt;&gt; np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1)
array([[4],
 [3]])
&gt;&gt;&gt; # Same as np.amax(x, axis=-1)
&gt;&gt;&gt; np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1).squeeze(axis=-1)
array([4, 3])
</pre></div>
<p>Setting <em>keepdims</em> to <em>True</em>,</p>
<div><pre>&gt;&gt;&gt; x = np.arange(24).reshape((2, 3, 4))
&gt;&gt;&gt; res = np.argmax(x, axis=1, keepdims=True)
&gt;&gt;&gt; res.shape
(2, 1, 4)
</pre></div>
</dd></dl>
</section>
 </article>||||I|||| * User Guide
 * API reference
 * Development
 * Release notes
 * Learn
 1.24
 * GitHub
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 * Array objects
 * Array API Standard Compatibility
 * Constants
 * Universal functions ( ufunc )
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 + Array creation routines
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 o numpy.sort
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 On this page
 * argmax

 numpy.argmax#

 numpy. argmax ( a , axis=None , out=None , * , keepdims=<no value> ) [source] #

 Returns the indices of the maximum values along an axis.

 Parameters : a array_like

 Input array.

 axis int, optional

 By default, the index is into the flattened array, otherwise along the specified axis.

 out array, optional

 If provided, the result will be inserted into this array. It should be of the appropriate shape and dtype.

 keepdims bool, optional

 If this is set to True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the array.

 New in version 1.22.0.

 Returns : index_array ndarray of ints

 Array of indices into the array. It has the same shape as a.shape with the dimension along axis removed. If keepdims is set to True, then the size of axis will be 1 with the resulting array having same shape as a.shape.

 See also

 ndarray.argmax , argmin amax

 The maximum value along a given axis.

 unravel_index

 Convert a flat index into an index tuple.

 take_along_axis

 Apply np.expand_dims(index_array, axis) from argmax to an array as if by calling max.

 Notes

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

 Examples

 >>> a = np.arange(6).reshape(2,3) + 10
 >>> a
 array([[10, 11, 12],
 [13, 14, 15]])
 >>> np.argmax(a)
 5
 >>> np.argmax(a, axis=0)
 array([1, 1, 1])
 >>> np.argmax(a, axis=1)
 array([2, 2])
 

 Indexes of the maximal elements of a N-dimensional array:

 >>> ind = np.unravel_index(np.argmax(a, axis=None), a.shape)
 >>> ind
 (1, 2)
 >>> a[ind]
 15
 
 >>> b = np.arange(6)
 >>> b[1] = 5
 >>> b
 array([0, 5, 2, 3, 4, 5])
 >>> np.argmax(b) # Only the first occurrence is returned.
 1
 
 >>> x = np.array([[4,2,3], [1,0,3]])
 >>> index_array = np.argmax(x, axis=-1)
 >>> # Same as np.amax(x, axis=-1, keepdims=True)
 >>> np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1)
 array([[4],
 [3]])
 >>> # Same as np.amax(x, axis=-1)
 >>> np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1).squeeze(axis=-1)
 array([4, 3])
 

 Setting keepdims to True,

 >>> x = np.arange(24).reshape((2, 3, 4))
 >>> res = np.argmax(x, axis=1, keepdims=True)
 >>> res.shape
 (2, 1, 4)
 

 previous

 numpy.argpartition

 next

 numpy.nanargmax

 © Copyright 2008-2022, NumPy Developers.

 Created using Sphinx 5.3.0.

numpy.argmax(a, axis=None, out=None, *, keepdims=&lt;no value&gt;)[source] # 
 Returns the indices of the maximum values along an axis.
Parameters:
 
 aarray_like  Input array.
 axisint, optional  By default, the index is into the flattened array, otherwise
along the specified axis.
 outarray, optional  If provided, the result will be inserted into this array. It should
be of the appropriate shape and dtype.
 keepdimsbool, optional  If this is set to True, the axes which are reduced are left
in the result as dimensions with size one. With this option,
the result will broadcast correctly against the array.
New in version 1.22.0.
Returns:
 
 index_arrayndarray of ints  Array of indices into the array. It has the same shape as a.shape
with the dimension along axis removed. If keepdims is set to True,
then the size of axis will be 1 with the resulting array having same
shape as a.shape.
Notes
In case of multiple occurrences of the maximum values, the indices
corresponding to the first occurrence are returned.
Examples
 &gt;&gt;&gt; a = np.arange(6).reshape(2,3) + 10
&gt;&gt;&gt; a
array([[10, 11, 12],
 [13, 14, 15]])
&gt;&gt;&gt; np.argmax(a)
5
&gt;&gt;&gt; np.argmax(a, axis=0)
array([1, 1, 1])
&gt;&gt;&gt; np.argmax(a, axis=1)
array([2, 2])
 
Indexes of the maximal elements of a N-dimensional array:
 &gt;&gt;&gt; ind = np.unravel_index(np.argmax(a, axis=None), a.shape)
&gt;&gt;&gt; ind
(1, 2)
&gt;&gt;&gt; a[ind]
15
 
 &gt;&gt;&gt; b = np.arange(6)
&gt;&gt;&gt; b[1] = 5
&gt;&gt;&gt; b
array([0, 5, 2, 3, 4, 5])
&gt;&gt;&gt; np.argmax(b) # Only the first occurrence is returned.
1
 
 &gt;&gt;&gt; x = np.array([[4,2,3], [1,0,3]])
&gt;&gt;&gt; index_array = np.argmax(x, axis=-1)
&gt;&gt;&gt; # Same as np.amax(x, axis=-1, keepdims=True)
&gt;&gt;&gt; np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1)
array([[4],
 [3]])
&gt;&gt;&gt; # Same as np.amax(x, axis=-1)
&gt;&gt;&gt; np.take_along_axis(x, np.expand_dims(index_array, axis=-1), axis=-1).squeeze(axis=-1)
array([4, 3])
 
Setting keepdims to True,
 &gt;&gt;&gt; x = np.arange(24).reshape((2, 3, 4))
&gt;&gt;&gt; res = np.argmax(x, axis=1, keepdims=True)
&gt;&gt;&gt; res.shape
(2, 1, 4)
numpy.argmax#
https://numpy.org/doc/stable/reference/generated/numpy.argmax.html
https://gist.github.com/pervognsen/f656075edf028e75b6c1158564387612
nice.py
2019-01-17
Pervognsen
<div><div>
 
 <div>
 
 
 
 
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 <a href="https://github.co/hiddenchars">Learn more about bidirectional Unicode characters</a>
 
 <div> <a href="{{ revealButtonHref }}"> Show hidden characters
</a>
</div>
</div>
 
 
 
 <table>
 <tbody><tr>
 <td></td>
 <td> A = blocks(A, (w, h))</td>
 </tr>
 <tr>
 <td></td>
 <td> B = blocks(B, (h, w))</td>
 </tr>
 <tr>
 <td></td>
 <td> C = blocks(out, (w, w))</td>
 </tr>
 <tr>
 <td></td>
 <td> for j in range(B.shape[0]):</td>
 </tr>
 <tr>
 <td></td>
 <td> for k in range(B.shape[1]):</td>
 </tr>
 <tr>
 <td></td>
 <td> np.copyto(column_panel, B[j, k])</td>
 </tr>
 <tr>
 <td></td>
 <td> for i in range(A.shape[0]):</td>
 </tr>
 <tr>
 <td></td>
 <td> np.copyto(row_panel, A[i, j])</td>
 </tr>
 <tr>
 <td></td>
 <td> microkernel(C[i, k], row_panel, column_panel)</td>
 </tr>
 </tbody></table>
</div>||||I|||| Skip to content
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 for j in range ( B . shape [ 0 ]): 
 for k in range ( B . shape [ 1 ]): 
 np . copyto ( column_panel , B [ j , k ]) 
 for i in range ( A . shape [ 0 ]): 
 np . copyto ( row_panel , A [ i , j ]) 
 microkernel ( C [ i , k ], row_panel , column_panel )
 
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  A = blocks(A, (w, h))
 
 
  
  B = blocks(B, (h, w))
 
 
  
  C = blocks(out, (w, w))
 
 
  
  for j in range(B.shape[0]):
 
 
  
  for k in range(B.shape[1]):
 
 
  
  np.copyto(column_panel, B[j, k])
 
 
  
  for i in range(A.shape[0]):
 
 
  
  np.copyto(row_panel, A[i, j])
 
 
  
  microkernel(C[i, k], row_panel, column_panel)
nice.py
https://gist.github.com/pervognsen/f656075edf028e75b6c1158564387612