.. Objectives

.. ----------

   .. By the end of this tutorial, you will be able to:

   ..   1. Access and change individual elements of arrays, both one

   ..   dimensional and multi-dimensional.

   ..   2. Access and change rows and columns of arrays. 

   ..   3. Access and change other chunks from an array, using slicing

   ..   and striding. 

   ..   4. Read images into arrays and perform processing on them, using

   ..   simple array manipulations. 

.. Prerequisites

.. -------------

..   1. getting started with arrays

.. Author              : Puneeth

   Internal Reviewer   : 

   External Reviewer   :

   Checklist OK?       : <put date stamp here, if OK> [2010-10-05]

Script

------

{{{ Screen shows welcome slide }}}

Welcome to the tutorial on accessing pieces of arrays

{{{ Show the outline for this tutorial }}} 

In this tutorial we shall learn to access individual elements of

arrays, get rows and columns and other chunks of arrays using

slicing and striding. 

{{{ switch back to the terminal }}}

As usual, we start IPython, using 

::

  ipython -pylab 

{{ Show the slide with the arrays, A and C }}

Let us have two arrays, A and C, as the sample arrays that we will

use to work through this tutorial. 

::

  A = array([12, 23, 34, 45, 56])

  C = array([[11, 12, 13, 14, 15],

             [21, 22, 23, 24, 25],

             [31, 32, 33, 34, 35],

             [41, 42, 43, 44, 45],

             [51, 52, 53, 54, 55]])

Pause the video here and make sure you have the arrays A and C,

typed in correctly.

Let us begin with the most elementary thing, accessing individual

elements. Also, let us first do it with the one-dimensional array

A, and then do the same thing with the two-dimensional array. 

To access, the element 34 in A, we say, 

::

  A[2]

Like lists, indexing starts from 0 in arrays, too. So, 34, the

third element has the index 2. 

Now, let us access the element 34 from C. To do this, we say

::

  C[2, 3]

34 is in the third row and the fourth column, and since indexing

begins from zero, the row index is 2 and column index is 3. 

Now, that we have accessed one element of the array, let us change

it. We shall change the 34 to -34 in both A and C. To do this, we

simply assign the new value after accessing the element. 

::

  A[2] = -34

  C[2, 3] = -34

Now that we have accessed and changed a single element, let us

access and change more than one element at a time; first rows and

then columns.

Let us access one row of C, say the third row. We do it by saying, 

::

  C[2] 

How do we access the last row of C? We could say,

::

  C[4] 

for the fifth row, or as with lists, use negative indexing and say

::

  C[-1]

Now, we could change the last row into all zeros, using either 

::

  C[-1] = [0, 0, 0, 0, 0]

or 

::

  C[-1] = 0

Now, how do we access one column of C? As with accessing individual

elements, the column is the second parameter to be specified (after

the comma). The first parameter, is replaced with a ``:``. This

specifies that we want all the elements of that dimension, instead of

just one particular element. We access the third column by

::

  C[:, 2]

Following is an exercise that you must do. 

{{ show slide containing Question 1}} 

%%1%% Change the last column of C to zeroes. 

Please, pause the video here. Do the exercises and then continue. 

::

  C[:, -1] = 0

Since A is one dimensional, rows and columns of A don't make much

sense. It has just one row and 

::

  A[:] 

gives the whole of A. 

Following is an exercise that you must do. 

{{ show slide containing Question 2}} 

%%2%% Change ``A`` to ``[11, 12, 13, 14, 15]``. 

Please, pause the video here. Do the exercises and then continue. 

To change A, we say

::

  A[:] = [11, 12, 13, 14, 15]

Now, that we know how to access, rows and columns of an array, we

shall learn how to access other pieces of an array. For this

purpose, we will be using image arrays. 

To read an image into an array, we use the ``imread`` command. We

shall use the image ``squares.png`` present in ``/home/fossee``. We

shall first navigate to that path in the OS and see what the image

contains. 

{{{ switch to the browser and show the image }}}

{{{ switch back to the ipython terminal }}}

Let us now read the data in ``squares.png`` into the array ``I``. 

::

  I = imread('/home/fossee/squares.png')

We can see the contents of the image, using the command

``imshow``. We say, 

::

  imshow(I) 

to see what has been read into ``I``. We do not see white and black

because, ``pylab`` has mapped white and black to different

colors. This can be changed by using a different colormap. 

To see that ``I`` is really, just an array, we say, 

::

  I 

at the prompt, and see that an array is displayed. 

To check the dimensions of any array, we can use ``.shape``. We say

::

  I.shape 

to get the dimensions of the image. As we can see, ``squares.png``

has the dimensions of 300x300. 

Our goal for this part of the tutorial would be to get the

top-left quadrant of the image. To do this, we need to access, a

few of the rows and a few of the columns of the array. 

To access, the third column of C, we said, ``C[:, 2]``. Essentially,

we are accessing all the rows in column three of C. Now, let us

modify this to access only the first three rows, of column three

of C. 

We say, 

::

  C[0:3, 2]

to get the elements of rows indexed from 0 to 3, 3 not included

and column indexed 2. Note that, the index before the colon is

we have obtained. This is very similar to the ``range`` function,

where ``range`` returns a list, in which the upper limit or stop

value is not included.

Now, if we wish to access the elements of row with index 2, and in

columns indexed 0 to 2 (included), we say, 

::

  C[2, 0:3]

Following is an exercise that you must do. 

{{ show slide containing Question 3 }} 

%%3%% First, obtain the elements [22, 23] from C. Then, obtain the

elements [11, 21, 31, 41] from C. Finally, obtain the elements [21,

31, 41, 0]. 

Please, pause the video here. Do the exercises and then continue. 

{{ show slide containing Solution 3 }}

::

  C[1, 1:3] 

gives the elements [22, 23]

::

  C[0:4, 0]

gives the elements [11, 21, 31, 41]

::

  C[1:5, 0]

gives the elements [21, 31, 41, 0]

Note that when specifying ranges, if you are starting from the

beginning or going up-to the end, the corresponding element may be

dropped. So, in the previous example to obtain [11, 21, 31, 41], we

could have simply said, ::

  C[:4, 0]

and 

::

  C[1:, 0]

gives the elements [21, 31, 41, 0]. If we skip both the indexes,

we get the slice from end to end, as we already know. 

Following is an exercise that you must do. 

{{ show slide containing Question 4 }} 

%%4%% Obtain the elements [[23, 24], [33, -34]] from C. 

Please, pause the video here. Do the exercises and then continue. 

{{ show slide containing Solution 4 }} 

::

  C[1:3, 2:4] 

gives us the elements, [[23, 24], [33, -34]]. 

Now, we wish to obtain the top left quarter of the image. How do

we go about doing it? Since, we know the shape of the image to be

300, we know that we need to get the first 150 rows and first 150

columns. 

::

  I[:150, :150]

gives us the top-left corner of the image. 

We use the ``imshow`` command to see the slice we obtained in the

form of an image and confirm. 

::

  imshow(I[:150, :150])

Following is an exercise that you must do. 

{{ show slide containing Question 5 }} 

%%5%% Obtain the square in the center of the image.

Please, pause the video here. Do the exercises and then continue. 

{{ show slide containing Solution 5 }} 

::

  imshow(I[75:225, 75:225])

Our next goal is to compress the image, using a very simple

technique to reduce the space that the image takes on disk while

not compromising too heavily on the image quality. The idea is to

drop alternate rows and columns of the image and save it. This way

we will be reducing the data to a fourth of the original data but

losing only so much of visual information. 

We shall first learn the idea of striding using the smaller array

C. Suppose we wish to access only the odd rows and columns (first,

third, fifth). We do this by, 

::

  C[0:5:2, 0:5:2]

if we wish to be explicit, or simply, 

::

  C[::2, ::2]

This is very similar to the step specified to the ``range``

function. It specifies, the jump or step in which to move, while

accessing the elements. If no step is specified, a default value

of 1 is assumed. 

::

  C[1::2, ::2] 

gives the elements, [[21, 23, 0], [41, 43, 0]]

{{ show slide containing Question 6 }} 

Following is an exercise that you must do. 

%%6%% Obtain the following. 

[[12, 0], [42, 0]]

[[12, 13, 14], [0, 0, 0]]

Please, pause the video here. Do the exercises and then continue. 

{{ show slide containing Solution 6 }} 

::

  C[::3, 1::3]

gives the elements [[12, 0], [42, 0]]

::

  C[::4, 1:4]

gives the elements [[12, 13, 14], [0, 0, 0]]

Now, that we know how to stride over an array, we can drop

alternate rows and columns out of the image in I. 

::

  I[::2, ::2]

To see this image, we say, 

::

  imshow(I[::2, ::2])

This does not have much data to notice any real difference, but

notice that the scale has reduced to show that we have dropped

alternate rows and columns. If you notice carefully, you will be

able to observe some blurring near the edges. To notice this

effect more clearly, increase the step to 4. 

::

  imshow(I[::4, ::4])

{{{ show summary slide }}}

That brings us to the end of this tutorial. In this tutorial, we

have learnt to access parts of arrays, specifically individual

elements, rows and columns and larger pieces of arrays. We have

also learnt how to modify arrays, element wise or in larger

pieces.

{{{ Show the "sponsored by FOSSEE" slide }}}

This tutorial was created as a part of FOSSEE project, NME ICT, MHRD India

Hope you have enjoyed and found it useful.

Thank you!

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