accessing-pieces-arrays/script.rst
author Puneeth Chaganti <punchagan@fossee.in>
Wed, 06 Oct 2010 19:28:46 +0530
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Added pre-reqs for accessing pieces of arrays LO.
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.. Objectives
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.. ----------
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   .. By the end of this tutorial, you will be able to:
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   ..   1. Access and change individual elements of arrays, both one
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   ..   dimensional and multi-dimensional.
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   ..   2. Access and change rows and columns of arrays. 
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   ..   3. Access and change other chunks from an array, using slicing
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   ..   and striding. 
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   ..   4. Read images into arrays and perform processing on them, using
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   ..   simple array manipulations. 
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.. Prerequisites
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.. -------------
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..   1. getting started with arrays
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.. Author              : Puneeth
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   Internal Reviewer   : 
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   External Reviewer   :
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   Checklist OK?       : <put date stamp here, if OK> [2010-10-05]
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Script
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------
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{{{ Screen shows welcome slide }}}
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Welcome to the tutorial on accessing pieces of arrays
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{{{ Show the outline for this tutorial }}} 
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In this tutorial we shall learn to access individual elements of
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arrays, get rows and columns and other chunks of arrays using
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slicing and striding. 
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{{{ switch back to the terminal }}}
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As usual, we start IPython, using 
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::
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  ipython -pylab 
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Let us have two arrays, A and C, as the sample arrays that we will
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use to work through this tutorial. 
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::
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  A = array([12, 23, 34, 45, 56])
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  C = array([[11, 12, 13, 14, 15],
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             [21, 22, 23, 24, 25],
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             [31, 32, 33, 34, 35],
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             [41, 42, 43, 44, 45],
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             [51, 52, 53, 54, 55]])
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Pause the video here and make sure you have the arrays A and C,
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typed in correctly.
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Let us begin with the most elementary thing, accessing individual
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elements. Also, let us first do it with the one-dimensional array
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A, and then do the same thing with the two-dimensional array. 
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To access, the element 34 in A, we say, 
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::
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  A[1]
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Like lists, indexing starts from 0 in arrays, too. So, 34, the
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third element has the index 2. 
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Now, let us access the element 34 from C. To do this, we say
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::
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  C[2, 3]
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34 is in the third row and the fourth column, and since indexing
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begins from zero, the row index is 2 and column index is 3. 
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Now, that we have accessed one element of the array, let us change
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it. We shall change the 34 to -34 in both A and C. To do this, we
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simply assign the new value after accessing the element. 
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::
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  A[2] = -34
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  C[2, 3] = -34
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Now that we have accessed and changed a single element, let us
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access and change more than one element at a time; first rows and
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then columns.
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Let us access one row of C, say the third row. We do it by saying, 
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  C[2] 
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How do we access the last row of C? We could say,
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  C[4] 
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for the fifth row, or as with lists, use negative indexing and say
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  C[-1]
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Now, we could change the last row into all zeros, using either 
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  C[-1] = [0, 0, 0, 0, 0]
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or 
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::
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  C[-1] = 0
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Now, how do we access one column of C? As with accessing
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individual elements, the column is the second parameter to be
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specified (after the comma). The first parameter, is now replaced
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with a ``:`` to say, that we want all the elements of that
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dimension, instead of one particular element. We access the third
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column by
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::
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  C[:, 2]
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Following is an exercise that you must do. 
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%%1%% Change the last column of C to zeroes. 
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Please, pause the video here. Do the exercises and then continue. 
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::
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  C[:, -1] = 0
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Since A is one dimensional, rows and columns of A don't make much
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sense. It has just one row and 
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::
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  A[:] 
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gives the whole of A. 
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Following is an exercise that you must do. 
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%%2%% Change ``A`` to ``[11, 12, 13, 14, 15]``. 
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Please, pause the video here. Do the exercises and then continue. 
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To change A, we say
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::
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  A[:] = [11, 12, 13, 14, 15]
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Now, that we know how to access, rows and columns of an array, we
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shall learn how to access other pieces of an array. For this
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purpose, we will be using image arrays. 
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To read an image into an array, we use the ``imread`` command. We
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shall use the image ``squares.png`` present in ``/home/fossee``. We
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shall first navigate to that path in the OS and see what the image
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contains. 
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{{{ switch to the browser and show the image }}}
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{{{ switch back to the ipython terminal }}}
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Let us now read the data in ``squares.png`` into the array ``I``. 
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::
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  I = imread('/home/fossee/squares.png')
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We can see the contents of the image, using the command
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``imshow``. We say, 
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::
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  imshow(I) 
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to see what has been read into ``I``. We do not see white and black
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because, ``pylab`` has mapped white and black to different
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colors. This can be changed by using a different colormap. 
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To see that ``I`` is really, just an array, we say, 
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::
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  I 
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at the prompt, and see that an array is displayed. 
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To check the dimensions of any array, we can use the method
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shape. We say
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::
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  I.shape 
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to get the dimensions of the image. As we can see, ``squares.png``
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has the dimensions of 300x300. 
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Our goal for this part of the tutorial would be to get the
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top-left quadrant of the image. To do this, we need to access, a
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few of the rows and a few of the columns of the array. 
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To access, the third column of C, we said, ``C[:, 2]``. Essentially,
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we are accessing all the rows in column three of C. Now, let us
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modify this to access only the first three rows, of column three
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of C. 
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We say, 
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::
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  C[0:3, 2]
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to get the elements of rows indexed from 0 to 3, 3 not included
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and column indexed 2. Note that, the index before the colon is
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included and the index after it is not included, in the slice that
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we have obtained. This is very similar to the ``range`` function,
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where ``range`` returns a list, in which the upper limit or stop
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value is not included.
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Now, if we wish to access the elements of row with index 2, and in
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columns indexed 0 to 2 (included), we say, 
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::
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  C[2, 0:3]
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Following is an exercise that you must do. 
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%%3%% First, obtain the elements [22, 23] from C. Then, obtain the
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elements [11, 21, 31, 41] from C. Finally, obtain the elements [21,
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31, 41, 0]. 
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Please, pause the video here. Do the exercises and then continue. 
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::
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  C[1, 1:3] 
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gives the elements [22, 23]
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::
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  C[0:4, 0]
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gives the elements [11, 21, 31, 41]
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::
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  C[1:5, 0]
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gives the elements [21, 31, 41, 0]
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Note that when specifying ranges, if you are starting from or
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going up-to the end, the corresponding element may be dropped. So,
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in the previous example to obtain [11, 21, 31, 41], we could have
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simply said, 
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::
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  C[:4, 0]
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b595f90016c5 Changed structure of my scripts.
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and 
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::
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  C[1:, 0]
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gives the elements [21, 31, 41, 0]. If we skip both the indexes,
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we get the slice from end to end, as we already know. 
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Following is an exercise that you must do. 
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%%4%% Obtain the elements [[23, 24], [33, -34]] from C. 
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Please, pause the video here. Do the exercises and then continue. 
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::
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  C[1:3, 2:4] 
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gives us the elements, [[23, 24], [33, -34]]. 
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Now, we wish to obtain the top left quarter of the image. How do
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we go about doing it? Since, we know the shape of the image to be
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300, we know that we need to get the first 150 rows and first 150
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columns. 
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parents:
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::
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  I[:150, :150]
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gives us the top-left corner of the image. 
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We use the ``imshow`` command to see the slice we obtained in the
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form of an image and confirm. 
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::
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  imshow(I[:150, :150])
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Following is an exercise that you must do. 
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%%5%% Pause the video here, and obtain the square in the center
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of the image. 
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Following is an exercise that you must do. 
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::
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  imshow(I[75:225, 75:225])
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Our next goal is to compress the image, using a very simple
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technique to reduce the space that the image takes on disk while
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not compromising too heavily on the image quality. The idea is to
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drop alternate rows and columns of the image and save it. This way
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we will be reducing the data to a fourth of the original data but
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losing only so much of visual information. 
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We shall first learn the idea of striding using the smaller array
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C. Suppose we wish to access only the odd rows and columns (first,
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third, fifth). We do this by, 
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::
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  C[0:5:2, 0:5:2]
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if we wish to be explicit, or simply, 
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::
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  C[::2, ::2]
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This is very similar to the step specified to the ``range``
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function. It specifies, the jump or step in which to move, while
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accessing the elements. If no step is specified, a default value
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of 1 is assumed. 
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::
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  C[1::2, ::2] 
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gives the elements, [[21, 23, 0], [41, 43, 0]]
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Following is an exercise that you must do. 
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%%6%% Obtain the following. 
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[[12, 0], [42, 0]]
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[[12, 13, 14], [0, 0, 0]]
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Please, pause the video here. Do the exercises and then continue. 
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::
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  C[::3, 1::3]
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gives the elements [[12, 0], [42, 0]]
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::
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  C[::4, 1:4]
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gives the elements [[12, 13, 14], [0, 0, 0]]
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Now, that we know how to stride over an image, we can drop
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alternate rows and columns out of the image in I. 
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::
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  I[::2, ::2]
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To see this image, we say, 
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::
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  imshow(I[::2, ::2])
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This does not have much data to notice any real difference, but
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notice that the scale has reduced to show that we have dropped
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alternate rows and columns. If you notice carefully, you will be
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able to observe some blurring near the edges. To notice this
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effect more clearly, increase the step to 4. 
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::
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  imshow(I[::4, ::4])
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{{{ show summary slide }}}
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That brings us to the end of this tutorial. In this tutorial, we
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have learnt to access parts of arrays, specifically individual
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elements, rows and columns and larger pieces of arrays. We have
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also learnt how to modify arrays, element wise or in larger
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pieces.
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Thank You!