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+++ b/accessing-pieces-arrays/script.rst Wed Oct 06 15:16:09 2010 +0530
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+.. Objectives
+.. ----------
+
+.. Clearly state the objectives of the LO (along with RBT level)
+
+.. Prerequisites
+.. -------------
+
+.. 1. Name of LO-1
+.. 2. Name of LO-2
+.. 3. Name of LO-3
+
+.. 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
+
+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[1]
+
+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 now replaced
+with a ``:`` to say, that we want all the elements of that
+dimension, instead of one particular element. We access the third
+column by
+
+::
+
+ C[:, 2]
+
+Following is an exercise that you must do.
+
+%%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.
+
+%%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 the method
+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
+included and the index after it is not included, in the slice that
+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.
+
+%%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.
+
+::
+
+ 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 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.
+
+%%4%% Obtain the elements [[23, 24], [33, -34]] from C.
+
+Please, pause the video here. Do the exercises and then continue.
+
+::
+
+ 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.
+
+%%5%% Pause the video here, and obtain the square in the center
+of the image.
+
+Following is an exercise that you must do.
+
+::
+
+ 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]]
+
+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.
+
+::
+
+ 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 image, 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.
+
+Thank You!