.. 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 : Language Reviewer : Bhanukiran Checklist OK? : <06-11-2010, Anand, OK> [2010-10-05] Script ------ .. L1 {{{ Show the first slide containing title, name of the production team along with the logo of MHRD }}} .. R1 Hello friends and Welcome to the tutorial on 'Accessing pieces of arrays'. .. L2 {{{ Show slide with objectives }}} .. R2 At the end of this tutorial, you will be able to, 1. Access and change individual elements of arrays, both one dimensional and multi-dimensional. #. Access and change rows and columns of arrays. #. Access and change other chunks from an array, using slicing and striding. #. Read images into arrays and perform processing on them, using simple array manipulations. .. L3 {{{ Switch to the pre-requisite slide }}} .. R3 Before beginning this tutorial,we would suggest you to complete the tutorial on "Getting started with arrays". .. R4 As usual, we start IPython, using .. L4 {{{ Open the terminal }}} :: ipython -pylab .. L5 {{ Show the slide with the arrays, A and C }} .. R5 Let us begin with the help of an example. Let us have two arrays, A and C, as the sample arrays that we will use to work through this tutorial. .. L6 :: 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]]) .. R6 Pause the recording and type the arrays A and C. also make sure that you have typed the arrays 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. .. R7 To access, the element 34 in array A, we say, A of 2, note that we are using square brackets. .. L7 :: A[2] .. R8 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 of 2,3. .. L8 :: C[2, 3] .. R9 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 34 to -34 in both A and C. To do this, we simply assign the new value after accessing the element. .. L9 :: A[2] = -34 C[2, 3] = -34 .. R10 Let us check our operations, .. L10 :: A[2] C[2,3] .. R11 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, .. L11 :: C[2] .. R12 How do we access the last row of C? We could say, .. L12 :: C[4] .. R13 or as with lists,we could use negative indexing and say, .. L13 :: C[-1] .. R14 Now, we could change the last row into all zeros, using either .. L14 :: C[-1] = [0, 0, 0, 0, 0] .. R15 or, we can use, .. L15 :: C[-1] = 0 .. R16 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 saying, .. L16 :: C[:, 2] .. R17 Pause the video here, try out the following exercise and resume the video. .. L17 .. L18 {{{ Show slide with exercise 1 }}} .. R18 Change the last column of C to zeros. .. R19 Switch to the terminal for solution.To change the entire last column of C to zeros, we simply say, .. L19 {{{ Continue from paused state }}} {{{ Switch to the terminal }}} :: C[:, -1] = 0 .. R20 Since A is one dimensional, rows and columns of A don't make much sense. It has just one row and A of colon gives the whole of A. .. L20 :: A[:] .. R21 Pause the video here, try out the following exercise and resume the video. .. L21 .. L22 {{{ show slide containing exercise 2 }}} .. R22 Change ``A`` to ``[11, 12, 13, 14, 15]``. .. R23 Switch to the terminal for solution. To change A, we say, .. L23 {{{continue from paused state}}} {{{ Switch to the terminal }}} :: A[:] = [11, 12, 13, 14, 15] .. R24 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 first navigate to that path in the OS and see what the image contains. .. L24 {{{ switch to the browser and show the image }}} {{{ switch back to the ipython terminal }}} .. R25 Let us now read the data in ``squares.png`` into the array ``I``. .. L25 :: I = imread('/home/fossee/squares.png') .. R26 We can see the contents of the image, using the command ``imshow``. We say, imshow(I) to see what has been read into ``I``. .. L26 :: imshow(I) .. R27 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 color map. To see that ``I`` is really, just an array, we say, I, at the prompt .. L27 :: I .. R28 We see that an array is displayed. To check the dimensions of any array, we can use ``.shape`` function. .. L28 :: I.shape .. R29 As we can see,we got the dimensions of the image.The image,``squares.png`` has the dimensions of 300x300. .. L29 .. L30 {{{ Switch to slide squares.png }}} {{{ Point at top-left quadrant of the image }}} .. R30 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, .. L31 :: C[0:3, 2] .. R31 C[0:3, 2] gives, 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. .. R32 Now, if we wish to access the elements of row with index 2, and in columns indexed 0 to 2 (included), we say, .. L32 :: C[2, 0:3] .. R33 Pause the video here, try out the following exercise and resume the video. .. L33 .. L34 {{{ show slide containing exercise 3 }}} .. R34 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]. Switch to the terminal for solution. .. L35 {{{continue from paused state}}} {{{ Switch to the terminal }}} :: C[1, 1:3] .. R35 C[1, 1:3] gives the elements [22, 23] .. L36 :: C[0:4, 0] .. R36 C[0:4, 0] gives the elements [11, 21, 31, 41] .. L37 :: C[1:5, 0] .. R37 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, .. L38 :: C[:4, 0] C[1:, 0] .. R38 We get the elements [21, 31, 41, 0]. If we skip both the indexes, we get the slice from end to end, as we already know. Pause the video here, try out the following exercise and resume the video. .. L39 {{{ show slide containing exercise 4 }}} .. R39 Obtain the elements [[23, 24], [33, -34]] from C. .. R40 Switch to the terminal for solution. .. L40 {{{continue from paused state}}} {{{ Switch to the terminal }}} :: C[1:3, 2:4] .. R41 C[1:3, 2:4] will give us the required elements. 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 is 300, we know that we need to get the first 150 rows and the first 150 columns. .. L41 :: I[:150, :150] .. R42 I[:150, :150] gives us the top-left corner of the image. .. R43 We use the ``imshow`` command to see the slice we obtained in the form of an image and confirm. .. L43 :: imshow(I[:150, :150]) .. R44 Pause the video here, try out the following exercise and resume the video. .. L44 .. L45 {{{ show slide containing exercise 5 }}} .. R45 Obtain the square in the center of the image. .. R46 Switch to the terminal for solution. .. L46 {{{continue from paused state}}} {{{ Switch to the terminal }}} :: imshow(I[75:225, 75:225]) .. R47 Hence, we get the center of the image. Our next goal is to compress the image, using a very simple technique, so as 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 one-fourth of the original data but losing only a little 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, .. L47 :: C[0:5:2, 0:5:2] .. R48 if we wish to be explicit, we say, .. L48 :: C[::2, ::2] .. R49 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. .. L49 :: C[1::2, ::2] .. R50 we get the elements, [[21, 23, 0], [41, 43, 0]] Pause the video here, try out the following exercise and resume the video. .. L50 .. L51 {{{ show slide containing exercise 6 }}} .. R51 Obtain the following. [[12, 0], [42, 0]] [[12, 13, 14], [0, 0, 0]] .. R52 The solution is on your screen. .. L52 {{{continue from paused state}}} {{{ show slide containing Solution 6 }}} .. R53 Now, that we know how to stride over an array, we can drop alternate rows and columns out of the image in I. .. L53 :: I[::2, ::2] .. R54 To see this image, we say, .. L54 :: imshow(I[::2, ::2]) .. R55 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. .. L55 :: imshow(I[::4, ::4]) .. L56 {{{ show summary slide }}} .. R56 This brings us to the end of this tutorial. In this tutorial, we have learnt to, 1. Manipulate single & multi dimensional arrays. #. Access and change individual elements by using their index numbers. #. Access and change rows and columns of arrays by specifying the row and column numbers. #. Slice and stride on arrays. #. Read images into arrays and manipulate them. .. L57 {{{Show self assessment questions slide}}} .. R57 Here are some self assessment questions for you to solve 1. Given the array, ``A = array([12, 15, 18, 21])``, how do we access the element ``18``? 2. Given the array, :: B = array([[10, 11, 12, 13], [20, 21, 22, 23], [30, 31, 32, 33], [40, 41, 42, 43]]) Obtain the elements, ``[[21, 22], [31, 32]]`` 3. Given the array, :: C = array([[10, 11, 12, 13], [20, 21, 22, 23]]) Change the array to :: C = array([[10, 11, 10, 11], [20, 21, 20, 21]]) .. L58 {{{solution of self assessment questions on slide}}} .. R58 And the answers, 1. The element 18 in array A has index number 2.Hence, we access it as A of 2 :: A[2] 2. To obtain the center four numbers in the array B, we say,B[1:3, 1:3] :: B[1:3, 1:3] 3. We can change the elements of array C,by using slicing and striding :: C[:2, 2:] = C[:2, :2] .. L59 {{{ Show the Thank you slide }}} .. R59 Hope you have enjoyed this tutorial and found it useful. Thank you!