diff options
| -rw-r--r-- | accessing_parts_of_arrays/script.rst | 92 | ||||
| -rw-r--r-- | getting_started_with_arrays/script.rst | 81 | ||||
| -rw-r--r-- | matrices/script.rst | 18 | ||||
| -rw-r--r-- | parsing_data/script.rst | 98 |
4 files changed, 148 insertions, 141 deletions
diff --git a/accessing_parts_of_arrays/script.rst b/accessing_parts_of_arrays/script.rst index c4a74fe..aeb1938 100644 --- a/accessing_parts_of_arrays/script.rst +++ b/accessing_parts_of_arrays/script.rst @@ -317,7 +317,7 @@ We can see the contents of the image, using the command 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. +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 @@ -460,6 +460,10 @@ Pause the video here, try out the following exercise and resume the video. Obtain the elements [[23, 24], [33, -34]] from C. +.. R40 + +Switch to the terminal for solution. + .. L40 {{{continue from paused state}}} @@ -468,10 +472,8 @@ Pause the video here, try out the following exercise and resume the video. C[1:3, 2:4] -.. R40 +.. R41 -Switch to the terminal for solution. -<Type the command> 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 @@ -484,35 +486,39 @@ columns. I[:150, :150] -.. R41 +.. R42 I[:150, :150] gives us the top-left corner of the image. -.. R42 +.. R43 We use the ``imshow`` command to see the slice we obtained in the form of an image and confirm. -.. L42 +.. L43 :: imshow(I[:150, :150]) -.. R43 +.. R44 Pause the video here, try out the following exercise and resume the video. -.. L43 - .. L44 +.. L45 + {{{ show slide containing exercise 5 }}} -.. R44 +.. R45 Obtain the square in the center of the image. -.. L45 +.. R46 + +Switch to the terminal for solution. + +.. L46 {{{continue from paused state}}} {{{ Switch to the terminal }}} @@ -520,10 +526,8 @@ Pause the video here, try out the following exercise and resume the video. imshow(I[75:225, 75:225]) -.. R45 +.. R47 -Switch to the terminal for solution. -<Type the command> Hence, we get the center of the image. Our next goal is to compress the image, using a very simple @@ -537,78 +541,78 @@ 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, -.. L46 +.. L47 :: C[0:5:2, 0:5:2] -.. R46 - -.. R47 +.. R48 if we wish to be explicit, we say, -.. L47 +.. L48 :: C[::2, ::2] -.. R48 +.. 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. -.. L48 +.. L49 :: C[1::2, ::2] -.. R49 +.. R50 we get the elements, [[21, 23, 0], [41, 43, 0]] Pause the video here, try out the following exercise and resume the video. -.. L49 - .. L50 +.. L51 + {{{ show slide containing exercise 6 }}} -.. R50 +.. R51 Obtain the following. [[12, 0], [42, 0]] [[12, 13, 14], [0, 0, 0]] -.. L51 +.. R52 + +The solution is on your screen. + +.. L52 {{{continue from paused state}}} {{{ show slide containing Solution 6 }}} -.. R51 - -The solution is on your screen. +.. R53 Now, that we know how to stride over an array, we can drop alternate rows and columns out of the image in I. -.. L52 +.. L53 :: I[::2, ::2] -.. R52 +.. R54 To see this image, we say, -.. L53 +.. L54 :: imshow(I[::2, ::2]) -.. R53 +.. 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 @@ -616,18 +620,16 @@ 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. -.. L54 +.. L55 :: imshow(I[::4, ::4]) -.. R54 - -.. L55 +.. L56 {{{ show summary slide }}} -.. R55 +.. R56 This brings us to the end of this tutorial. In this tutorial, we have learnt to, @@ -639,11 +641,11 @@ have learnt to, #. Slice and stride on arrays. #. Read images into arrays and manipulate them. -.. L56 +.. L57 {{{Show self assessment questions slide}}} -.. R56 +.. R57 Here are some self assessment questions for you to solve @@ -675,11 +677,11 @@ Change the array to B = array([[10, 11, 10, 11], [20, 21, 20, 21]]) -.. L57 +.. L58 {{{solution of self assessment questions on slide}}} -.. R57 +.. R58 And the answers, @@ -699,11 +701,11 @@ And the answers, B[:2, 2:] = B[:2, :2] -.. L58 +.. L59 {{{ Show the Thank you slide }}} -.. R58 +.. R59 Hope you have enjoyed this tutorial and found it useful. Thank you! diff --git a/getting_started_with_arrays/script.rst b/getting_started_with_arrays/script.rst index 242ac8e..84c1f6b 100644 --- a/getting_started_with_arrays/script.rst +++ b/getting_started_with_arrays/script.rst @@ -192,15 +192,14 @@ Pause the video here, try out the following exercise and resume the video. Find out the shape of the other arrays i.e. a1, a3, ar that we have created. -.. L15 - -{{{ Continue from paused state }}} - .. R15 -It can be done as, +Switch to the terminal for solution -.. L16 +.. L15 + +{{{ Continue from paused state }}} +{{{ Switch to the terminal }}} :: a1.shape @@ -212,7 +211,7 @@ It can be done as, Now let us try to create a new array with a mix of elements and see what will happen, -.. L17 +.. L16 :: a4 = array([1,2,3,'a string']) @@ -224,7 +223,7 @@ arrays handle elements with the same datatype, but it didn't raise an error. Let us check the values in the new array created. Type a4 in the terminal, -.. L18 +.. L17 :: a4 @@ -240,6 +239,8 @@ Also,if you have noticed,we got something like 'dtype S8' in the output. dtype is nothing but the datatype which is the minimum type required to hold the objects in the sequence. +.. L18 + .. L19 {{{ switch to the next slide, identity & zeros methods }}} @@ -257,6 +258,8 @@ The function ``identity()`` takes an integer argument which specifies the size of the desired matrix, .. L20 + +{{{ Switch to the terminal }}} :: identity(3) @@ -289,19 +292,18 @@ Pause the video here, try out the following exercise and resume the video. .. R22 -We learned two functions ``identity()`` and ``zeros()``, find out more -about the functions ``zeros_like()``, ``ones()``, ``ones_like()``. +Find out about the functions + - zeros_like() + - ones() + - ones_like() -.. L23 - -{{{ continue from paused state }}} -{{{ Switch to the terminal }}} +< pause for some time and then continue > .. R23 Try the following, first check the value of a1, -.. L24 +.. L23 :: a1 @@ -311,16 +313,17 @@ Try the following, first check the value of a1, We see that ``a1`` is a single dimensional array, Let us now try a1*2 -.. L25 +.. L24 :: a1 * 2 .. R25 + It returned a new array with all the elements multiplied by 2. Now let us again check the contents of a1 -.. L26 +.. L25 :: a1 @@ -329,17 +332,15 @@ Now let us again check the contents of a1 note that the value of a1 still remains the same. -.. R27 - Similarly with addition, -.. L27 +.. L26 :: a1 + 2 a1 -.. R28 +.. R27 it returns a new array, with all the elements summed with two. But again notice that the value of a1 has not been changed. @@ -347,53 +348,53 @@ again notice that the value of a1 has not been changed. You may change the value of a1 by simply assigning the newly returned array as, -.. L28 +.. L27 :: a1 += 2 -.. R29 +.. R28 -Notice the change in elements of a, +Notice the change in elements of a by typing 'a' -.. L29 +.. L28 :: a -.. R30 +.. R29 We can use all the mathematical operations with arrays, Now let us try this -.. L30 +.. L29 :: a1 = array([1,2,3,4]) a2 = array([1,2,3,4]) a1 + a2 -.. R31 +.. R30 This returns an array with element by element addition -.. L31 +.. L30 :: a1 * a2 -.. R32 +.. R31 a1*a2 returns an array with element by element multiplication, notice that it does not perform matrix multiplication. -.. L32 +.. L31 -.. L33 +.. L32 {{{ switch to summary slide }}} -.. R33 +.. R32 This brings us to the end of the end of this tutorial.In this tutorial, we have learnt to, @@ -408,13 +409,13 @@ we have learnt to, - zeros() & zeros_like() - ones() & ones_like() -.. L34 +.. L33 {{{Show self assessment questions slide}}} -.. R34 +.. R33 -Here are some self assessment questionss for you to solve +Here are some self assessment questions for you to solve 1. ``x = array([1, 2, 3], [5, 6, 7])`` is a valid statement @@ -435,11 +436,11 @@ Here are some self assessment questionss for you to solve - Both statement A and B are correct. - Both statement A and B are incorrect. -.. L35 +.. L34 {{{solution of self assessment questions on slide}}} -.. R35 +.. R34 And the answers, @@ -453,11 +454,11 @@ And the answers, 2. The function ``ones_like()`` returns an array of ones with the same shape and type as a given array. -.. L36 +.. L35 {{{ switch to thank you slide }}} -.. R36 +.. R35 Hope you have enjoyed this tutorial and found it useful. Thank you! diff --git a/matrices/script.rst b/matrices/script.rst index 36bc001..ac8c455 100644 --- a/matrices/script.rst +++ b/matrices/script.rst @@ -144,10 +144,10 @@ m3 can be created as, .. R11 -Let us now move to matrix matrix operations. +Let us now move to matrix operations. We can do matrix addition and subtraction easily. m3+m2 does element by element addition, that is matrix addition. -Note that both the matrices are of the same order. +Note that both the matrices should be of the same order. .. L11 :: @@ -187,8 +187,8 @@ Matrix multiplication in matrices are done using the function ``dot()`` .. R15 -Due to size mismatch the multiplication could not be done and it -returned an error, +Due to size mismatch, the multiplication could not be done and it +returned an error. Now let us see an example for matrix multiplication. For doing matrix multiplication we need to have two matrices of the order n by m and m @@ -306,7 +306,7 @@ And the Frobenius norm of the matrix ``im5`` can be found out as, .. R25 -Thus we have successfully obtained the frobenius norm of the matrix m5 +Thus we have successfully obtained the Frobenius norm of the matrix m5 Pause the video here, try out the following exercise and resume the video. @@ -355,7 +355,7 @@ The norm of a matrix can be found out using the method .. R30 -Inorder to find out the Frobenius norm of the matrix im5, +In order to find out the Frobenius norm of the matrix im5, we do, .. L30 @@ -377,7 +377,7 @@ And to find out the Infinity norm of the matrix im5, we do, .. R32 This is easier when compared to the code we wrote. Read the documentation -of ``norm`` to read up more about ord and the possible type of norms +of ``norm`` to read up more about ``ord`` and the possible type of norms the norm function produces. Now let us find out the determinant of a the matrix m5. @@ -545,10 +545,10 @@ And the answers, 2. False. ``eig(A)[0]`` and ``eigvals(A)`` are same, that is both will give the - eigen values of matrrix A. + eigen values of matrix A. 3. ``norm(A,ord='fro')`` and ``norm(A)`` are same, since the order='fro' - stands for frobenius norm. Hence true. + stands for Frobenius norm. Hence true. .. L45 diff --git a/parsing_data/script.rst b/parsing_data/script.rst index fdbdc35..87a337f 100644 --- a/parsing_data/script.rst +++ b/parsing_data/script.rst @@ -57,7 +57,7 @@ Invoke the ipython interpreter by typing ipython on your terminal. .. L4 -{{{ Open the tutorial }}} +{{{ Open the terminal }}} :: ipython @@ -73,8 +73,6 @@ corresponds to a student. {{{ Open the file sslc.txt and show }}} -.. R5 - .. L6 {{{ show the slide 'Data set' }}} @@ -88,14 +86,13 @@ language,first language, maths, science and social and total marks. Our job is to calculate the arithmetic mean of all the maths marks in the region "B". -Now let us understand, what is meant by 'parsing data'. - .. L7 {{{ Open the file sslc.txt and show }}} .. R7 +Now let us understand, what is meant by 'parsing data'. From the input file, we can see that the data we have is in the form of text. Parsing this data is all about reading it and converting it into a form which can be used for computations -- in our case,it will be a @@ -135,8 +132,8 @@ all the spaces are treated as one big space. .. R11 -the function ``split`` can also split on a string of our choice. -This is acheived by passing that as an argument. But first lets define +The function ``split`` can also split on a string of our choice. +This is achieved by passing that as an argument. But first lets define a sample record from the file. .. L11 @@ -167,18 +164,18 @@ Pause the video here, try out the following exercise and resume the video Split the variable line using a space as argument. Is it same as splitting without an argument ? +.. R14 + +Switch to terminal for the solution + .. L14 {{{ continue from paused state }}} - {{{ Switch to the terminal }}} - :: record.split() -.. R14 - .. L15 {{{ Show slide with Solution 1 }}} @@ -186,7 +183,7 @@ Pause the video here, try out the following exercise and resume the video .. R15 We see that when we split on space, multiple whitespaces are not clubbed -as one and there is an empty string everytime there are two consecutive +as one and there is an empty string every time there are two consecutive spaces. .. L16 @@ -213,7 +210,7 @@ a string by typing .. R17 -We can see that strip removes all the whitespace around the sentence +We can see that strip removes all the whitespace around the sentence. Pause the video here, try out the following exercise and resume the video @@ -225,23 +222,27 @@ Pause the video here, try out the following exercise and resume the video What happens to the white space inside the sentence when it is stripped +.. R19 + +Switch to the terminal for solution + .. L19 {{{ continue from paused state }}} - {{{ Switch to the terminal }}} - :: a_str = " white space " a_str.strip() -.. R19 +.. R20 We see that, the whitespace inside the sentence is only removed and the rest remains unaffected. -.. R20 +.. L20 + +.. R21 By now we know enough to separate fields from the record and to strip out any white space. The only road block we now have, is conversion of @@ -253,14 +254,14 @@ and mathematical operations are not possible on them. We must convert them into numbers (integers or floats), before we can perform mathematical operations on them. -.. L20 +.. L21 -.. R21 +.. R22 -We shall look at converting strings into floats. We define a float string +We shall now look at converting strings into floats. We define a float string first. Type -.. L21 +.. L22 :: mark_str = "1.25" @@ -268,39 +269,42 @@ first. Type type(mark_str) type(mark) -.. R22 +.. R23 We can see that string is converted to float. We can perform mathematical operations on them now. Pause the video here, try out the following exercise and resume the video -.. L22 +.. L23 -.. R23 +.. L24 {{{ Show slide with Exercise 3 }}} -.. L23 +.. R24 What happens if you do int("1.25") -.. L24 +.. R25 -{{{ continue from paused state }}} +Switch to the terminal for solution +.. L25 + +{{{ continue from paused state }}} {{{ Switch to the terminal }}} :: int("1.25") -.. R24 +.. R26 It raises an error since converting a float string into integer directly is not possible. It involves an intermediate step of converting to float. Hence we will have to do the following conversions. -.. L25 +.. L26 :: dcml_str = "1.25" @@ -309,10 +313,10 @@ Hence we will have to do the following conversions. number = int(flt) number -.. R25 - .. R26 +.. R27 + Using ``int``, it is also possible to convert float into integers. Now that we have all the machinery required to parse the file, let us @@ -320,9 +324,9 @@ solve the problem. We first read the file line by line and parse each record. We then see if the region code is B and store the marks accordingly. -.. L26 - .. L27 + +.. L28 :: math_marks_B = [] # an empty list to store the marks @@ -338,46 +342,46 @@ accordingly. if region_code == "B": math_marks_B.append(math_mark) -.. R27 - .. R28 +.. R29 + Now we have all the math marks of region "B" in the list math_marks_B. To get the mean, we just have to sum the marks and divide by the length. -.. L28 +.. L29 :: math_marks_mean = sum(math_marks_B) / len(math_marks_B) math_marks_mean -.. R29 +.. R30 Hence we get our final output. This is how we split and read such a huge data and perform computations on it. -.. L29 - .. L30 +.. L31 + {{{ Show summary slide }}} -.. R30 +.. R31 This brings us to the end of the tutorial. In this tutorial, we have learnt to, 1. Tokenize a string using various delimiters like semi-colons. - #. Split a data seperated by delimiters by using the function ``split()``. + #. Split a data separated by delimiters by using the function ``split()``. #. Get rid of extra white spaces around using the ``strip()`` function. #. Convert datatypes of numbers from one type to another. #. Parse input data and perform computations on it. -.. L31 +.. L32 {{{Show self assessment questions slide}}} -.. R31 +.. R32 Here are some self assessment questions for you to solve @@ -394,11 +398,11 @@ Here are some self assessment questions for you to solve - Error - "20" -.. L32 +.. L33 {{{solution of self assessment questions on slide}}} -.. R32 +.. R33 And the answers, @@ -411,11 +415,11 @@ And the answers, 3. int("20.0") will give an error, because converting a float string, 20.0, directly into integer is not possible. -.. L33 +.. L34 {{{ Show the Thank you slide }}} -.. R33 +.. R34 Hope you have enjoyed this tutorial and found it useful. Thank you. |