author | amit |
Wed, 27 Oct 2010 15:13:17 +0530 | |
changeset 362 | a77a27916f81 |
parent 341 | 9f7eb1ed0e08 |
child 499 | fff4a90b2310 |
permissions | -rw-r--r-- |
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.. Objectives |
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.. ---------- |
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.. Plotting a least square fit line |
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.. Prerequisites |
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.. ------------- |
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.. 1. Basic Plotting |
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.. 2. Arrays |
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.. 3. Loading data from files |
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.. Author : Nishanth Amuluru |
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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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.. #[Puneeth: Add pre-requisites.] |
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Script |
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------ |
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Hello friends and welcome to the tutorial on Least Square Fit |
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{{{ Show the slide containing title }}} |
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{{{ Show the slide containing the outline slide }}} |
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In this tutorial, we shall look at generating the least square fit line for a |
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given set of points. |
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First let us have a look at the problem. |
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{{{ Show the slide containing problem statement. }}} |
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We have an input file generated from a simple pendulum experiment. |
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It contains two columns of data. The first column is the length of the |
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pendulum and the second is the corresponding time period of the pendulum. |
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As we know, the square of time period of a pendulum is directly proportional to |
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its length, we shall plot l vs t^2 and verify this. |
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.. #[Puneeth:] removed the explanation about loadtxt and unpack |
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.. option. It's been done in another LO already. simple dependency |
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.. should work? |
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To read the input file and parse the data, we are going to use the |
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loadtxt function. Type |
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:: |
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l, t = loadtxt("/home/fossee/pendulum.txt", unpack=True) |
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l |
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t |
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We can see that l and t are two sequences containing length and time values |
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correspondingly. |
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Let us first plot l vs t^2. Type |
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:: |
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tsq = t * t |
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plot(l, tsq, 'bo') |
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{{{ switch to the plot window }}} |
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.. #[Puneeth:] Moved explanation of least square fit here. seems more |
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.. apt. |
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We can see that there is a visible linear trend, but we do not get a |
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straight line connecting them. We shall, therefore, generate a least |
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square fit line. |
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{{{ show the slide containing explanation on least square fit }}} |
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As shown in the slide, we are first going to generate the two matrices |
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tsq and A. Then we are going to use the ``lstsq`` function to find the |
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values of m and c. |
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let us now generate the A matrix with l values. |
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We shall first generate a 2 x 90 matrix with the first row as l values and the |
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second row as ones. Then take the transpose of it. Type |
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:: |
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inter_mat = array((l, ones_like(l))) |
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inter_mat |
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We see that we have intermediate matrix. Now we need the transpose. Type |
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:: |
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A = inter_mat.T |
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A |
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Now we have both the matrices A and tsq. We only need to use the ``lstsq`` |
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Type |
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:: |
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result = lstsq(A, tsq) |
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The result is a sequence of values. The first item in this sequence, |
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is the matrix p i.e., the values of m and c. Hence, |
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:: |
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m, c = result[0] |
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m |
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c |
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Now that we have m and c, we need to generate the fitted values of t^2. Type |
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:: |
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tsq_fit = m * l + c |
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plot(l, tsq, 'bo') |
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plot(l, tsq_fit, 'r') |
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We get the least square fit of l vs t^2 |
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{{{ Pause here and try out the following exercises }}} |
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%% 2 %% change the label on y-axis to "y" and save the lines of code |
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accordingly |
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{{{ continue from paused state }}} |
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{{{ Show summary slide }}} |
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This brings us to the end of the tutorial. |
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we have learnt |
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|
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* how to generate a least square fit |
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|
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{{{ Show the "sponsored by FOSSEE" slide }}} |
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.. #[Nishanth]: Will add this line after all of us fix on one. |
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.. This tutorial was created as a part of FOSSEE project, NME ICT, MHRD India |
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|
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Hope you have enjoyed and found it useful. |
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Thank you |
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|
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.. |
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Local Variables: |
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mode: rst |
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indent-tabs-mode: nil |
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sentence-end-double-space: nil |
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fill-column: 75 |
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End: |