diff options
| -rw-r--r-- | lstsq.rst | 24 |
1 files changed, 16 insertions, 8 deletions
@@ -27,7 +27,8 @@ As shown in the slide, we are first going to generate the two matrices tsq and A. Then we are going to use the =lstsq= function to find the values of m and c. To read the input file and parse the data, we are going to loadtxt function. -Type:: +Type +:: data = loadtxt("/home/fossee/pendulum.txt") data @@ -40,7 +41,8 @@ all the time values. Hence we have to use the unpack option with loadtxt. It unpacks the data into sequences depending on the structure of data. -Type:: +Type +:: l, t = loadtxt("/home/fossee/pendulum.txt", unpack=True) l @@ -49,7 +51,8 @@ Type:: We can see that l and t are two sequences containing length and time values correspondingly. -Let us first plot l vs t^2. Type:: +Let us first plot l vs t^2. Type +:: tsq = t * t plot(l, tsq, 'bo') @@ -61,29 +64,34 @@ We can see that there is a visible linear trend. let us now generate the A matrix with l values. We shall first generate a 2 x 90 matrix with the first row as l values and the -second row as ones. Then take the transpose of it. Type:: +second row as ones. Then take the transpose of it. Type +:: inter_mat = array((l, ones_like(l))) inter_mat -We see that we have intermediate matrix. Now we need the transpose.Type:: +We see that we have intermediate matrix. Now we need the transpose.Type +:: A = inter_mat.T A Now we have both the matrices A and tsq. We only need to use the =lstsq= -Type:: +Type +:: result = lstsq(A, tsq) The result is a sequence of values. The first item is the matrix p or in simple -words, the values of m and c. Hence, :: +words, the values of m and c. Hence, +:: m, c = result[0] m c -Now that we have m and c, we need to generate the fitted values of t^2. Type:: +Now that we have m and c, we need to generate the fitted values of t^2. Type +:: tsq_fit = m * l + c plot(l, tsq, 'bo') |