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ML4DS_week1_lab_with_solutions.ipynb
- ML4DS Lab: week 1: Simple linear regression lab
- Aims
- Tasks
- Task 1: Download and import the olympic data
- Task 2: Plot the data
- Task 3. Fit a straightline using `LinearRegression` in `sklean`. Plot the model with the data and prediction at 2012.
- Task 5: Fit the model with the Least Square solution. Plot the model with the data and prediction at 2012.
//lab/week1/
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- ML4DS_week1_lab_with_solutions.ipynb
<h3>ML4DS Lab: week 1: Simple linear regression lab</h3>
#### Aims
* Do a simple linear regression on the Olympic 100~m data in Python
* Practise numpy, matplotlib, and sklearn
#### Tasks
* Download the data (‘olympic100m.txt’) from the Moodle page
* Plot Olympic year against winning time
* Plot the loss function in 1D and 2D
* Fit a model with sklearn functions
* Fit a model with using provided expressions to compute $w_0$ and $w_1$.
* Create a new plot that includes the data and the function defined by $w_0$ and $w_1$
* Make a prediction at 2012
ML4DS Lab: week 1: Simple linear regression lab¶
ML4DS 实验：第 1 周：简单线性回归实验¶
Aims¶ 目标¶
- Do a simple linear regression on the Olympic 100~m data in Python
在 Python 中对奥林匹克 100~m 数据进行简单的线性回归 - Practise numpy, matplotlib, and sklearn
练习 numpy、matplotlib 和 sklearn
Tasks¶ 任务¶
- Download the data (‘olympic100m.txt’) from the Moodle page
从 Moodle 页面下载数据 （'olympic100m.txt'） - Plot Olympic year against winning time
绘制奥林匹克年与获胜时间 - Plot the loss function in 1D and 2D
在 1D 和 2D 中绘制损失函数 - Fit a model with sklearn functions
使用 sklearn 函数拟合模型 - Fit a model with using provided expressions to compute
使用提供的表达式拟合模型以计算 - Create a new plot that includes the data and the function defined by
创建一个新绘图，其中包含由 - Make a prediction at 2012
在 2012 年做出预测
#### Task 1: Download and import the olympic data
Task 1: Download and import the olympic data¶
任务 1：下载并导入奥林匹克数据¶
[2]:
import numpy as np
data = np.loadtxt('olympic100m.txt', delimiter=',') # load olympic data
data
[2]:
array([[1896. , 12. ],
[1900. , 11. ],
[1904. , 11. ],
[1906. , 11.2 ],
[1908. , 10.8 ],
[1912. , 10.8 ],
[1920. , 10.8 ],
[1924. , 10.6 ],
[1928. , 10.8 ],
[1932. , 10.3 ],
[1936. , 10.3 ],
[1948. , 10.3 ],
[1952. , 10.4 ],
[1956. , 10.5 ],
[1960. , 10.2 ],
[1964. , 10. ],
[1968. , 9.95],
[1972. , 10.14],
[1976. , 10.06],
[1980. , 10.25],
[1984. , 9.99],
[1988. , 9.92],
[1992. , 9.96],
[1996. , 9.84],
[2000. , 9.87],
[2004. , 9.85],
[2008. , 9.69]])#### Task 2: Plot the data
Task 2: Plot the data¶
任务 2：绘制数据¶
[9]:
%matplotlib inline
import pylab as plt
x = data[:,0]
t = data[:,1]
plt.scatter(x,t) # draw a scatter plot
plt.xlabel('Years') # always label x&y-axis
plt.ylabel('Time (seconds)') # always label x&y-axis
(27,)
[9]:
Text(0, 0.5, 'Time (seconds)')
#### Task 3. Fit a straightline using `LinearRegression` in `sklean`. Plot the model with the data and prediction at 2012.
Task 3. Fit a straightline using LinearRegression in sklean. Plot the model with the data and prediction at 2012.¶
任务 3.在 sklean 中使用 LinearRegression 拟合直线。使用 2012 年的数据和预测绘制模型。
[4]:
from sklearn.linear_model import LinearRegression # import
x = x[:,None] # 27 x 1 array
t = t[:,None] # 27 x 1 array
print(t.shape)
print(x.shape)
reg = LinearRegression().fit(x, t)
print([reg.intercept_, reg.coef_])
(27, 1) (27, 1) [array([36.4164559]), array([[-0.01333089]])]
[5]:
x_test = np.linspace(1880,2020, 100) # generate new x to plot the fitted line. Note better not to use the original x !
plt.plot(x_test,reg.predict(x_test[:,None]),'b-',linewidth=2) # plot the fitted data
plt.plot(2012,reg.predict(np.array([[2012]])),'ro') # plot the fitted data
plt.scatter(x,t) # draw a scatter plot
plt.xlabel('Years') # always label x&y-axis
plt.ylabel('Time (seconds)') # always label x&y-axis
[5]:
Text(0, 0.5, 'Time (seconds)')
#### Task 5: Fit the model with the Least Square solution. Plot the model with the data and prediction at 2012.
Let's fit a model with an analytical solution to the problem of finding the parameters with the minimum average loss.
Recall that the average loss is $$ L(w_0, w_1) = \frac{1}{N}\sum_{n=1}^N (t_n - w_0 - w_1x_n)^2 $$
$L$ is a function of $w_0$ and $w_1$. All $x_n$ and $t_n$ are given.
The procedure to find the analytical expression of the optimal parameters is the following:
- Solving $$\frac{\partial L(w_0, w_1)}{\partial w_0} = 0, \quad \frac{\partial L(w_0, w_1)}{\partial w_1} = 0 $$
- the average loss is minimised:
$$ w_1 = \frac{\bar{x}\bar{t} - \bar{xt}}{\bar{x}\bar{x} - \bar{x^2}} $$
and
$$ w_0 = \bar{t} - w_1\bar{x} $$
where $\bar{z} = \frac{1}{N}\sum_{n=1}^N z_n$.
You are encouraged to derive these yourself
Task 5: Fit the model with the Least Square solution. Plot the model with the data and prediction at 2012.¶
任务 5：使用最小二乘法拟合模型。使用 2012 年的数据和预测绘制模型。
Let's fit a model with an analytical solution to the problem of finding the parameters with the minimum average loss.
让我们用解析解拟合一个模型，以解决找到平均损失最小的参数的问题。
Recall that the average loss is
回想一下，平均损失为
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The procedure to find the analytical expression of the optimal parameters is the following:
求最佳参数的解析表达式的过程如下：
- Solving
- the average loss is minimised:
平均损失最小化：
You are encouraged to derive these yourself
我们鼓励您自己推导这些
[6]:
xbar = x.mean()
tbar = t.mean()
xxbar = np.multiply(x,x).mean()
xtbar = np.multiply(x,t).mean()
print(xbar)
print(tbar)
print(xxbar)
print(xtbar)
1952.3703703703704 10.389629629629631 3812975.5555555555 20268.06814814815
[7]:
w1 = (tbar*xbar - xtbar)/(xbar*xbar - xxbar)
w0 = tbar-w1*xbar
print(w0)
print(w1)
36.41645590250286 -0.013330885710960602
[8]:
x_test = np.linspace(1880,2020, 100) # generate new x to plot the fitted line. Note better not to use the original x !
plt.plot(x_test,w0+w1*x_test,'b-',linewidth=2) # plot the fitted data
plt.plot(2012,w0+w1*2012,'ro') # plot the fitted data
plt.scatter(x,t) # draw a scatter plot
plt.xlabel('Years') # always label x&y-axis
plt.ylabel('Time (seconds)') # always label x&y-axis
[8]:
Text(0, 0.5, 'Time (seconds)')
[ ]: