### Ridge Regression

In this notebook, we will run ridge regression multiple times with different L2 penalties to see which one produces the best fit. We will revisit the example of polynomial regression as a means to see the effect of L2 regularization.

outline for this notebook

• we will use a pre-built implementation of regression (GraphLab Create) to run polynomial regression
• we will use matplotlib to visualize polynomial regressions
• we will use a pre-built implementation of regression (GraphLab Create) to run polynomial regression, this time with L2 penalty
• we will use matplotlib to visualize polynomial regressions under L2 regularization
• we will choose best L2 penalty using cross-validation.
• we aill assess the final fit using test data.

We will continue to use the House data from previous notebooks. (In the next programming assignment for this module, you will implement your own ridge regression learning algorithm using gradient descent.)

### import library

%matplotlib inline
import warnings
warnings.filterwarnings('ignore')
import pandas as pd
import numpy as np
import seaborn as sns
sns.set(color_codes=True)
from sklearn.linear_model import Ridge
import matplotlib.pyplot as plt


Now, we will read data that we will use in this notebook. This data is house sales in King County, the region where the city of Seattle, WA is located.

df = pd.read_csv("kc_house_data.csv")
colname_lst = list(df.columns.values)
coltype_lst =  [str, str, float, float, float, float, int, str, int, int, int, int, int, int, int, int, str, float, float, float, float]
col_type_dict = dict(zip(colname_lst, coltype_lst))

id date price bedrooms bathrooms sqft_living sqft_lot floors waterfront view ... grade sqft_above sqft_basement yr_built yr_renovated zipcode lat long sqft_living15 sqft_lot15
0 7129300520 20141013T000000 221900 3 1.00 1180 5650 1 0 0 ... 7 1180 0 1955 0 98178 47.5112 -122.257 1340 5650
1 6414100192 20141209T000000 538000 3 2.25 2570 7242 2 0 0 ... 7 2170 400 1951 1991 98125 47.7210 -122.319 1690 7639
2 5631500400 20150225T000000 180000 2 1.00 770 10000 1 0 0 ... 6 770 0 1933 0 98028 47.7379 -122.233 2720 8062
3 2487200875 20141209T000000 604000 4 3.00 1960 5000 1 0 0 ... 7 1050 910 1965 0 98136 47.5208 -122.393 1360 5000
4 1954400510 20150218T000000 510000 3 2.00 1680 8080 1 0 0 ... 8 1680 0 1987 0 98074 47.6168 -122.045 1800 7503

5 rows × 21 columns

we will also sort our data by “price” and “sqft_living”. This will help us with better visualisation.

data = df.sort(['sqft_living', 'price'], ascending=[1, 1])

id date price bedrooms bathrooms sqft_living sqft_lot floors waterfront view ... grade sqft_above sqft_basement yr_built yr_renovated zipcode lat long sqft_living15 sqft_lot15
19452 3980300371 20140926T000000 142000 0 0.00 290 20875 1 0 0 ... 1 290 0 1963 0 98024 47.5308 -121.888 1620 22850
15381 2856101479 20140701T000000 276000 1 0.75 370 1801 1 0 0 ... 5 370 0 1923 0 98117 47.6778 -122.389 1340 5000
860 1723049033 20140620T000000 245000 1 0.75 380 15000 1 0 0 ... 5 380 0 1963 0 98168 47.4810 -122.323 1170 15000
18379 1222029077 20141029T000000 265000 0 0.75 384 213444 1 0 0 ... 4 384 0 2003 0 98070 47.4177 -122.491 1920 224341
4868 6896300380 20141002T000000 228000 0 1.00 390 5900 1 0 0 ... 4 390 0 1953 0 98118 47.5260 -122.261 2170 6000

5 rows × 21 columns

### polynomial regression function

Now we will create a polynomial function for later use. This function will create the polynomial of the terget feature up to the given degree.

def polynomial_dataframe(feature, degree):
# assume that degree >= 1
# and set polynomial_df['power_1'] equal to the passed feature
# use deep copy here. otherwise, it will do shallow copy.
polynomial_df = feature.copy(deep=True)
polynomial_df.columns = ["power_1"]
# first check if degree > 1
if degree > 1:
# then loop over the remaining degrees:
# range usually starts at 0 and stops at the endpoint-1. We want it to start at 2 and stop at degree
for power in range(2, degree+1):
# first we'll give the column a name:
name = 'power_' + str(power)
# then assign polynomial_df[name] to the appropriate power of feature
polynomial_df[name]=feature.apply(lambda x: x**power)
return polynomial_df


We generate polynomial features up to degree 15 using polynomial_dataframe() and fit a model with these features. When fitting the model, we will use an L2 penalty of 1e-5:

l2_small_penalty = 1e-5


Note: When we have so many features and so few data points, the solution can become highly numerically unstable, which can sometimes lead to strange unpredictable results. Thus, rather than using no regularization, we will introduce a tiny amount of regularization (l2_penalty=1e-5) to make the solution numerically stable.

With the L2 penalty specified above, we will fit the model and print out the learned weights.

def fit15_deg_poly(data, l2_penalty):
poly15_data = polynomial_dataframe(pd.DataFrame(data["sqft_living"]), 15)
features15 = list(poly15_data.columns.values) # get the name of the features
poly15_data["price"] = data["price"] # add price to the data since it's the target
# Create linear regression object
reg15 = Ridge(alpha=l2_penalty, solver='svd')
#train model
reg15.fit(poly15_data[features15], poly15_data.iloc[:,(len(poly15_data.columns)-1)].to_frame())
print("intercept "+str(reg15.intercept_))
print("coefficient "+str(reg15.coef_))
#let's make the prediction first, then plot prediction
poly15_data["predicted"] = reg15.predict(poly15_data[features15])
plt.plot(poly15_data["power_1"],poly15_data["price"],"b.",
poly15_data["power_1"], poly15_data["predicted"],"c-")

fit15_deg_poly(data, l2_small_penalty)

intercept [ 156830.73690457]
coefficient [[  1.35096850e+02  -1.09078329e-02   1.16379466e-05  -7.32516116e-10
7.86140736e-16   7.25121286e-16  -2.67247443e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


### observe overfitting

The polynomial fit of degree 15 changed wildly whenever the data changed. In particular, when we split the data and fit the model of degree 15, the result came out to be very different for each subset. The model had a high variance. This is where ridge regression kicks in because it reduces such variance. But first, we will reproduce such cases.

First, split the data into split the data into four subsets of roughly equal size and call them set_1, set_2, set_3, and set_4.

idx = np.random.rand(len(data))<0.5
semi_split1 = data[idx]; semi_split2 = data[~idx]
idx = np.random.rand(len(semi_split1))<0.5
set_1 = semi_split1[idx]; set_2 = semi_split1[~idx]
idx = np.random.rand(len(semi_split2))<0.5
set_3 = semi_split2[idx]; set_4 = semi_split2[~idx]


Next, fit a 15th degree polynomial on set_1, set_2, set_3, and set_4, using ‘sqft_living’ to predict prices. Print the weights and make a plot of the resulting model.

#set 1
fit15_deg_poly(set_1, l2_small_penalty)

intercept [ 214847.27784093]
coefficient [[  8.79395504e+01  -8.91792713e-03   1.73505356e-05  -1.63867300e-09
-9.52928073e-15   3.41267009e-16  -1.09453289e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 2
fit15_deg_poly(set_2, l2_small_penalty)

intercept [ 117345.07723391]
coefficient [[  2.19657759e+02  -5.45342928e-02   2.00656617e-05  -1.20189180e-09
-1.04445880e-14   3.91844475e-16  -1.28687054e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 3
fit15_deg_poly(set_3, l2_small_penalty)

intercept [ 274777.57344825]
coefficient [[ -1.79365225e+02   1.48652979e-01  -2.01886970e-05   1.32309772e-09
2.69638258e-14   5.23834725e-16  -1.12635400e-14   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 4
fit15_deg_poly(set_4, l2_small_penalty)

intercept [ 275790.92011891]
coefficient [[ -9.87019267e+01   1.04674177e-01  -9.05370886e-06   4.31084780e-10
-1.97719200e-14   3.26295135e-16  -5.90832396e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


The four curves should differ from one another a lot, as should the coefficients you learned.

### ridge regression comes to rescue

Generally, whenever we see weights change so much in response to change in data, we believe the variance of our estimate to be large. Ridge regression aims to address this issue by penalising “large” weights. Note that weights of model15 looked quite small, but they are not that small because ‘sqft_living’ input is in the order of thousands.

With the argument l2_penalty=1e5, fit a 15th-order polynomial model on set_1, set_2, set_3, and set_4. Other than the change in the l2_penalty parameter, the code should be the same as the experiment above.

#define new l2 penaly
l2_penalty = 1e10

#set 1
fit15_deg_poly(set_1, l2_penalty)

intercept [ 250818.50246403]
coefficient [[  8.81891273e-02   3.30201687e-02   9.44020901e-06  -1.14140943e-09
-8.46120591e-15  -1.61276169e-17  -3.47763588e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 2
fit15_deg_poly(set_2, l2_penalty)

intercept [ 215546.17905305]
coefficient [[  4.51670991e-01   2.93502739e-02   8.46473490e-06  -7.19484779e-10
-1.95592769e-14  -1.39990461e-15  -8.48861065e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 3
fit15_deg_poly(set_3, l2_penalty)

intercept [ 187052.22395021]
coefficient [[ -2.49140949e-01   7.12380348e-02  -7.53020689e-06   6.58253134e-10
3.17578659e-14   1.49678286e-15  -7.20349010e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 4
fit15_deg_poly(set_4, l2_penalty)

intercept [ 229258.33434574]
coefficient [[ -1.74749728e-01   6.50438805e-02  -3.20023745e-06   1.65131759e-10
-1.60408116e-14   9.99168000e-16  -3.15793547e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


#set 4
fit15_deg_poly(set_4, l2_small_penalty)

intercept [ 275790.92011891]
coefficient [[ -9.87019267e+01   1.04674177e-01  -9.05370886e-06   4.31084780e-10
-1.97719200e-14   3.26295135e-16  -5.90832396e-15   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00   0.00000000e+00
0.00000000e+00   0.00000000e+00   0.00000000e+00]]


### selecting an L2 penalty via cross-validation

As seen, with randon L2 parameter, we cannot see obvious improvement in our dataset. The L2 penalty is a “magic” parameter we need to select. We could use the validation set approach but that approach has a major disadvantage: it leaves fewer observations available for training. Cross-validation seeks to overcome this issue by using all of the training set in a smart way.

We will implement a kind of cross-validation called k-fold cross-validation. The method gets its name because it involves dividing the training set into k segments of roughtly equal size. Similar to the validation set method, we measure the validation error with one of the segments designated as the validation set. The major difference is that we repeat the process k times as follows:

Set aside segment 0 as the validation set, and fit a model on rest of data, and evalutate it on this validation set
Set aside segment 1 as the validation set, and fit a model on rest of data, and evalutate it on this validation set

Set aside segment k-1 as the validation set, and fit a model on rest of data, and evalutate it on this validation set

After this process, we compute the average of the k validation errors, and use it as an estimate of the generalization error. Notice that all observations are used for both training and validation, as we iterate over segments of data.

To estimate the generalization error well, it is crucial to shuffle the training data before dividing them into segments.

train_valid_shuffled = data.iloc[np.random.permutation(len(data))]

id date price bedrooms bathrooms sqft_living sqft_lot floors waterfront view ... grade sqft_above sqft_basement yr_built yr_renovated zipcode lat long sqft_living15 sqft_lot15
21183 5528600005 20150327T000000 272167 2 2.50 1620 3795 2.0 0 0 ... 7 1620 0 2014 0 98027 47.5321 -122.034 1620 6000
8673 4151800530 20141028T000000 1090000 4 2.50 2780 6837 2.0 0 0 ... 9 2780 0 2004 0 98033 47.6660 -122.201 1160 6837
9277 8129700644 20140703T000000 715000 3 4.00 2080 2250 3.0 0 4 ... 8 2080 0 1997 0 98103 47.6598 -122.355 2080 2250
6677 2795000080 20140919T000000 535100 3 2.25 2070 7207 1.0 0 0 ... 8 1720 350 1973 0 98177 47.7735 -122.371 2350 7980
1369 3374300070 20140623T000000 334000 4 1.50 1150 9360 1.5 0 0 ... 6 1150 0 1970 0 98034 47.7197 -122.173 1480 8155

5 rows × 21 columns

Once the data is shuffled, we divide it into equal segments. Each segment should receive n/k elements, where n is the number of observations in the training set and k is the number of segments. Since the segment 0 starts at index 0 and contains n/k elements, it ends at index (n/k)-1. The segment 1 starts where the segment 0 left off, at index (n/k). With n/k elements, the segment 1 ends at index (n*2/k)-1. Continuing in this fashion, we deduce that the segment i starts at index (n*i/k) and ends at (n*(i+1)/k)-1.

With this pattern in mind, we write a short loop that prints the starting and ending indices of each segment, just to make sure you are getting the splits right.

n = len(train_valid_shuffled)
k = 10 # 10-fold cross-validation

for i in xrange(k):
start = (n*i)/k
end = (n*(i+1))/k-1
print i, (start, end)

0 (0, 2160)
1 (2161, 4321)
2 (4322, 6482)
3 (6483, 8644)
4 (8645, 10805)
5 (10806, 12966)
6 (12967, 15128)
7 (15129, 17289)
8 (17290, 19450)
9 (19451, 21612)


Let’s familiarise ourselves with array slicing with dataframe. To extract a continuous slice from an data, use colon in square brackets. For instance, the following cell extracts rows 0 to 9 of train_valid_shuffled. Notice that the first index (0) is included in the slice but the last index (10) is omitted.

tmp_df = train_valid_shuffled[0:10] # rows 0 to 9
tmp_df

id date price bedrooms bathrooms sqft_living sqft_lot floors waterfront view ... grade sqft_above sqft_basement yr_built yr_renovated zipcode lat long sqft_living15 sqft_lot15
21183 5528600005 20150327T000000 272167 2 2.50 1620 3795 2.0 0 0 ... 7 1620 0 2014 0 98027 47.5321 -122.034 1620 6000
8673 4151800530 20141028T000000 1090000 4 2.50 2780 6837 2.0 0 0 ... 9 2780 0 2004 0 98033 47.6660 -122.201 1160 6837
9277 8129700644 20140703T000000 715000 3 4.00 2080 2250 3.0 0 4 ... 8 2080 0 1997 0 98103 47.6598 -122.355 2080 2250
6677 2795000080 20140919T000000 535100 3 2.25 2070 7207 1.0 0 0 ... 8 1720 350 1973 0 98177 47.7735 -122.371 2350 7980
1369 3374300070 20140623T000000 334000 4 1.50 1150 9360 1.5 0 0 ... 6 1150 0 1970 0 98034 47.7197 -122.173 1480 8155
13586 1226059101 20140701T000000 502000 3 2.25 1600 45613 2.0 0 0 ... 8 1600 0 1983 0 98072 47.7523 -122.117 2320 43005
16297 9346900170 20140922T000000 615000 4 2.25 2330 7020 1.0 0 0 ... 8 1450 880 1973 0 98006 47.5620 -122.139 2330 8500
1101 2197600451 20141105T000000 631000 5 2.00 2270 2400 2.0 0 0 ... 7 2270 0 1905 0 98122 47.6051 -122.319 1320 2400
15516 579000595 20140906T000000 724000 2 1.00 1560 5000 1.5 0 1 ... 7 1560 0 1942 0 98117 47.7006 -122.386 2620 5400
3013 3423049209 20150318T000000 200450 3 1.00 970 9130 1.0 0 0 ... 6 970 0 1957 0 98188 47.4369 -122.272 1000 8886

10 rows × 21 columns

Now let’s extract individual segments with array slicing. we should consider the scenario where we group the houses in the train_valid_shuffled dataframe into k=10 segments of roughly equal size, with starting and ending indices computed as above. Just for fun, let’s extract the fourth segment (segment 3) and assign it to a variable called validation4.

Now we are ready to implement k-fold cross-validation. We will write a function that computes k validation errors by designating each of the k segments as the validation set. It will accept parameters (i) k, (ii) l2_penalty, (iii) dataframe, (iv) name of output column (e.g. price) and (v) list of feature names. The function will return the average validation error using k segments as validation sets.

• For each i in [0, 1, …, k-1]:
• Compute starting and ending indices of segment i and call ‘start’ and ‘end’
• Form validation set by taking a slice (start:end+1) from the data.
• Form training set by appending slice (end+1:n) to the end of slice (0:start).
• Train a linear model using training set just formed, with a given l2_penalty
• Compute validation error using validation set just formed
def k_fold_cross_validation(k, l2_penalty, data, output, features_list):
errors = []
n = len(data)
for i in xrange(k):
start = (n*i)/k
end = (n*(i+1))/k-1
# get first fold
valid = data[start:end]
first_fold = data[0:start-1]
remainder_fold = data[end+1:]
train = first_fold.append(remainder_fold)
# train model
# create linear regression object
reg = Ridge(alpha=l2_penalty, solver='svd')
# train model
reg.fit(train[features_list], train.iloc[:,(len(train.columns)-1)].to_frame())
# prediction
predicted = reg.predict(valid[features_list])
sse = (predicted- valid[output])**2
errors.append(np.mean(sse))
mse = np.mean(list(errors))
#print("l2_penalty: %s, \n\t Average MSE: \$%.6f" % (l2_penalty, mse))
return mse


Now that we have a function to compute the average validation error for a model, we can write a loop to find the model that minimizes the average validation error. Now let’s write a loop that does the following:

• We will again be aiming to fit a 15th-order polynomial model using the sqft_living input
• For l2_penalty in [10^1, 10^1.5, 10^2, 10^2.5, …, 10^7] (to get this in Python, you can use this Numpy function: np.logspace(1, 7, num=13).)
• Run 10-fold cross-validation with l2_penalty
• Investigate which L2 penalty produced the lowest average validation error.

Note: since the degree of the polynomial is now fixed to 15, to make things faster, we should generate polynomial features in advance and re-use them throughout the loop. Note2: make sure to use train_valid_shuffled when generating polynomial features!

poly15_data = polynomial_dataframe(pd.DataFrame(train_valid_shuffled["sqft_living"]), 15)
fifteen_features = list(poly15_data.columns.values)# get the name of the features
poly15_data['price'] = train_valid_shuffled['price'] # add price to the data since it's the target

results = []

for l2_penalty in np.logspace(1, 7, num=13):
average_error = k_fold_cross_validation(10, l2_penalty, poly15_data, ['price'], fifteen_features)
results.append((l2_penalty, average_error))
mse_df = pd.DataFrame(results, columns=["penalty", "error"])
mse_df

penalty error
0 10.000000 6.630754e+10
1 31.622777 6.630754e+10
2 100.000000 6.630754e+10
3 316.227766 6.630754e+10
4 1000.000000 6.630753e+10
5 3162.277660 6.630750e+10
6 10000.000000 6.630742e+10
7 31622.776602 6.630716e+10
8 100000.000000 6.630633e+10
9 316227.766017 6.630372e+10
10 1000000.000000 6.629560e+10
11 3162277.660168 6.627113e+10
12 10000000.000000 6.620413e+10

it will be useful to plot the k-fold cross-validation errors you have obtained to better understand the behavior of the method.

plt.plot(mse_df['penalty'],mse_df['error'],'b.-')
plt.xscale('log')
plt.xlabel('log(l2_penalty)')
plt.ylabel('average_error')
plt.title('k-fold cross-validation errors')

<matplotlib.text.Text at 0x1c672d68>


last edited: 29/10/2016

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