Python - How to cross check the obtained W and B intercepts in SGD for Linear Regression?
I have manually implemented SGD for Linear Regression using the partial derivates. I am working on Boston housing prices dataset from SKlearn. The input to my UDF is a (train) dataframe with standardised data except the target column.
X_train, X_test, y_train, y_test = train_test_split(df.loc[:, df.columns != 'target'], df.target, test_size=0.15, random_state=42)
dt_scaler = StandardScaler().fit(X_train)
scaler_data = dt_scaler.transform(X_train)
final_ds = pd.DataFrame(scaler_data, columns= boston.feature_names)
final_ds['target'] = y_train
final_ds.target = final_ds.target.fillna((df.target.mean()))
Now my UDF is,
def best_w_b(data):
w0 = np.random.normal(0, 1, (boston.data.shape[1],)).T
b0 = np.random.normal(0, 1)
r = 1
i = 1
while(1):
k = data.sample(n=150 ,replace = True)
for i in range(0,150,1):
der_w= np.zeros(boston.data.shape[1]) ;der_b = 0
der_w += np.dot(-2 * k.iloc[i][k.columns !='target'].T ,(k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0))
der_b += (-2 * (k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0 ))
w1 = np.subtract(w0,(r * der_w/150))
b1 = b0 - (r * der_b/150)
w_dist = np.linalg.norm(w0-w1)
b_dist = np.linalg.norm(b0-b1)
if (w0==w1).all():
return w0,b0
else:
w0 = w1
b0 = b1
r = r/2
i = i + 1
After running this method I have obtained the W and B values so if I use Summation 0 to n-1 (Y-Y_hat)^2 /n on the train data should not I be getting a value around 0.
Here is the code:
error = 0
for i in range(0,X_train.shape[0],1):
error = error + (final_ds.iloc[i].target - (np.dot(optimal_w,final_ds.iloc[i][final_ds.columns !='target']) + optimal_b))**2
print(error/X_train.shape[0])
I am getting an error around 582. Is it the right way to test?
PS:
Optimal W is : [-0.22178286, -1.30943816, -0.61933446, 1.07290039, -0.96299363,-0.59459475, -1.4094494 , 0.2022922 , -1.45901487, 0.00561458,-0.31858595, -0.71790656, -0.97285501]
Optimal B is : 1.179334612848228
python-3.x linear-regression stochastic
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
add a comment |
I have manually implemented SGD for Linear Regression using the partial derivates. I am working on Boston housing prices dataset from SKlearn. The input to my UDF is a (train) dataframe with standardised data except the target column.
X_train, X_test, y_train, y_test = train_test_split(df.loc[:, df.columns != 'target'], df.target, test_size=0.15, random_state=42)
dt_scaler = StandardScaler().fit(X_train)
scaler_data = dt_scaler.transform(X_train)
final_ds = pd.DataFrame(scaler_data, columns= boston.feature_names)
final_ds['target'] = y_train
final_ds.target = final_ds.target.fillna((df.target.mean()))
Now my UDF is,
def best_w_b(data):
w0 = np.random.normal(0, 1, (boston.data.shape[1],)).T
b0 = np.random.normal(0, 1)
r = 1
i = 1
while(1):
k = data.sample(n=150 ,replace = True)
for i in range(0,150,1):
der_w= np.zeros(boston.data.shape[1]) ;der_b = 0
der_w += np.dot(-2 * k.iloc[i][k.columns !='target'].T ,(k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0))
der_b += (-2 * (k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0 ))
w1 = np.subtract(w0,(r * der_w/150))
b1 = b0 - (r * der_b/150)
w_dist = np.linalg.norm(w0-w1)
b_dist = np.linalg.norm(b0-b1)
if (w0==w1).all():
return w0,b0
else:
w0 = w1
b0 = b1
r = r/2
i = i + 1
After running this method I have obtained the W and B values so if I use Summation 0 to n-1 (Y-Y_hat)^2 /n on the train data should not I be getting a value around 0.
Here is the code:
error = 0
for i in range(0,X_train.shape[0],1):
error = error + (final_ds.iloc[i].target - (np.dot(optimal_w,final_ds.iloc[i][final_ds.columns !='target']) + optimal_b))**2
print(error/X_train.shape[0])
I am getting an error around 582. Is it the right way to test?
PS:
Optimal W is : [-0.22178286, -1.30943816, -0.61933446, 1.07290039, -0.96299363,-0.59459475, -1.4094494 , 0.2022922 , -1.45901487, 0.00561458,-0.31858595, -0.71790656, -0.97285501]
Optimal B is : 1.179334612848228
python-3.x linear-regression stochastic
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
add a comment |
I have manually implemented SGD for Linear Regression using the partial derivates. I am working on Boston housing prices dataset from SKlearn. The input to my UDF is a (train) dataframe with standardised data except the target column.
X_train, X_test, y_train, y_test = train_test_split(df.loc[:, df.columns != 'target'], df.target, test_size=0.15, random_state=42)
dt_scaler = StandardScaler().fit(X_train)
scaler_data = dt_scaler.transform(X_train)
final_ds = pd.DataFrame(scaler_data, columns= boston.feature_names)
final_ds['target'] = y_train
final_ds.target = final_ds.target.fillna((df.target.mean()))
Now my UDF is,
def best_w_b(data):
w0 = np.random.normal(0, 1, (boston.data.shape[1],)).T
b0 = np.random.normal(0, 1)
r = 1
i = 1
while(1):
k = data.sample(n=150 ,replace = True)
for i in range(0,150,1):
der_w= np.zeros(boston.data.shape[1]) ;der_b = 0
der_w += np.dot(-2 * k.iloc[i][k.columns !='target'].T ,(k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0))
der_b += (-2 * (k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0 ))
w1 = np.subtract(w0,(r * der_w/150))
b1 = b0 - (r * der_b/150)
w_dist = np.linalg.norm(w0-w1)
b_dist = np.linalg.norm(b0-b1)
if (w0==w1).all():
return w0,b0
else:
w0 = w1
b0 = b1
r = r/2
i = i + 1
After running this method I have obtained the W and B values so if I use Summation 0 to n-1 (Y-Y_hat)^2 /n on the train data should not I be getting a value around 0.
Here is the code:
error = 0
for i in range(0,X_train.shape[0],1):
error = error + (final_ds.iloc[i].target - (np.dot(optimal_w,final_ds.iloc[i][final_ds.columns !='target']) + optimal_b))**2
print(error/X_train.shape[0])
I am getting an error around 582. Is it the right way to test?
PS:
Optimal W is : [-0.22178286, -1.30943816, -0.61933446, 1.07290039, -0.96299363,-0.59459475, -1.4094494 , 0.2022922 , -1.45901487, 0.00561458,-0.31858595, -0.71790656, -0.97285501]
Optimal B is : 1.179334612848228
python-3.x linear-regression stochastic
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
I have manually implemented SGD for Linear Regression using the partial derivates. I am working on Boston housing prices dataset from SKlearn. The input to my UDF is a (train) dataframe with standardised data except the target column.
X_train, X_test, y_train, y_test = train_test_split(df.loc[:, df.columns != 'target'], df.target, test_size=0.15, random_state=42)
dt_scaler = StandardScaler().fit(X_train)
scaler_data = dt_scaler.transform(X_train)
final_ds = pd.DataFrame(scaler_data, columns= boston.feature_names)
final_ds['target'] = y_train
final_ds.target = final_ds.target.fillna((df.target.mean()))
Now my UDF is,
def best_w_b(data):
w0 = np.random.normal(0, 1, (boston.data.shape[1],)).T
b0 = np.random.normal(0, 1)
r = 1
i = 1
while(1):
k = data.sample(n=150 ,replace = True)
for i in range(0,150,1):
der_w= np.zeros(boston.data.shape[1]) ;der_b = 0
der_w += np.dot(-2 * k.iloc[i][k.columns !='target'].T ,(k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0))
der_b += (-2 * (k.iloc[i].target - np.dot(w0,k.iloc[i][k.columns !='target'].values) - b0 ))
w1 = np.subtract(w0,(r * der_w/150))
b1 = b0 - (r * der_b/150)
w_dist = np.linalg.norm(w0-w1)
b_dist = np.linalg.norm(b0-b1)
if (w0==w1).all():
return w0,b0
else:
w0 = w1
b0 = b1
r = r/2
i = i + 1
After running this method I have obtained the W and B values so if I use Summation 0 to n-1 (Y-Y_hat)^2 /n on the train data should not I be getting a value around 0.
Here is the code:
error = 0
for i in range(0,X_train.shape[0],1):
error = error + (final_ds.iloc[i].target - (np.dot(optimal_w,final_ds.iloc[i][final_ds.columns !='target']) + optimal_b))**2
print(error/X_train.shape[0])
I am getting an error around 582. Is it the right way to test?
PS:
Optimal W is : [-0.22178286, -1.30943816, -0.61933446, 1.07290039, -0.96299363,-0.59459475, -1.4094494 , 0.2022922 , -1.45901487, 0.00561458,-0.31858595, -0.71790656, -0.97285501]
Optimal B is : 1.179334612848228
python-3.x linear-regression stochastic
python-3.x linear-regression stochastic
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
asked Nov 25 '18 at 8:16
Sanjeev RamSanjeev Ram
94
94
add a comment |
add a comment |
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