It's in the documentation that verbose=True will output time elapsed, but it is not doing so for me:
from sklearn.ensemble import VotingClassifier
voting_c_all = VotingClassifier(
estimators=[
('random_forest', gs_forest2),
('grid_search', gs),
],
voting='soft',
verbose=True,
n_jobs=-1
)
voting_c_all.fit(X_res, y_res)
Using the example from the manual:
import numpy as np
from sklearn.linear_model import LogisticRegression
from sklearn.ensemble import RandomForestClassifier, VotingClassifier
clf1 = LogisticRegression(multi_class='multinomial', random_state=1)
clf2 = RandomForestClassifier(n_estimators=50, random_state=1)
X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y = np.array([1, 1, 1, 2, 2, 2])
eclf1 = VotingClassifier(estimators=[('lr', clf1), ('rf', clf2)], voting='soft',verbose=True)
eclf1 = eclf1.fit(X, y)
[Voting] ....................... (1 of 2) Processing lr, total= 0.0s
[Voting] ....................... (2 of 2) Processing rf, total= 0.1s
But once you set n_jobs to be more than 1, the job should be sent to other cores and you don't see the print, hence doesn't keep track of the time:
eclf1 = VotingClassifier(estimators=[('lr', clf1), ('rf', clf2)], voting='soft',verbose=True,n_jobs=2)
eclf1 = eclf1.fit(X, y)
Related
I have the following very simple code trying to model a simple dataset:
from sklearn.pipeline import Pipeline
from sklearn.impute import SimpleImputer
from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import GridSearchCV
data = {'Feature_A': [1, 2, 3, 4], 'Feature_B': [7, 8, 9, 10], 'Feature_C': [2, 3, 4, 5], 'Label': [7, 7, 8, 9]}
data = pd.DataFrame(data)
data_labels = data['Label']
data = data.drop(columns=['Label'])
pipeline = Pipeline([('imputer', SimpleImputer()),
('std_scaler', StandardScaler())])
data_prepared = pipeline.fit_transform(data)
lin_reg = LinearRegression()
lin_grid = {"n_jobs": [20, 50]}
error = "max_error"
grid_search = GridSearchCV(lin_reg, param_grid=lin_grid, verbose=3, cv=2, refit=True, scoring=error, return_train_score=True)
grid_search.fit(data_prepared, data_labels)
print(grid_search.best_estimator_.coef_)
print(grid_search.best_estimator_.intercept_)
print(list(data_labels))
print(list(grid_search.best_estimator_.predict(data_prepared)))
That gives me the following results:
[0.2608746 0.2608746 0.2608746]
7.75
[7, 7, 8, 9]
[6.7, 7.4, 8.1, 8.799999999999999]
From there, is there a way of computing the values of the features that would give me the maximum label, within the boundaries of the dataset?
If I understand your question correctly, this should work:
import numpy as np
id_max = np.argmax(grid_search.predict(data)) # find id of the maximum predicted label
print(data.loc[id_max])
I have a data-set X with 260 unique observations.
when running x_train,x_test,_,_=test_train_split(X,y,test_size=0.2) I would assume that
[p for p in x_test if p in x_train] would be empty, but it is not. Actually it turns out that only two observations in x_test is not in x_train.
Is that intended or...?
EDIT (posted the data I am using):
from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split as split
import numpy as np
DATA=load_breast_cancer()
X=DATA.data
y= DATA.target
y=np.array([1 if p==0 else 0 for p in DATA.target])
x_train,x_test,y_train,y_test=split(X,y,test_size=0.2,stratify=y,random_state=42)
len([p for p in x_test if p in x_train]) #is not 0
EDIT 2.0: Showing that the test works
a=np.array([[1,2,3],[4,5,6]])
b=np.array([[1,2,3],[11,12,13]])
len([p for p in a if p in b]) #1
This is not a bug with the implementation of train_test_split in sklearn, but a weird peculiarity of how the in operator works on numpy arrays. The in operator first does an elementwise comparison between two arrays, and returns True if ANY of the elements match.
import numpy as np
a = np.array([[1, 2, 3], [4, 5, 6]])
b = np.array([[6, 7, 8], [5, 5, 5]])
a in b # True
The correct way to test for this kind of overlap is using the equality operator and np.all and np.any. As a bonus, you also get the indices that overlap for free.
import numpy as np
a = np.array([[1, 2, 3], [4, 5, 6]])
b = np.array([[6, 7, 8], [5, 5, 5], [7, 8, 9]])
a in b # True
z = np.any(np.all(a == b[:, None, :], -1)) # False
a = np.array([[1, 2, 3], [4, 5, 6]])
b = np.array([[6, 7, 8], [1, 2, 3], [7, 8, 9]])
a in b # True
overlap = np.all(a == b[:, None, :], -1)
z = np.any(overlap) # True
indices = np.nonzero(overlap) # (1, 0)
You need to check using the following:
from sklearn.datasets import load_breast_cancer
from sklearn.model_selection import train_test_split as split
import numpy as np
DATA=load_breast_cancer()
X=DATA.data
y= DATA.target
y=np.array([1 if p==0 else 0 for p in DATA.target])
x_train,x_test,y_train,y_test=split(X,y,test_size=0.2,stratify=y,random_state=42)
len([p for p in x_test.tolist() if p in x_train.tolist()])
0
Using x_test.tolist() the in operator will work as intended.
Reference: testing whether a Numpy array contains a given row
I have found an inconsistency in the predict function of the SVM model for multiclass problems. I have trained a model with SKlearn SVM.SVC function for a multiclass prediction problem (see plot below).
But on some occasions, the predict functions gives me different results when I did the prediction instead with the argmax of the decision function. One can see that the inconsistency is close to the decision boundary.
This inconsistency vanishes when I use the OneVsRestClassifier directly. Does the predict function of the SVM.SVC classes some corrections or why does it differ from the argmax prediction?
Here is the code to reproduce the result:
import numpy as np
from sklearn import svm, datasets
from sklearn.multiclass import OneVsRestClassifier
from scipy.linalg import cho_solve, cho_factor
def create_data(n_samples, noise):
# 4 gaussian blobs with different means and variances
sample_per_cls = np.int(n_samples/4)
sample_per_cls_rest = sample_per_cls + n_samples - 4*sample_per_cls #puts the rest of the samples into the last class
x1 = np.random.multivariate_normal([20, 18], np.array([[2, 3], [3, 7]])*4*noise, sample_per_cls, 'warn')
x2 = np.random.multivariate_normal([13, 27], np.array([[10, 3], [3, 2]])*4*noise, sample_per_cls, 'warn')
x3 = np.random.multivariate_normal([9, 13], np.array([[6, 1], [1, 5]])*4*noise, sample_per_cls, 'warn')
x4 = np.random.multivariate_normal([14, 20], np.array([[4, 0.2], [0.2, 7]])*4*noise, sample_per_cls_rest, 'warn')
X = np.vstack([x1,x2,x3,x4])
#define the labels for each class
Y = np.empty([n_samples], dtype=np.int)
Y[0:sample_per_cls] = 0
Y[sample_per_cls:2*sample_per_cls] = 1
Y[2*sample_per_cls:3*sample_per_cls] = 2
Y[3*sample_per_cls:] = 3
#shuffle the data set
rand_int = np.arange(n_samples)
np.random.shuffle(rand_int)
X = X[rand_int]
Y = Y[rand_int]
return X, Y
X, Y = create_data(n_samples=800, noise=0.15)
clf = svm.SVC(C=0.5, kernel='rbf', gamma=0.1, decision_function_shape='ovr', cache_size=8000)
#the classifier below is consistent
#clf = OneVsRestClassifier(svm.SVC(C=0.5, kernel='rbf', gamma=0.1, decision_function_shape='ovr', cache_size=8000))
clf.fit(X,Y)
Xs = np.linspace(np.min(X[:,0] - 1), np.max(X[:,0] + 1), 150)
Ys = np.linspace(np.min(X[:,1] - 1), np.max(X[:,1] + 1), 150)
XX, YY = np.meshgrid(Xs, Ys)
test_set = np.stack([XX, YY], axis=2).reshape(-1,2)
#prediction via argmax of the decision function
pred = np.argmax(clf.decision_function(test_set), axis=1)
#prediction with sklearn function
pred_1 = clf.predict(test_set)
diff = np.equal(pred, pred_1)
error = np.where(diff == False)[0]
print(error)
import matplotlib.pyplot as plt
plt.rcParams['figure.figsize'] = [16, 10]
plt.contourf(XX, YY, pred_1.reshape(XX.shape), alpha=0.5, cmap='seismic')
plt.colorbar()
plt.scatter(X[:,0], X[:,1], c=Y, s=20, marker='o', edgecolors='k')
plt.scatter(test_set[error, 0], test_set[error, 1], c=pred_1[error], s=120, marker='^', edgecolors='k')
plt.show()
Triangles are marking the inconsistent points:
I'm trying to find best values of C & gamma for SVR() estimator using GridSearchCV() but I get this error
TypeError: 'KFold' object is not iterable
This the code
from sklearn.grid_search import GridSearchCV
from sklearn.model_selection import KFold
C_range = np.logspace(-2, 10, 13)
gamma_range = np.logspace(-9, 3, 13)
param_grid = dict(gamma=gamma_range, C=C_range)
cv = KFold(n_splits=5, shuffle=False, random_state=None)
grid = GridSearchCV(SVR(kernel='rbf'), param_grid=param_grid, cv=cv)
grid.fit(X, y)
print("The best parameters are %s with a score of %0.2f"
% (grid.best_params_, grid.best_score_))
Similar problem solved by:
Replacing:
from sklearn.grid_search import GridSearchCV
with
from sklearn.model_selection import GridSearchCV
cv is an object in your case.
You should use KFold in order to create batches of data, and pass these batches to GridSearchCV in cv argument.
Here an example on how to create splits, using KFold:
>>> from sklearn.model_selection import KFold
>>> X = np.array([[1, 2], [3, 4], [1, 2], [3, 4]])
>>> y = np.array([1, 2, 3, 4])
>>> kf = KFold(n_splits=2)
>>> kf.get_n_splits(X)
2
>>> print(kf)
KFold(n_splits=2, random_state=None, shuffle=False)
>>> for train_index, test_index in kf.split(X):
... print("TRAIN:", train_index, "TEST:", test_index)
... X_train, X_test = X[train_index], X[test_index]
... y_train, y_test = y[train_index], y[test_index]
TRAIN: [2 3] TEST: [0 1]
TRAIN: [0 1] TEST: [2 3]
Is there any way of automatically selecting the 'training samples' from the collection of features for better fit of the model (DT or SVM)? I know about selecting the 'features'. But I am talking about selecting the 'samples' after selecting the features.
There are a couple different ways to split your set into training, testing, and cross validation sets. Check out sklearn.cross_validation.train_test_split. But also take a look at the plethora of advanced splitting methods that are also available in SK-Learn.
Here's an example with test_train_split:
In:
import numpy as np
from sklearn.cross_validation import train_test_split
a, b = np.arange(10).reshape((5, 2)), range(5)
a
Out:
array([[0, 1],
[2, 3],
[4, 5],
[6, 7],
[8, 9]])
In:
list(b)
Out:
[0, 1, 2, 3, 4]
In:
a_train, a_test, b_train, b_test = train_test_split(a, b, test_size=0.33, random_state=42)
a_train
Out:
array([[4, 5],
[0, 1],
[6, 7]])
In:
b_train
Out:
[2, 0, 3]
In:
a_test
Out:
array([[2, 3],
[8, 9]])
In:
b_test
Out:
[1, 4]
There are generally two ways to do feature selections: Univariate Feature Selection and L1-based Sparse Feature Selection.
from sklearn.datasets import make_classification
from sklearn.feature_selection import f_classif, SelectKBest
from sklearn.svm import LinearSVC
import matplotlib.pyplot as plt
import numpy as np
# simulate some artificial data: 2000 obs, features: 1000-dim
# but only 2 out 1000 features are informative, the rest 998 features are noises
X, y = make_classification(n_samples=2000, n_features=1000, n_informative=2, random_state=0)
X.shape
Out[153]: (2000, 1000)
# Univariate Feature Selection: select 20 best from 1000 features
# ==========================================================================
# classification F-test
X_selected = SelectKBest(f_classif, k=20).fit_transform(X, y)
X_selected.shape
# or to visualize each f-score/p-value of 1000 features
X_f_scores, X_f_pval = f_classif(X, y)
fig, ax = plt.subplots(figsize=(8,6))
ax.plot(X_f_scores)
ax.set_title('Univariate Feature Selection: Classification F-Score')
ax.set_xlabel('features')
ax.set_ylabel('F-score')
# which features are most important: top 10
np.argsort(X_f_scores)[-10:] # argsort is from smallest to largest
Out[154]: array([940, 163, 574, 969, 994, 977, 360, 291, 838, 524])
# L1-based Sparse Feature Selection: any algo implementation penalty 'l1'
# ==========================================================================
# use LinearSVC for example here
# other popular choices: logistic regression, Lasso (for regression)
feature_selector = LinearSVC(C=0.01, penalty='l1', dual=False)
feature_selector.fit(X, y)
# get features with non-zero coefficients: exactly 2
(feature_selector.coef_ != 0.0).sum()
Out[155]: 2
X_selected_l1 = feature_selector.transform(X)
# or X[:, feature_selector.coef_ != 0.0]