3.1. Cross-validation_ evaluating estimator performance — scikit-learn 1.4.1 documentation
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3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
1/12
X_test, y_test
train_test_split
C
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
train_test_split
>>> from
sklearn
import
datasets
>>> from
sklearn
import
svm
>>> X, y = datasets
.
load_iris(return_X_y
=
True
)
>>> X
.
shape, y
.
shape
((150, 4), (150,))
>>> X_train, X_test, y_train, y_test = train_test_split(
... X, y, test_size
=
0.4
, random_state
=
0
)
>>> X_train
.
shape, y_train
.
shape
((90, 4), (90,))
>>> X_test
.
shape, y_test
.
shape
((60, 4), (60,))
>>> clf = svm
.
SVC(kernel
=
'linear'
, C
=
1
)
.
fit(X_train, y_train)
>>> clf
.
score(X_test, y_test)
0.96...
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
2/12
cross_val_score
score
cv
cross_val_score
KFold
StratifiedKFold
ClassifierMixin
>>> from
sklearn.model_selection
import
cross_val_score
>>> clf = svm
.
SVC(kernel
=
'linear'
, C
=
1
, random_state
=
42
)
>>> scores = cross_val_score(clf, X, y, cv
=
5
)
>>> scores
array([0.96..., 1. , 0.96..., 0.96..., 1. ])
>>> print
(
"
%0.2f
accuracy with a standard deviation of %0.2f
" % (scores
.
mean(), scores
.
std()))
0.98 accuracy with a standard deviation of 0.02
>>> from
sklearn
import
metrics
>>> scores = cross_val_score(
... clf, X, y, cv
=
5
, scoring
=
'f1_macro'
)
>>> scores
array([0.96..., 1. ..., 0.96..., 0.96..., 1. ])
>>> from
sklearn.model_selection
import
ShuffleSplit
>>> n_samples = X
.
shape[
0
]
>>> cv = ShuffleSplit(n_splits
=
5
, test_size
=
0.3
, random_state
=
0
)
>>> cross_val_score(clf, X, y, cv
=
cv)
array([0.977..., 0.977..., 1. ..., 0.955..., 1. ])
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
3/12
Pipeline
cross_validate
cross_val_score
['test_score', 'fit_time',
'score_time']
['test_<scorer1_name>', 'test_<scorer2_name>',
'test_<scorer...>', 'fit_time', 'score_time']
return_train_score
False
True
return_estimator=True
re‐
turn_indices=True
>>> def
custom_cv_2folds
(X):
... n = X
.
shape[
0
]
... i = 1
... while
i <= 2
:
... idx = np
.
arange(n * (i - 1
) / 2
, n * i / 2
, dtype
=
int
)
... yield
idx, idx
... i += 1
...
>>> custom_cv = custom_cv_2folds(X)
>>> cross_val_score(clf, X, y, cv
=
custom_cv)
array([1. , 0.973...])
>>> from
sklearn
import
preprocessing
>>> X_train, X_test, y_train, y_test = train_test_split(
... X, y, test_size
=
0.4
, random_state
=
0
)
>>> scaler = preprocessing
.
StandardScaler()
.
fit(X_train)
>>> X_train_transformed = scaler
.
transform(X_train)
>>> clf = svm
.
SVC(C
=
1
)
.
fit(X_train_transformed, y_train)
>>> X_test_transformed = scaler
.
transform(X_test)
>>> clf
.
score(X_test_transformed, y_test)
0.9333...
>>> from
sklearn.pipeline
import
make_pipeline
>>> clf = make_pipeline(preprocessing
.
StandardScaler(), svm
.
SVC(C
=
1
))
>>> cross_val_score(clf, X, y, cv
=
cv)
array([0.977..., 0.933..., 0.955..., 0.933..., 0.977...])
>>> from
sklearn.model_selection
import
cross_validate
>>> from
sklearn.metrics
import
recall_score
>>> scoring = [
'precision_macro'
, 'recall_macro'
]
>>> clf = svm
.
SVC(kernel
=
'linear'
, C
=
1
, random_state
=
0
)
>>> scores = cross_validate(clf, X, y, scoring
=
scoring)
>>> sorted
(scores
.
keys())
['fit_time', 'score_time', 'test_precision_macro', 'test_recall_macro']
>>> scores[
'test_recall_macro'
]
array([0.96..., 1. ..., 0.96..., 0.96..., 1. ])
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3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
4/12
cross_validate
cross_val_predict
cross_val_score
cross_val_predict
cross_val_score
cross_val_score
cross_val_predict
cross_val_predict
cross_val_predict
KFold
>>> from
sklearn.metrics
import
make_scorer
>>> scoring = {
'prec_macro'
: 'precision_macro'
,
... 'rec_macro'
: make_scorer(recall_score, average
=
'macro'
)}
>>> scores = cross_validate(clf, X, y, scoring
=
scoring,
... cv
=
5
, return_train_score
=
True
)
>>> sorted
(scores
.
keys())
['fit_time', 'score_time', 'test_prec_macro', 'test_rec_macro',
'train_prec_macro', 'train_rec_macro']
>>> scores[
'train_rec_macro'
]
array([0.97..., 0.97..., 0.99..., 0.98..., 0.98...])
>>> scores = cross_validate(clf, X, y,
... scoring
=
'precision_macro'
, cv
=
5
,
... return_estimator
=
True
)
>>> sorted
(scores
.
keys())
['estimator', 'fit_time', 'score_time', 'test_score']
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
5/12
KFold
RepeatedKFold
KFold
RepeatedStratifiedKFold
LeaveOneOut
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
KFold
>>> X = [
"a"
, "b"
, "c"
, "d"
]
>>> kf = KFold(n_splits
=
2
)
>>> for
train, test in
kf
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[2 3] [0 1]
[0 1] [2 3]
>>> X = np
.
array([[
0.
, 0.
], [
1.
, 1.
], [
-
1.
, -
1.
], [
2.
, 2.
]])
>>> y = np
.
array([
0
, 1
, 0
, 1
])
>>> X_train, X_test, y_train, y_test = X[train], X[test], y[train], y[test]
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
RepeatedKFold
>>> X = np
.
array([[
1
, 2
], [
3
, 4
], [
1
, 2
], [
3
, 4
]])
>>> random_state = 12883823
>>> rkf = RepeatedKFold(n_splits
=
2
, n_repeats
=
2
, random_state
=
random_state)
>>> for
train, test in
rkf
.
split(X):
... print
(
"
%s
%s
" % (train, test))
...
[2 3] [0 1]
[0 1] [2 3]
[0 2] [1 3]
[1 3] [0 2]
>>> from
sklearn.model_selection
import
LeaveOneOut
>>> X = [
1
, 2
, 3
, 4
]
>>> loo = LeaveOneOut()
>>> for
train, test in
loo
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[1 2 3] [0]
[0 2 3] [1]
[0 1 3] [2]
[0 1 2] [3]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
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LeavePOut
LeaveOneOut
LeaveOneOut
KFold
ShuffleSplit
random_state
ShuffleSplit
ShuffleSplit
KFold
>>> from
sklearn.model_selection
import
LeavePOut
>>> X = np
.
ones(
4
)
>>> lpo = LeavePOut(p
=
2
)
>>> for
train, test in
lpo
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[2 3] [0 1]
[1 3] [0 2]
[1 2] [0 3]
[0 3] [1 2]
[0 2] [1 3]
[0 1] [2 3]
>>> from
sklearn.model_selection
import
ShuffleSplit
>>> X = np
.
arange(
10
)
>>> ss = ShuffleSplit(n_splits
=
5
, test_size
=
0.25
, random_state
=
0
)
>>> for
train_index, test_index in
ss
.
split(X):
... print
(
"
%s
%s
" % (train_index, test_index))
[9 1 6 7 3 0 5] [2 8 4]
[2 9 8 0 6 7 4] [3 5 1]
[4 5 1 0 6 9 7] [2 3 8]
[2 7 5 8 0 3 4] [6 1 9]
[4 1 0 6 8 9 3] [5 2 7]
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3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
7/12
StratifiedKFold
StratifiedShuffleSplit
StratifiedKFold
KFold
StratifiedKFold
RepeatedStratifiedKFold
StratifiedShuffleSplit
>>> from
sklearn.model_selection
import
StratifiedKFold, KFold
>>> import
numpy
as
np
>>> X, y = np
.
ones((
50
, 1
)), np
.
hstack(([
0
] * 45
, [
1
] * 5
))
>>> skf = StratifiedKFold(n_splits
=
3
)
>>> for
train, test in
skf
.
split(X, y):
... print
(
'train - {}
| test - {}
'
.
format(
... np
.
bincount(y[train]), np
.
bincount(y[test])))
train - [30 3] | test - [15 2]
train - [30 3] | test - [15 2]
train - [30 4] | test - [15 1]
>>> kf = KFold(n_splits
=
3
)
>>> for
train, test in
kf
.
split(X, y):
... print
(
'train - {}
| test - {}
'
.
format(
... np
.
bincount(y[train]), np
.
bincount(y[test])))
train - [28 5] | test - [17]
train - [28 5] | test - [17]
train - [34] | test - [11 5]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
8/12
groups
GroupKFold
GroupKFold
StratifiedGroupKFold
KFold
GroupKFold
KFold
GroupKFold
KFold
shuffle=True
StratifiedGroupKFold
StratifiedKFold
GroupKFold
GroupKFold
>>> from
sklearn.model_selection
import
GroupKFold
>>> X = [
0.1
, 0.2
, 2.2
, 2.4
, 2.3
, 4.55
, 5.8
, 8.8
, 9
, 10
]
>>> y = [
"a"
, "b"
, "b"
, "b"
, "c"
, "c"
, "c"
, "d"
, "d"
, "d"
]
>>> groups = [
1
, 1
, 1
, 2
, 2
, 2
, 3
, 3
, 3
, 3
]
>>> gkf = GroupKFold(n_splits
=
3
)
>>> for
train, test in
gkf
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[0 1 2 3 4 5] [6 7 8 9]
[0 1 2 6 7 8 9] [3 4 5]
[3 4 5 6 7 8 9] [0 1 2]
>>> from
sklearn.model_selection
import
StratifiedGroupKFold
>>> X = list
(
range
(
18
))
>>> y = [
1
] * 6 + [
0
] * 12
>>> groups = [
1
, 2
, 3
, 3
, 4
, 4
, 1
, 1
, 2
, 2
, 3
, 4
, 5
, 5
, 5
, 6
, 6
, 6
]
>>> sgkf = StratifiedGroupKFold(n_splits
=
3
)
>>> for
train, test in
sgkf
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[ 0 2 3 4 5 6 7 10 11 15 16 17] [ 1 8 9 12 13 14]
[ 0 1 4 5 6 7 8 9 11 12 13 14] [ 2 3 10 15 16 17]
[ 1 2 3 8 9 10 12 13 14 15 16 17] [ 0 4 5 6 7 11]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
9/12
GroupKFold
LeaveOneGroupOut
LeavePGroupsOut
n_groups=1
GroupKFold
n_splits
groups
LeaveOneGroupOut
LeavePGroupsOut
LeaveOneGroupOut
>>> from
sklearn.model_selection
import
LeaveOneGroupOut
>>> X = [
1
, 5
, 10
, 50
, 60
, 70
, 80
]
>>> y = [
0
, 1
, 1
, 2
, 2
, 2
, 2
]
>>> groups = [
1
, 1
, 2
, 2
, 3
, 3
, 3
]
>>> logo = LeaveOneGroupOut()
>>> for
train, test in
logo
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[2 3 4 5 6] [0 1]
[0 1 4 5 6] [2 3]
[0 1 2 3] [4 5 6]
>>> from
sklearn.model_selection
import
LeavePGroupsOut
>>> X = np
.
arange(
6
)
>>> y = [
1
, 1
, 1
, 2
, 2
, 2
]
>>> groups = [
1
, 1
, 2
, 2
, 3
, 3
]
>>> lpgo = LeavePGroupsOut(n_groups
=
2
)
>>> for
train, test in
lpgo
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[4 5] [0 1 2 3]
[2 3] [0 1 4 5]
[0 1] [2 3 4 5]
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3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
10/12
GroupShuffleSplit
ShuffleSplit
LeavePGroupsOut
LeavePGroupsOut
GroupShuffleSplit
LeavePGroupsOut
PredefinedSplit
test_fold
train_test_split
ShuffleSplit
split()
>>> from
sklearn.model_selection
import
GroupShuffleSplit
>>> X = [
0.1
, 0.2
, 2.2
, 2.4
, 2.3
, 4.55
, 5.8
, 0.001
]
>>> y = [
"a"
, "b"
, "b"
, "b"
, "c"
, "c"
, "c"
, "a"
]
>>> groups = [
1
, 1
, 2
, 2
, 3
, 3
, 4
, 4
]
>>> gss = GroupShuffleSplit(n_splits
=
4
, test_size
=
0.5
, random_state
=
0
)
>>> for
train, test in
gss
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
...
[0 1 2 3] [4 5 6 7]
[2 3 6 7] [0 1 4 5]
[2 3 4 5] [0 1 6 7]
[4 5 6 7] [0 1 2 3]
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
GroupShuffleSplit
>>> X = np
.
array([
0.1
, 0.2
, 2.2
, 2.4
, 2.3
, 4.55
, 5.8
, 0.001
])
>>> y = np
.
array([
"a"
, "b"
, "b"
, "b"
, "c"
, "c"
, "c"
, "a"
])
>>> groups = np
.
array([
1
, 1
, 2
, 2
, 3
, 3
, 4
, 4
])
>>> train_indx, test_indx = next
(
... GroupShuffleSplit(random_state
=
7
)
.
split(X, y, groups)
... )
>>> X_train, X_test, y_train, y_test = \
... X[train_indx], X[test_indx], y[train_indx], y[test_indx]
>>> X_train
.
shape, X_test
.
shape
((6,), (2,))
>>> np
.
unique(groups[train_indx]), np
.
unique(groups[test_indx])
(array([1, 2, 4]), array([3]))
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
11/12
KFold
ShuffleSplit
TimeSeriesSplit
TimeSeriesSplit
KFold
cv=some_integer
cross_val_score
train_test_split
random_state
None
KFold(..., shuffle=True)
GridSearchCV
fit
random_state
>>> from
sklearn.model_selection
import
TimeSeriesSplit
>>> X = np
.
array([[
1
, 2
], [
3
, 4
], [
1
, 2
], [
3
, 4
], [
1
, 2
], [
3
, 4
]])
>>> y = np
.
array([
1
, 2
, 3
, 4
, 5
, 6
])
>>> tscv = TimeSeriesSplit(n_splits
=
3
)
>>> print
(tscv)
TimeSeriesSplit(gap=0, max_train_size=None, n_splits=3, test_size=None)
>>> for
train, test in
tscv
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[0 1 2] [3]
[0 1 2 3] [4]
[0 1 2 3 4] [5]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
12/12
permutation_test_score
permutation_test_score
n_permutations
n_permutations
cv
permutation_test_score
permutation_test_score
(n_permutations + 1) * n_cv
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E MASTER - AP Statistics 20-21, section APStats (Sem 2) SP21,
* INSTRUCTOR
Section 8.2: Testing Mean with Known SD (Homework)
Justin Thomas
Springfield Public Schools.MO
A Print Assignment
Current Score
Due Date
QUESTION
5
SAT, MAY 29, 2021
1
3
4
TOTAL SCORE
11:59 PM CDT
POINTS
1/1
1/1
1/1
6/8
4/4
13/15
86.7%
+ Request Extension
Assignment Submission & Scoring
Assignment Submission
For this assignment, you submit answers by question parts. The number of submissions…
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Interpreting technology: The following display from the TI-84 Plus calculator presents the least-squares regression line for predicting the price of a certain stock y from the prime interest rate in percent x
LinReg
y=+a+bx
a=2.29525776
b=0.38970413
r^2=0.4319044662
r=0.65719439
Predict the price when the prime interest rate is 6%. Round the answer to at least four decimal places.
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I ONLY NEED PART C,D, and E answered please thanks
A regression was run to determine if there is a relationship between the happiness index (y) and life expectancy in years of a given country (x).The results of the regression were:
ˆyy^=a+bxa=-1.102b=0.074
(a) Write the equation of the Least Squares Regression line of the formˆyy^= + x(b) Which is a possible value for the correlation coefficient, rr?
-0.858
-1.07
1.07
0.858
(c) If a country increases its life expectancy, the happiness index will
decrease
increase
(d) If the life expectancy is increased by 2.5 years in a certain country, how much will the happiness index change? Round to two decimal places._____(e) Use the regression line to predict the happiness index of a country with a life expectancy of 63 years. Round to two decimal places._______Use the space below to type your answer AND/OR to upload a picture of your work for all the questions in this problem.
arrow_forward
Please answer as many as your allowed too. Thank you :)
A regression was run to determine if there is a relationship between the happiness index (y) and life expectancy in years of a given country (x).The results of the regression were:
ˆyy^=a+bxa=-1.68b=0.168
(a) Write the equation of the Least Squares Regression line of the formˆyy^= + x(b) Which is a possible value for the correlation coefficient, rr?
-1.417
1.417
0.702
-0.702
(c) If a country increases its life expectancy, the happiness index will
increase
decrease
(d) If the life expectancy is increased by 0.5 years in a certain country, how much will the happiness index change? Round to two decimal places.(e) Use the regression line to predict the happiness index of a country with a life expectancy of 69 years. Round to two decimal places.
arrow_forward
Use the maximum likehood method in deriving the error variance in regression
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Benign prostatic hyperplasia is a noncancerous
enlargement of the prostate gland that adversely
affects the quality of life (QoL) of millions of men. A
study of minimally invasive procedures for the
treatment for this condition looked at pretreatment
QoL (qol_base) and quality of life after 3 month on
treatment (qol_3mo)
The baseline data for 10 patients and their 3 month
follow-up data is presented below:
MAXFLO_B = maximum urine flow at baseline (urine
flow measurement scale misplaced)
MAXFLO3M = maximum urine flow after 3 months of
treatment
maxflo_b
maxflo3m
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9
16
11
8
4
9
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10
6
8
14
10
13
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