Do I need to encode the target variable for sklearn logistic regression
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I'm trying to get familiar with the sklearn library, and now I'm trying to implement logistic regression for a dataframe containing numerical and categorical values to predict a binary target variable.
While reading some documentation I found the logistic regression should be used to predict binary variables presented by 0 and 1.
My target variable is "YES" and "NO", should I code it to 0 and 1 for the algorithm to work properly, or there is no difference?
Maybe I just didn't get the idea but can someone confirm this to me.
scikit-learn logistic-regression
asked 2 days ago
GreenGreen
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I'm trying to get familiar with the sklearn library, and now I'm trying to implement logistic regression for a dataframe containing numerical and categorical values to predict a binary target variable.
While reading some documentation I found the logistic regression should be used to predict binary variables presented by 0 and 1.
My target variable is "YES" and "NO", should I code it to 0 and 1 for the algorithm to work properly, or there is no difference?
Maybe I just didn't get the idea but can someone confirm this to me.
scikit-learn logistic-regression
asked 2 days ago
GreenGreen
84
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm trying to get familiar with the sklearn library, and now I'm trying to implement logistic regression for a dataframe containing numerical and categorical values to predict a binary target variable.
While reading some documentation I found the logistic regression should be used to predict binary variables presented by 0 and 1.
My target variable is "YES" and "NO", should I code it to 0 and 1 for the algorithm to work properly, or there is no difference?
Maybe I just didn't get the idea but can someone confirm this to me.
scikit-learn logistic-regression
asked 2 days ago
GreenGreen
84
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm trying to get familiar with the sklearn library, and now I'm trying to implement logistic regression for a dataframe containing numerical and categorical values to predict a binary target variable.
While reading some documentation I found the logistic regression should be used to predict binary variables presented by 0 and 1.
My target variable is "YES" and "NO", should I code it to 0 and 1 for the algorithm to work properly, or there is no difference?
Maybe I just didn't get the idea but can someone confirm this to me.
scikit-learn logistic-regression
scikit-learn logistic-regression
asked 2 days ago
GreenGreen
84
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
GreenGreen
84
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
GreenGreen
84
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
GreenGreen
84
asked 2 days ago
GreenGreen
84
84
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Green is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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The string labels work just fine, here is an example:
from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
import numpy
X, y = load_iris(return_X_y=True)
y_string = numpy.array(['YES' if label == 1 else 'NO' for label in y])
clf = LogisticRegression(random_state=0, solver='lbfgs', multi_class='multinomial').fit(X, y_string)
y_pred = clf.predict(X[50:100, :])
print(y_pred)
Output:
['NO' 'NO' 'NO' 'YES' 'NO' 'YES' 'NO' 'YES' 'NO' 'NO' 'YES' 'NO' 'YES'
'NO' 'NO' 'NO' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'YES' 'YES' 'NO' 'NO'
'YES' 'NO' 'NO' 'YES' 'YES' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO'
'YES' 'YES' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO']
Yo can replace y_string to y for the numerical example.
answered 2 days ago
EsmailianEsmailian
1,096112
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The string labels work just fine, here is an example:
from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
import numpy
X, y = load_iris(return_X_y=True)
y_string = numpy.array(['YES' if label == 1 else 'NO' for label in y])
clf = LogisticRegression(random_state=0, solver='lbfgs', multi_class='multinomial').fit(X, y_string)
y_pred = clf.predict(X[50:100, :])
print(y_pred)
Output:
['NO' 'NO' 'NO' 'YES' 'NO' 'YES' 'NO' 'YES' 'NO' 'NO' 'YES' 'NO' 'YES'
'NO' 'NO' 'NO' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'YES' 'YES' 'NO' 'NO'
'YES' 'NO' 'NO' 'YES' 'YES' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO'
'YES' 'YES' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO']
Yo can replace y_string to y for the numerical example.
answered 2 days ago
EsmailianEsmailian
1,096112
$endgroup$
add a comment |
$begingroup$
The string labels work just fine, here is an example:
from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
import numpy
X, y = load_iris(return_X_y=True)
y_string = numpy.array(['YES' if label == 1 else 'NO' for label in y])
clf = LogisticRegression(random_state=0, solver='lbfgs', multi_class='multinomial').fit(X, y_string)
y_pred = clf.predict(X[50:100, :])
print(y_pred)
Output:
['NO' 'NO' 'NO' 'YES' 'NO' 'YES' 'NO' 'YES' 'NO' 'NO' 'YES' 'NO' 'YES'
'NO' 'NO' 'NO' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'YES' 'YES' 'NO' 'NO'
'YES' 'NO' 'NO' 'YES' 'YES' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO'
'YES' 'YES' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO']
Yo can replace y_string to y for the numerical example.
answered 2 days ago
EsmailianEsmailian
1,096112
$endgroup$
add a comment |
$begingroup$
The string labels work just fine, here is an example:
from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
import numpy
X, y = load_iris(return_X_y=True)
y_string = numpy.array(['YES' if label == 1 else 'NO' for label in y])
clf = LogisticRegression(random_state=0, solver='lbfgs', multi_class='multinomial').fit(X, y_string)
y_pred = clf.predict(X[50:100, :])
print(y_pred)
Output:
['NO' 'NO' 'NO' 'YES' 'NO' 'YES' 'NO' 'YES' 'NO' 'NO' 'YES' 'NO' 'YES'
'NO' 'NO' 'NO' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'YES' 'YES' 'NO' 'NO'
'YES' 'NO' 'NO' 'YES' 'YES' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO'
'YES' 'YES' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO']
Yo can replace y_string to y for the numerical example.
answered 2 days ago
EsmailianEsmailian
1,096112
$endgroup$
The string labels work just fine, here is an example:
from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
import numpy
X, y = load_iris(return_X_y=True)
y_string = numpy.array(['YES' if label == 1 else 'NO' for label in y])
clf = LogisticRegression(random_state=0, solver='lbfgs', multi_class='multinomial').fit(X, y_string)
y_pred = clf.predict(X[50:100, :])
print(y_pred)
Output:
['NO' 'NO' 'NO' 'YES' 'NO' 'YES' 'NO' 'YES' 'NO' 'NO' 'YES' 'NO' 'YES'
'NO' 'NO' 'NO' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'YES' 'YES' 'NO' 'NO'
'YES' 'NO' 'NO' 'YES' 'YES' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO'
'YES' 'YES' 'NO' 'YES' 'YES' 'YES' 'NO' 'NO' 'NO' 'YES' 'NO']
Yo can replace y_string to y for the numerical example.
answered 2 days ago
EsmailianEsmailian
1,096112
answered 2 days ago
EsmailianEsmailian
1,096112
answered 2 days ago
EsmailianEsmailian
1,096112
answered 2 days ago
EsmailianEsmailian
1,096112
1,096112
add a comment |
add a comment |
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Green is a new contributor. Be nice, and check out our Code of Conduct.
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