About sklearn.metrics.average_precision_score documentationHow to interpret the AUC score in this case?Where to find statistically relevant documentation of common Python packages?Irregular Precision-Recall CurveNeed a Work-around for OneHotEncoder Issue in SKLearn PreprocessingSci-kit learn function to select threshold for higher recall than precisionXGBoost: Quantifying Feature Importancessklearn nmf - question about its usemodel.predict in Keras, Python errorDifference between sklearn’s “log_loss” and “LogisticRegression”?Not sure if over-fitting
About sklearn.metrics.average_precision_score documentation
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About sklearn.metrics.average_precision_score documentation
How to interpret the AUC score in this case?Where to find statistically relevant documentation of common Python packages?Irregular Precision-Recall CurveNeed a Work-around for OneHotEncoder Issue in SKLearn PreprocessingSci-kit learn function to select threshold for higher recall than precisionXGBoost: Quantifying Feature Importancessklearn nmf - question about its usemodel.predict in Keras, Python errorDifference between sklearn’s “log_loss” and “LogisticRegression”?Not sure if over-fitting
$begingroup$
There is a example in sklearn.metrics.average_precision_score documentation.
import numpy as np
from sklearn.metrics import average_precision_score
y_true = np.array([0, 0, 1, 1])
y_scores = np.array([0.1, 0.4, 0.35, 0.8])
average_precision_score(y_true, y_scores)
0.83
But when I plot precision_recall_curve
precision, recall, _ = precision_recall_curve(y_true, y_scores)
plt.plot( recall,precision)
I got the picture:
why the area under the precision_recall_curve is not 0.83?
python scikit-learn scoring
edited 2 hours ago
Ben Reiniger
664214
asked 6 hours ago
disney82231disney82231
352
$endgroup$
add a comment |
$begingroup$
There is a example in sklearn.metrics.average_precision_score documentation.
import numpy as np
from sklearn.metrics import average_precision_score
y_true = np.array([0, 0, 1, 1])
y_scores = np.array([0.1, 0.4, 0.35, 0.8])
average_precision_score(y_true, y_scores)
0.83
But when I plot precision_recall_curve
precision, recall, _ = precision_recall_curve(y_true, y_scores)
plt.plot( recall,precision)
I got the picture:
why the area under the precision_recall_curve is not 0.83?
python scikit-learn scoring
edited 2 hours ago
Ben Reiniger
664214
asked 6 hours ago
disney82231disney82231
352
$endgroup$
add a comment |
$begingroup$
There is a example in sklearn.metrics.average_precision_score documentation.
import numpy as np
from sklearn.metrics import average_precision_score
y_true = np.array([0, 0, 1, 1])
y_scores = np.array([0.1, 0.4, 0.35, 0.8])
average_precision_score(y_true, y_scores)
0.83
But when I plot precision_recall_curve
precision, recall, _ = precision_recall_curve(y_true, y_scores)
plt.plot( recall,precision)
I got the picture:
why the area under the precision_recall_curve is not 0.83?
python scikit-learn scoring
edited 2 hours ago
Ben Reiniger
664214
asked 6 hours ago
disney82231disney82231
352
$endgroup$
There is a example in sklearn.metrics.average_precision_score documentation.
import numpy as np
from sklearn.metrics import average_precision_score
y_true = np.array([0, 0, 1, 1])
y_scores = np.array([0.1, 0.4, 0.35, 0.8])
average_precision_score(y_true, y_scores)
0.83
But when I plot precision_recall_curve
precision, recall, _ = precision_recall_curve(y_true, y_scores)
plt.plot( recall,precision)
I got the picture:
why the area under the precision_recall_curve is not 0.83?
python scikit-learn scoring
python scikit-learn scoring
edited 2 hours ago
Ben Reiniger
664214
asked 6 hours ago
disney82231disney82231
352
edited 2 hours ago
Ben Reiniger
664214
asked 6 hours ago
disney82231disney82231
352
edited 2 hours ago
Ben Reiniger
664214
edited 2 hours ago
Ben Reiniger
664214
edited 2 hours ago
Ben Reiniger
664214
664214
asked 6 hours ago
disney82231disney82231
352
asked 6 hours ago
disney82231disney82231
352
asked 6 hours ago
disney82231disney82231
352
352
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
According to the documentation, the value is not exactly the area under curve, it is
$$textAP = sum_n(R_n - R_n-1)P_n.$$
which is a rectangular approximation.
For your specific example, it is calculated as
$$beginalign*
textAP & = overbrace(0.5 - 0.0)times1.0^(R_2 - R_1)P_2 + overbrace(0.5 - 0.5)times 0.5^(R_3 - R_2)P_3 + overbrace(1.0 - 0.5) times0.66^(R_4 - R_3)P_4 \
&= 0.5 + 0.00+ 0.33 = 0.83
endalign*$$
which is the area under the red curve as illustrated below:
answered 2 hours ago
EsmailianEsmailian
4,519422
$endgroup$
3
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to the documentation, the value is not exactly the area under curve, it is
$$textAP = sum_n(R_n - R_n-1)P_n.$$
which is a rectangular approximation.
For your specific example, it is calculated as
$$beginalign*
textAP & = overbrace(0.5 - 0.0)times1.0^(R_2 - R_1)P_2 + overbrace(0.5 - 0.5)times 0.5^(R_3 - R_2)P_3 + overbrace(1.0 - 0.5) times0.66^(R_4 - R_3)P_4 \
&= 0.5 + 0.00+ 0.33 = 0.83
endalign*$$
which is the area under the red curve as illustrated below:
answered 2 hours ago
EsmailianEsmailian
4,519422
$endgroup$
3
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
add a comment |
$begingroup$
According to the documentation, the value is not exactly the area under curve, it is
$$textAP = sum_n(R_n - R_n-1)P_n.$$
which is a rectangular approximation.
For your specific example, it is calculated as
$$beginalign*
textAP & = overbrace(0.5 - 0.0)times1.0^(R_2 - R_1)P_2 + overbrace(0.5 - 0.5)times 0.5^(R_3 - R_2)P_3 + overbrace(1.0 - 0.5) times0.66^(R_4 - R_3)P_4 \
&= 0.5 + 0.00+ 0.33 = 0.83
endalign*$$
which is the area under the red curve as illustrated below:
answered 2 hours ago
EsmailianEsmailian
4,519422
$endgroup$
3
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
add a comment |
$begingroup$
According to the documentation, the value is not exactly the area under curve, it is
$$textAP = sum_n(R_n - R_n-1)P_n.$$
which is a rectangular approximation.
For your specific example, it is calculated as
$$beginalign*
textAP & = overbrace(0.5 - 0.0)times1.0^(R_2 - R_1)P_2 + overbrace(0.5 - 0.5)times 0.5^(R_3 - R_2)P_3 + overbrace(1.0 - 0.5) times0.66^(R_4 - R_3)P_4 \
&= 0.5 + 0.00+ 0.33 = 0.83
endalign*$$
which is the area under the red curve as illustrated below:
answered 2 hours ago
EsmailianEsmailian
4,519422
$endgroup$
According to the documentation, the value is not exactly the area under curve, it is
$$textAP = sum_n(R_n - R_n-1)P_n.$$
which is a rectangular approximation.
For your specific example, it is calculated as
$$beginalign*
textAP & = overbrace(0.5 - 0.0)times1.0^(R_2 - R_1)P_2 + overbrace(0.5 - 0.5)times 0.5^(R_3 - R_2)P_3 + overbrace(1.0 - 0.5) times0.66^(R_4 - R_3)P_4 \
&= 0.5 + 0.00+ 0.33 = 0.83
endalign*$$
which is the area under the red curve as illustrated below:
answered 2 hours ago
EsmailianEsmailian
4,519422
answered 2 hours ago
EsmailianEsmailian
4,519422
answered 2 hours ago
EsmailianEsmailian
4,519422
answered 2 hours ago
EsmailianEsmailian
4,519422
4,519422
3
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
add a comment |
3
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
3
3
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
$begingroup$
Beat me to it; in particular, the documentation states "This implementation is not interpolated and is different from computing the area under the precision-recall curve with the trapezoidal rule, which uses linear interpolation and can be too optimistic."
$endgroup$
– Ben Reiniger
2 hours ago
add a comment |
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