Micro F1 score in Scikit-Learn with Class imbalance
I have some class imbalance and a simple baseline classifier that assigns the majority class to every sample:
from sklearn.metrics import precision_score, recall_score, confusion_matrix
y_true = [0,0,0,1]
y_pred = [0,0,0,0]
confusion_matrix(y_true, y_pred)
This yields
[[3, 0],
[1, 0]]
This means TP=3, FP=1, FN=0.
So far, so good. Now I want to calculate the micro average of precision and recall.
precision_score(y_true, y_pred, average='micro') # yields 0.75
recall_score(y_true, y_pred, average='micro') # yields 0.75
I am Ok with the precision, but why is recall not 1.0? How can they ever be the same in this example, given that FP > 0 and FN == 0? I know it must have to do with the micro averaging, but I can't wrap my head around this one.
scikit-learn precision-recall
asked Nov 20 '18 at 14:16
TobyToby
5901719
add a comment |
I have some class imbalance and a simple baseline classifier that assigns the majority class to every sample:
from sklearn.metrics import precision_score, recall_score, confusion_matrix
y_true = [0,0,0,1]
y_pred = [0,0,0,0]
confusion_matrix(y_true, y_pred)
This yields
[[3, 0],
[1, 0]]
This means TP=3, FP=1, FN=0.
So far, so good. Now I want to calculate the micro average of precision and recall.
precision_score(y_true, y_pred, average='micro') # yields 0.75
recall_score(y_true, y_pred, average='micro') # yields 0.75
I am Ok with the precision, but why is recall not 1.0? How can they ever be the same in this example, given that FP > 0 and FN == 0? I know it must have to do with the micro averaging, but I can't wrap my head around this one.
scikit-learn precision-recall
asked Nov 20 '18 at 14:16
TobyToby
5901719
add a comment |
I have some class imbalance and a simple baseline classifier that assigns the majority class to every sample:
from sklearn.metrics import precision_score, recall_score, confusion_matrix
y_true = [0,0,0,1]
y_pred = [0,0,0,0]
confusion_matrix(y_true, y_pred)
This yields
[[3, 0],
[1, 0]]
This means TP=3, FP=1, FN=0.
So far, so good. Now I want to calculate the micro average of precision and recall.
precision_score(y_true, y_pred, average='micro') # yields 0.75
recall_score(y_true, y_pred, average='micro') # yields 0.75
I am Ok with the precision, but why is recall not 1.0? How can they ever be the same in this example, given that FP > 0 and FN == 0? I know it must have to do with the micro averaging, but I can't wrap my head around this one.
scikit-learn precision-recall
asked Nov 20 '18 at 14:16
TobyToby
5901719
I have some class imbalance and a simple baseline classifier that assigns the majority class to every sample:
from sklearn.metrics import precision_score, recall_score, confusion_matrix
y_true = [0,0,0,1]
y_pred = [0,0,0,0]
confusion_matrix(y_true, y_pred)
This yields
[[3, 0],
[1, 0]]
This means TP=3, FP=1, FN=0.
So far, so good. Now I want to calculate the micro average of precision and recall.
precision_score(y_true, y_pred, average='micro') # yields 0.75
recall_score(y_true, y_pred, average='micro') # yields 0.75
I am Ok with the precision, but why is recall not 1.0? How can they ever be the same in this example, given that FP > 0 and FN == 0? I know it must have to do with the micro averaging, but I can't wrap my head around this one.
scikit-learn precision-recall
scikit-learn precision-recall
asked Nov 20 '18 at 14:16
TobyToby
5901719
asked Nov 20 '18 at 14:16
TobyToby
5901719
asked Nov 20 '18 at 14:16
TobyToby
5901719
asked Nov 20 '18 at 14:16
TobyToby
5901719
asked Nov 20 '18 at 14:16
TobyToby
5901719
5901719
add a comment |
add a comment |
1 Answer
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Yes, its because of micro-averaging. See the documentation here to know how its calculated:
Note that if all labels are included, “micro”-averaging in a
multiclass setting will produce precision, recall and f-score that are all
identical to accuracy.
As you can see in the above linked page, both precision and recall are defined as:
where R(y, y-hat) is:
So in your case, Recall-micro will be calculated as
R = number of correct predictions / total predictions = 3/4 = 0.75
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
1
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
add a comment |
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votes
Yes, its because of micro-averaging. See the documentation here to know how its calculated:
Note that if all labels are included, “micro”-averaging in a
multiclass setting will produce precision, recall and f-score that are all
identical to accuracy.
As you can see in the above linked page, both precision and recall are defined as:
where R(y, y-hat) is:
So in your case, Recall-micro will be calculated as
R = number of correct predictions / total predictions = 3/4 = 0.75
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
1
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
add a comment |
Yes, its because of micro-averaging. See the documentation here to know how its calculated:
Note that if all labels are included, “micro”-averaging in a
multiclass setting will produce precision, recall and f-score that are all
identical to accuracy.
As you can see in the above linked page, both precision and recall are defined as:
where R(y, y-hat) is:
So in your case, Recall-micro will be calculated as
R = number of correct predictions / total predictions = 3/4 = 0.75
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
1
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
add a comment |
Yes, its because of micro-averaging. See the documentation here to know how its calculated:
Note that if all labels are included, “micro”-averaging in a
multiclass setting will produce precision, recall and f-score that are all
identical to accuracy.
As you can see in the above linked page, both precision and recall are defined as:
where R(y, y-hat) is:
So in your case, Recall-micro will be calculated as
R = number of correct predictions / total predictions = 3/4 = 0.75
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
Yes, its because of micro-averaging. See the documentation here to know how its calculated:
Note that if all labels are included, “micro”-averaging in a
multiclass setting will produce precision, recall and f-score that are all
identical to accuracy.
As you can see in the above linked page, both precision and recall are defined as:
where R(y, y-hat) is:
So in your case, Recall-micro will be calculated as
R = number of correct predictions / total predictions = 3/4 = 0.75
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
answered Nov 21 '18 at 10:37
Vivek KumarVivek Kumar
16.3k42055
16.3k42055
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
1
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
add a comment |
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
1
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
Thanks. The paragraph you linked is weird though. The sentence "Note that if all labels are included, “micro”-averaging in a multiclass setting will produce precision, recall and that are all identical to accuracy" is in line with what you say, but that paragraph does not say what A and B actually are, and weirdly, it defines y-hat as the set of true labels and y as the set of predicted labels, which is unconventional, no? My conclusion is to not use micro-averaging for binary classification and to think some more about this.
– Toby
Nov 28 '18 at 8:24
1
1
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
Ah, forget the thing I said about A, B not being defined, I get it. But too late to edit my comment. Thanks!
– Toby
Nov 28 '18 at 8:32
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
@Toby Yes, that is unconventional. But all the following things match with this, so no worries. If you want, you may put an issue regarding this usage on scikit-learn github page
– Vivek Kumar
Nov 28 '18 at 8:33
add a comment |
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