How to use MLP mistakes in a certain class to improve it's precision in that particular class?
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Suppose we are developing an app which is supposed to predict a dog's breed by it's picture. We trained a classifier (in my case an MLP) using some dataset and shipped the app to users. Now suppose some user comes and takes a picture of a friend's dog and the app tells her there is 90% chance that this dog is an X. The user knows that this is not true, but she doesn't know what is the dog's breed (if she knew, why would she use our app in the first place?). So we get a feedback which tells us "this is a picture of a dog which is not an X". This sample could be a sample of some other class or a new class or not a dog at all.
I'm looking for a way to use this feedback, to improve the precision of my MLP in class X without touching other classes.
mlp
asked 2 days ago
MehrabanMehraban
1134
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add a comment |
$begingroup$
Suppose we are developing an app which is supposed to predict a dog's breed by it's picture. We trained a classifier (in my case an MLP) using some dataset and shipped the app to users. Now suppose some user comes and takes a picture of a friend's dog and the app tells her there is 90% chance that this dog is an X. The user knows that this is not true, but she doesn't know what is the dog's breed (if she knew, why would she use our app in the first place?). So we get a feedback which tells us "this is a picture of a dog which is not an X". This sample could be a sample of some other class or a new class or not a dog at all.
I'm looking for a way to use this feedback, to improve the precision of my MLP in class X without touching other classes.
mlp
asked 2 days ago
MehrabanMehraban
1134
New contributor
Mehraban 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 |
$begingroup$
Suppose we are developing an app which is supposed to predict a dog's breed by it's picture. We trained a classifier (in my case an MLP) using some dataset and shipped the app to users. Now suppose some user comes and takes a picture of a friend's dog and the app tells her there is 90% chance that this dog is an X. The user knows that this is not true, but she doesn't know what is the dog's breed (if she knew, why would she use our app in the first place?). So we get a feedback which tells us "this is a picture of a dog which is not an X". This sample could be a sample of some other class or a new class or not a dog at all.
I'm looking for a way to use this feedback, to improve the precision of my MLP in class X without touching other classes.
mlp
asked 2 days ago
MehrabanMehraban
1134
New contributor
Mehraban is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose we are developing an app which is supposed to predict a dog's breed by it's picture. We trained a classifier (in my case an MLP) using some dataset and shipped the app to users. Now suppose some user comes and takes a picture of a friend's dog and the app tells her there is 90% chance that this dog is an X. The user knows that this is not true, but she doesn't know what is the dog's breed (if she knew, why would she use our app in the first place?). So we get a feedback which tells us "this is a picture of a dog which is not an X". This sample could be a sample of some other class or a new class or not a dog at all.
I'm looking for a way to use this feedback, to improve the precision of my MLP in class X without touching other classes.
mlp
mlp
asked 2 days ago
MehrabanMehraban
1134
New contributor
Mehraban 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
MehrabanMehraban
1134
New contributor
Mehraban 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
MehrabanMehraban
1134
New contributor
Mehraban 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
MehrabanMehraban
1134
asked 2 days ago
MehrabanMehraban
1134
1134
New contributor
Mehraban is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mehraban is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mehraban is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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1 Answer
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This can be accomplished by a modification to multi-class cross-entropy.
We are faced with two types of supervision. First type is "data $i$ belongs to class $k$" denoted by $y_{ik}=1$, and second type is "data $i$ does not belong to class $k$" denoted by $bar{y}_{ik}=1$. For example, for 3 classes, $y_i=(1, 0, 0)$ denotes that point $i$ belongs to class $1$, and $bar{y}_{i}=(0, 0, 1)$ denotes that point $i$ does not belong to class $3$. Let $y'_{ik} in [0, 1]$ denote the model prediction. The original cross-entropy for $K$ classes is:
$$H_y(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})$$.
This objective assigns loss $-log(y'_{ik})$ to $y_{ik} = 1$ to encourage the model to output $y'_{ik} rightarrow 1$ leading to $-log(y'_{ik})rightarrow 0$.
On the other hand, for the second supervision $bar{y}_{ik}=1$, we want to encourage the model to output $y'_{ik} rightarrow 0$. Therefore, loss $-log(1- y'_{ik})$ can be used to have $-log(1- y'_{ik})rightarrow 0$.
Accordingly, second supervision can be combined with first one as follows:
$$H_{(y,bar{y})}(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})+bar{y}_{ik}log(1-y'_{ik})$$
Note that supervision "data $i$ does not belong to classes $1$ and $2$" is also supported. For example, $bar{y}_{i}=(1, 1, 0,...)$ activates both $-log(1 - y'_{i1})$ and $-log(1 - y'_{i2})$ to encourage the model to output less probabilities for classes $1$ and $2$, i.e. $y'_{i1} rightarrow 0$, and $y'_{i2} rightarrow 0$.
answered yesterday
EsmailianEsmailian
3865
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$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
add a comment |
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1 Answer
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$begingroup$
This can be accomplished by a modification to multi-class cross-entropy.
We are faced with two types of supervision. First type is "data $i$ belongs to class $k$" denoted by $y_{ik}=1$, and second type is "data $i$ does not belong to class $k$" denoted by $bar{y}_{ik}=1$. For example, for 3 classes, $y_i=(1, 0, 0)$ denotes that point $i$ belongs to class $1$, and $bar{y}_{i}=(0, 0, 1)$ denotes that point $i$ does not belong to class $3$. Let $y'_{ik} in [0, 1]$ denote the model prediction. The original cross-entropy for $K$ classes is:
$$H_y(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})$$.
This objective assigns loss $-log(y'_{ik})$ to $y_{ik} = 1$ to encourage the model to output $y'_{ik} rightarrow 1$ leading to $-log(y'_{ik})rightarrow 0$.
On the other hand, for the second supervision $bar{y}_{ik}=1$, we want to encourage the model to output $y'_{ik} rightarrow 0$. Therefore, loss $-log(1- y'_{ik})$ can be used to have $-log(1- y'_{ik})rightarrow 0$.
Accordingly, second supervision can be combined with first one as follows:
$$H_{(y,bar{y})}(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})+bar{y}_{ik}log(1-y'_{ik})$$
Note that supervision "data $i$ does not belong to classes $1$ and $2$" is also supported. For example, $bar{y}_{i}=(1, 1, 0,...)$ activates both $-log(1 - y'_{i1})$ and $-log(1 - y'_{i2})$ to encourage the model to output less probabilities for classes $1$ and $2$, i.e. $y'_{i1} rightarrow 0$, and $y'_{i2} rightarrow 0$.
answered yesterday
EsmailianEsmailian
3865
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
add a comment |
$begingroup$
This can be accomplished by a modification to multi-class cross-entropy.
We are faced with two types of supervision. First type is "data $i$ belongs to class $k$" denoted by $y_{ik}=1$, and second type is "data $i$ does not belong to class $k$" denoted by $bar{y}_{ik}=1$. For example, for 3 classes, $y_i=(1, 0, 0)$ denotes that point $i$ belongs to class $1$, and $bar{y}_{i}=(0, 0, 1)$ denotes that point $i$ does not belong to class $3$. Let $y'_{ik} in [0, 1]$ denote the model prediction. The original cross-entropy for $K$ classes is:
$$H_y(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})$$.
This objective assigns loss $-log(y'_{ik})$ to $y_{ik} = 1$ to encourage the model to output $y'_{ik} rightarrow 1$ leading to $-log(y'_{ik})rightarrow 0$.
On the other hand, for the second supervision $bar{y}_{ik}=1$, we want to encourage the model to output $y'_{ik} rightarrow 0$. Therefore, loss $-log(1- y'_{ik})$ can be used to have $-log(1- y'_{ik})rightarrow 0$.
Accordingly, second supervision can be combined with first one as follows:
$$H_{(y,bar{y})}(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})+bar{y}_{ik}log(1-y'_{ik})$$
Note that supervision "data $i$ does not belong to classes $1$ and $2$" is also supported. For example, $bar{y}_{i}=(1, 1, 0,...)$ activates both $-log(1 - y'_{i1})$ and $-log(1 - y'_{i2})$ to encourage the model to output less probabilities for classes $1$ and $2$, i.e. $y'_{i1} rightarrow 0$, and $y'_{i2} rightarrow 0$.
answered yesterday
EsmailianEsmailian
3865
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
add a comment |
$begingroup$
This can be accomplished by a modification to multi-class cross-entropy.
We are faced with two types of supervision. First type is "data $i$ belongs to class $k$" denoted by $y_{ik}=1$, and second type is "data $i$ does not belong to class $k$" denoted by $bar{y}_{ik}=1$. For example, for 3 classes, $y_i=(1, 0, 0)$ denotes that point $i$ belongs to class $1$, and $bar{y}_{i}=(0, 0, 1)$ denotes that point $i$ does not belong to class $3$. Let $y'_{ik} in [0, 1]$ denote the model prediction. The original cross-entropy for $K$ classes is:
$$H_y(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})$$.
This objective assigns loss $-log(y'_{ik})$ to $y_{ik} = 1$ to encourage the model to output $y'_{ik} rightarrow 1$ leading to $-log(y'_{ik})rightarrow 0$.
On the other hand, for the second supervision $bar{y}_{ik}=1$, we want to encourage the model to output $y'_{ik} rightarrow 0$. Therefore, loss $-log(1- y'_{ik})$ can be used to have $-log(1- y'_{ik})rightarrow 0$.
Accordingly, second supervision can be combined with first one as follows:
$$H_{(y,bar{y})}(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})+bar{y}_{ik}log(1-y'_{ik})$$
Note that supervision "data $i$ does not belong to classes $1$ and $2$" is also supported. For example, $bar{y}_{i}=(1, 1, 0,...)$ activates both $-log(1 - y'_{i1})$ and $-log(1 - y'_{i2})$ to encourage the model to output less probabilities for classes $1$ and $2$, i.e. $y'_{i1} rightarrow 0$, and $y'_{i2} rightarrow 0$.
answered yesterday
EsmailianEsmailian
3865
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This can be accomplished by a modification to multi-class cross-entropy.
We are faced with two types of supervision. First type is "data $i$ belongs to class $k$" denoted by $y_{ik}=1$, and second type is "data $i$ does not belong to class $k$" denoted by $bar{y}_{ik}=1$. For example, for 3 classes, $y_i=(1, 0, 0)$ denotes that point $i$ belongs to class $1$, and $bar{y}_{i}=(0, 0, 1)$ denotes that point $i$ does not belong to class $3$. Let $y'_{ik} in [0, 1]$ denote the model prediction. The original cross-entropy for $K$ classes is:
$$H_y(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})$$.
This objective assigns loss $-log(y'_{ik})$ to $y_{ik} = 1$ to encourage the model to output $y'_{ik} rightarrow 1$ leading to $-log(y'_{ik})rightarrow 0$.
On the other hand, for the second supervision $bar{y}_{ik}=1$, we want to encourage the model to output $y'_{ik} rightarrow 0$. Therefore, loss $-log(1- y'_{ik})$ can be used to have $-log(1- y'_{ik})rightarrow 0$.
Accordingly, second supervision can be combined with first one as follows:
$$H_{(y,bar{y})}(y')=-sum_{i}sum_{k=1}^{K}y_{ik}log(y'_{ik})+bar{y}_{ik}log(1-y'_{ik})$$
Note that supervision "data $i$ does not belong to classes $1$ and $2$" is also supported. For example, $bar{y}_{i}=(1, 1, 0,...)$ activates both $-log(1 - y'_{i1})$ and $-log(1 - y'_{i2})$ to encourage the model to output less probabilities for classes $1$ and $2$, i.e. $y'_{i1} rightarrow 0$, and $y'_{i2} rightarrow 0$.
answered yesterday
EsmailianEsmailian
3865
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
EsmailianEsmailian
3865
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
EsmailianEsmailian
3865
answered yesterday
EsmailianEsmailian
3865
3865
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Esmailian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
add a comment |
$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
What is $y'$? Did you mix $y'$ and $bar y$?
$endgroup$
– Mehraban
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
$begingroup$
It denotes the model prediction. No they are not mixed.
$endgroup$
– Esmailian
yesterday
add a comment |
Mehraban is a new contributor. Be nice, and check out our Code of Conduct.
Mehraban is a new contributor. Be nice, and check out our Code of Conduct.
Mehraban is a new contributor. Be nice, and check out our Code of Conduct.
Mehraban is a new contributor. Be nice, and check out our Code of Conduct.
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