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Explore Stack InternalYou verballyalmost solved the problem in the last paragraph. Expressed more formally, your cost function could be
$$\frac{1}{N} \sum_{i,j} c_{i,j} y_{i,j} \log x_{i,j}$$
where $i$ runs over items/documents, and $j$ runs over classes, $x$ is your prediction, $y$ is the binary label (1 if item $i$ has class $j$), and $0 < c < 1$ is your confidence. This is a simple modification of the cross entropy. When the confidence $c$ is low, the value of the prediction matters less, as desired.
You verbally solved the problem in the last paragraph. Expressed more formally, your cost function could be
$$\frac{1}{N} \sum_{i,j} c_{i,j} y_{i,j} \log x_{i,j}$$
where $i$ runs over items/documents, and $j$ runs over classes, $x$ is your prediction, $y$ is the binary label (1 if item $i$ has class $j$), and $0 < c < 1$ is your confidence. This is a simple modification of the cross entropy. When the confidence $c$ is low, the value of the prediction matters less, as desired.
You almost solved the problem in the last paragraph. Expressed more formally, your cost function could be
$$\frac{1}{N} \sum_{i,j} c_{i,j} y_{i,j} \log x_{i,j}$$
where $i$ runs over items/documents, and $j$ runs over classes, $x$ is your prediction, $y$ is the binary label (1 if item $i$ has class $j$), and $0 < c < 1$ is your confidence. This is a simple modification of the cross entropy. When the confidence $c$ is low, the value of the prediction matters less.
You verbally solved the problem in the last paragraph. Expressed more formally, your cost function could be
$$\frac{1}{N} \sum_{i,j} c_{i,j} y_{i,j} \log x_{i,j}$$
where $i$ runs over items/documents, and $j$ runs over classes, $x$ is your prediction, $y$ is the binary label (1 if item $i$ has class $j$), and $0 < c < 1$ is your confidence. This is a simple modification of the cross entropy. When the confidence $c$ is low, the value of the prediction matters less, as desired.
