I have some images which look like this one:


They exist of 3 possible characters (A-C) and a length of 4.

Now, I would like to run a neural network, which recognizes each character in the picture without any segmentation before that; I would like to use a similar approach as the authors in this paper did: https://medusa.fit.vutbr.cz/traffic/research-topics/general-traffic-analysis/holistic-recognition-of-low-quality-license-plates-by-cnn-using-track-annotated-data-iwt4s-avss-2017/.

Now, I am wondering which kind of classification problem I am facing and how to solve that?

Is this a multi-label classification problem? Or a multi-class problem? Or is it a sequential data labeling problem (I assume this)? Or something completely different?

To differentiate the different problems: Multi-label classification would mean: First class is 'AAAA', second class 'AAAB', ... and the last class 'CCCC'. That means, we would have 3x3x3x3 = 81 possible classes. So, we would have one column for each label which contains a string label.

Multi-label classification would mean: The labels are split in parts for every character. First position: A-D. Second: A-D, Third: A-D, Fourth: A-D. So, we would have 4 columns for each label with the appropriate character.

I think, both are not the approach the authors used. If I understood correctly the paper I linked above, they handled the data as sequential data, and I do not have to split the labels but I can use the whole label. Futhermore, the convolutional neural network is able to learn the spatial positions automatically. Because of that, I do not have to split the labels but can them use as they are. If this is right, then I think the target shape of my output should be something like (n, 4, 3), with n as the number of images, 4 as the length of the text labels and 3 as the possibilities for each character.

Is this correct?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.