Comments (4)
Aha, that makes a lot of sense, thank you!
I was confused because another paper cited your work but didn't mention the label-conditioned part in their approach. They didn't open source their code so I came here.
Thanks for your explanation!
from const_layout.
Thank you for your question! And sorry for the late response.
This function assumes that the two input layouts have the same label set and that _bi
and _bj
are bbox sequences of the same length, i.e., both have the same number of a specific type of elements. The code for creating two sets of layouts that have the same label set can be found here.
from const_layout.
Thanks for the explanation!
But since the code took an intersection between the 2 key sets (line 130), it seems like only the shared labels and their bounding boxes are considered in computing the max iou? My question is largely, what happens when the generated result has different quantity of keys/bounding boxes?
from const_layout.
Yes, we can only compute the max iou with the shared labels and their bounding boxes. Since our problem setting is label-conditioned layout generation, we evaluated the models with given labels, i.e., using the same labels as the test set (see generation code).
For different quantities of keys (layouts?) or bounding boxes, the max iou becomes ill-defined and would cause unwanted behavior such as ignoring partial boxes.
from const_layout.
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from const_layout.