I have never seen such a great course video! Thank you very much
@lucywang86433 жыл бұрын
did such a great job in explaining! Thank you!
@osmanjant4 жыл бұрын
You are aswesome teacher. Don't stop uploading new videos. Thanks.
@zdx45715 ай бұрын
谢谢老师让我了解新知识~
@ritaxu793424 күн бұрын
感谢老师!终于懂了为什么positional embedding可以直接加上来
@mengyuge33693 жыл бұрын
Thank you very much. Great video
@sollunaliu71333 жыл бұрын
史上最强transfermer视频讲解,支持老师~
@StevenLiuX4 жыл бұрын
感谢老师! 终于看懂了 恨不能多点几个赞!
@wolfmib4 жыл бұрын
for 13:20: we could consider two Vector with D dimension : 1. When doing the inner product of the vector pairs , we can think : ___ the same vector with different dimension shall meet equal relationship as we expected:___ such as A(1,1,1), with B(1,1,1) D=3 a(1,1,1,1) with b (1,1,1,1) , D = 4 This two pair of vector (A,B ) , (a,b) shall has the same attention value: So take the inner product for both of two paris: A * B = 3 a * b = 4 and we found 3 != 4, so we divide by the square of the dimension: A*B / sqrt(3) = 3 / 1.7320 ~ 1.7320 a* b / sqrt(4) = 4 / 2 ~ 2 by this approximate : indeed, A*B is closer to a*b (even it's not exactly equal) , but it definitely is a better solution instead of taking the inner product without divide anything.
26:40 Is the head-split happening at ai? or at qi/ki/vi?
@jkrong87965 жыл бұрын
这么好的视频还有人踩的咯
@xiangliu27672 ай бұрын
这个encode 和decode动画哪里可以看到,地址是什么
@user-cg5ke1pv1z3 жыл бұрын
给力
@user-kt8nc4xd1u2 жыл бұрын
At 40:03, is b=[b^1; b^2;...;b^4] or b is one of b^1, b^2, b^3, and b^4?
@beandog54454 жыл бұрын
you save my ass
@castaway16974 жыл бұрын
Thanks a lot!
@user-bz3ie9jm3s5 жыл бұрын
课程很棒,请问老师后面会有 BERT 的讲解吗
@HungyiLeeNTU5 жыл бұрын
會有的
@mitddxia47993 жыл бұрын
第33分钟左右的矩阵分块有点问题,结果不是直接相加,而是上下的关系才对
@gourmetdata9713 жыл бұрын
有一个问题:positional encoding 应该是直接加上去而不是concatenate吧。原文是:The positional encodings have the same dimension d_model as the embeddings, so that the two can be summed