How to do the sum for 2 gradient objects in the deep learning toolbox?

조회 수: 4(최근 30일)
SC 2019년 11월 27일
답변: Sourav Bairagya 2019년 12월 10일
Hi,
I have gradients1 and gradients2 which have exactly same structure but different numerical values. How can I do the sum? Current I tried gradients1+gradients2 but I got error.
Thanks!
My code:
rng(123); % seed
X_ori=[4,163,80;5,164,75]; % data; #(number) = 2; #(features) = 3;
X=permute(X_ori,[3,4,2,1]);
dlX = dlarray(X, 'SSCB');
Y_ori=[0, 0, 0, 1; 0, 1, 0, 0]; % data labels (i.e. one-hot vectors for 4 classes)
myModel = [
imageInputLayer([1 1 3],'Normalization','none','Name','in')
fullyConnectedLayer(7,'Name','Layer 1')
fullyConnectedLayer(4,'Name','Layer 2')];
MyLGraph = layerGraph(myModel);
myDLnet = dlnetwork(MyLGraph);
CorrectLabels_transpose=transpose(CorrectLabels);
[modelOutput,state] = forward(myModel,modelInput);
loss = -31*sum(sum(CorrectLabels_transpose.*log(sigmoid(modelOutput/100))));
end
CorrectLabels_transpose=transpose(CorrectLabels);
[modelOutput,state] = forward(myModel,modelInput);
loss = -42*sum(sum(CorrectLabels_transpose.*log(sigmoid(modelOutput/100))));
end
댓글 수: 1표시 없음숨기기 없음
SC 2019년 11월 30일
편집: SC 2019년 11월 30일
I can change the line "gradients_sum = gradients1+gradients2;" to the followings and then I can get the sum. But I still want to know if there are some more efficient ways to do so.
for i=1:num_layers
end
end

댓글을 달려면 로그인하십시오.

답변(1개)

Sourav Bairagya 2019년 12월 10일
As in this case, 'gradients1.Value' and 'gradients2.Value' both are cell arrays and each one contains another cell arrays as elements within it, hence, direct conversion of these two cell arrays into matrices using 'cell2mat' or direct addition of them using '+' operator is not possible. Hence, you have to access each element individually and add them.

댓글을 달려면 로그인하십시오.

범주

Find more on Image Data Workflows in Help Center and File Exchange

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by