Why is this the correct way to do a cost function for a neural network?Neural Network Cost Function in MATLABUnderstanding Neural Network BackpropagationRole of Bias in Neural NetworksEpoch vs Iteration when training neural networksWhat are advantages of Artificial Neural Networks over Support Vector Machines?In Neural network prediction(not classification), What Cost function shall i use?Neural Network Cost Function in MATLABNeural Network not fitting XORNeural network backpropagation with RELUCalculation of Cost function in Neural Network Getting NaN or InfNeural Network Cost function returning Nan in Matlab

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Why is this the correct way to do a cost function for a neural network?


Neural Network Cost Function in MATLABUnderstanding Neural Network BackpropagationRole of Bias in Neural NetworksEpoch vs Iteration when training neural networksWhat are advantages of Artificial Neural Networks over Support Vector Machines?In Neural network prediction(not classification), What Cost function shall i use?Neural Network Cost Function in MATLABNeural Network not fitting XORNeural network backpropagation with RELUCalculation of Cost function in Neural Network Getting NaN or InfNeural Network Cost function returning Nan in Matlab






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








-1















So after beating my head against the wall for a few hours, I looked online for a solution to my problem, and it worked great. I just want to know what caused the issue with the way I was originally going about it.



enter image description here



enter image description here



here are some more details. The input is a 20x20px image from the MNIST datset, and there are 5000 samples, so X, or A1 is 5000x400. There are 25 nodes in the single hidden layer. The output is a one hot vector of 0-9 digits. y (not Y, which is the one hot encoding of y) is a 5000x1 vector with the value of 1-10.



enter image description here



Here was my original code for the cost function:



Y = zeros(m, num_labels);
for i = 1:m
Y(i, y(i)) = 1;
endfor
H = sigmoid(Theta2*[ones(1,m);sigmoid(Theta1*[ones(m, 1) X]'))
J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))')))


But then I found this:



A1 = [ones(m, 1) X];
Z2 = A1 * Theta1';
A2 = [ones(size(Z2, 1), 1) sigmoid(Z2)];
Z3 = A2*Theta2';
H = A3 = sigmoid(Z3);

J = (1/m)*sum(sum((-Y).*log(H) - (1-Y).*log(1-H), 2));


I see that this may be slightly cleaner, but what functionally causes my original code to get 304.88 and the other to get ~ 0.25? Is it the element wise multiplication?



FYI, this is the same problem as this question if you need the formal equation written out.



Thanks for any help I can get! I really want to understand where I'm going wrong










share|improve this question
























  • MCVE ? So that we can see the two codes actually in action?

    – tryman
    Mar 25 at 4:15












  • @tryman Do those two screen grabs help clear it up? The whole problem is just writing a cost function for a neural network in Octave

    – Carson P
    Mar 25 at 4:19











  • @tryman I added a lot more details. Hopefully that will clear everything up

    – Carson P
    Mar 25 at 4:24











  • With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).

    – tryman
    Mar 25 at 4:36












  • @tryman thanks! What I’m really trying to figure out is how I was supposed to know to do an element wise multiply instead of matrix wise, just from those equations and figures above. Since In that equation, y and log don’t have some kind of * does that technically mean it’s element wise? Or is there something else here I’m not getting

    – Carson P
    Mar 25 at 4:38


















-1















So after beating my head against the wall for a few hours, I looked online for a solution to my problem, and it worked great. I just want to know what caused the issue with the way I was originally going about it.



enter image description here



enter image description here



here are some more details. The input is a 20x20px image from the MNIST datset, and there are 5000 samples, so X, or A1 is 5000x400. There are 25 nodes in the single hidden layer. The output is a one hot vector of 0-9 digits. y (not Y, which is the one hot encoding of y) is a 5000x1 vector with the value of 1-10.



enter image description here



Here was my original code for the cost function:



Y = zeros(m, num_labels);
for i = 1:m
Y(i, y(i)) = 1;
endfor
H = sigmoid(Theta2*[ones(1,m);sigmoid(Theta1*[ones(m, 1) X]'))
J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))')))


But then I found this:



A1 = [ones(m, 1) X];
Z2 = A1 * Theta1';
A2 = [ones(size(Z2, 1), 1) sigmoid(Z2)];
Z3 = A2*Theta2';
H = A3 = sigmoid(Z3);

J = (1/m)*sum(sum((-Y).*log(H) - (1-Y).*log(1-H), 2));


I see that this may be slightly cleaner, but what functionally causes my original code to get 304.88 and the other to get ~ 0.25? Is it the element wise multiplication?



FYI, this is the same problem as this question if you need the formal equation written out.



Thanks for any help I can get! I really want to understand where I'm going wrong










share|improve this question
























  • MCVE ? So that we can see the two codes actually in action?

    – tryman
    Mar 25 at 4:15












  • @tryman Do those two screen grabs help clear it up? The whole problem is just writing a cost function for a neural network in Octave

    – Carson P
    Mar 25 at 4:19











  • @tryman I added a lot more details. Hopefully that will clear everything up

    – Carson P
    Mar 25 at 4:24











  • With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).

    – tryman
    Mar 25 at 4:36












  • @tryman thanks! What I’m really trying to figure out is how I was supposed to know to do an element wise multiply instead of matrix wise, just from those equations and figures above. Since In that equation, y and log don’t have some kind of * does that technically mean it’s element wise? Or is there something else here I’m not getting

    – Carson P
    Mar 25 at 4:38














-1












-1








-1








So after beating my head against the wall for a few hours, I looked online for a solution to my problem, and it worked great. I just want to know what caused the issue with the way I was originally going about it.



enter image description here



enter image description here



here are some more details. The input is a 20x20px image from the MNIST datset, and there are 5000 samples, so X, or A1 is 5000x400. There are 25 nodes in the single hidden layer. The output is a one hot vector of 0-9 digits. y (not Y, which is the one hot encoding of y) is a 5000x1 vector with the value of 1-10.



enter image description here



Here was my original code for the cost function:



Y = zeros(m, num_labels);
for i = 1:m
Y(i, y(i)) = 1;
endfor
H = sigmoid(Theta2*[ones(1,m);sigmoid(Theta1*[ones(m, 1) X]'))
J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))')))


But then I found this:



A1 = [ones(m, 1) X];
Z2 = A1 * Theta1';
A2 = [ones(size(Z2, 1), 1) sigmoid(Z2)];
Z3 = A2*Theta2';
H = A3 = sigmoid(Z3);

J = (1/m)*sum(sum((-Y).*log(H) - (1-Y).*log(1-H), 2));


I see that this may be slightly cleaner, but what functionally causes my original code to get 304.88 and the other to get ~ 0.25? Is it the element wise multiplication?



FYI, this is the same problem as this question if you need the formal equation written out.



Thanks for any help I can get! I really want to understand where I'm going wrong










share|improve this question
















So after beating my head against the wall for a few hours, I looked online for a solution to my problem, and it worked great. I just want to know what caused the issue with the way I was originally going about it.



enter image description here



enter image description here



here are some more details. The input is a 20x20px image from the MNIST datset, and there are 5000 samples, so X, or A1 is 5000x400. There are 25 nodes in the single hidden layer. The output is a one hot vector of 0-9 digits. y (not Y, which is the one hot encoding of y) is a 5000x1 vector with the value of 1-10.



enter image description here



Here was my original code for the cost function:



Y = zeros(m, num_labels);
for i = 1:m
Y(i, y(i)) = 1;
endfor
H = sigmoid(Theta2*[ones(1,m);sigmoid(Theta1*[ones(m, 1) X]'))
J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))')))


But then I found this:



A1 = [ones(m, 1) X];
Z2 = A1 * Theta1';
A2 = [ones(size(Z2, 1), 1) sigmoid(Z2)];
Z3 = A2*Theta2';
H = A3 = sigmoid(Z3);

J = (1/m)*sum(sum((-Y).*log(H) - (1-Y).*log(1-H), 2));


I see that this may be slightly cleaner, but what functionally causes my original code to get 304.88 and the other to get ~ 0.25? Is it the element wise multiplication?



FYI, this is the same problem as this question if you need the formal equation written out.



Thanks for any help I can get! I really want to understand where I'm going wrong







matlab neural-network vectorization octave backpropagation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 25 at 4:24







Carson P

















asked Mar 25 at 4:11









Carson PCarson P

836




836












  • MCVE ? So that we can see the two codes actually in action?

    – tryman
    Mar 25 at 4:15












  • @tryman Do those two screen grabs help clear it up? The whole problem is just writing a cost function for a neural network in Octave

    – Carson P
    Mar 25 at 4:19











  • @tryman I added a lot more details. Hopefully that will clear everything up

    – Carson P
    Mar 25 at 4:24











  • With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).

    – tryman
    Mar 25 at 4:36












  • @tryman thanks! What I’m really trying to figure out is how I was supposed to know to do an element wise multiply instead of matrix wise, just from those equations and figures above. Since In that equation, y and log don’t have some kind of * does that technically mean it’s element wise? Or is there something else here I’m not getting

    – Carson P
    Mar 25 at 4:38


















  • MCVE ? So that we can see the two codes actually in action?

    – tryman
    Mar 25 at 4:15












  • @tryman Do those two screen grabs help clear it up? The whole problem is just writing a cost function for a neural network in Octave

    – Carson P
    Mar 25 at 4:19











  • @tryman I added a lot more details. Hopefully that will clear everything up

    – Carson P
    Mar 25 at 4:24











  • With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).

    – tryman
    Mar 25 at 4:36












  • @tryman thanks! What I’m really trying to figure out is how I was supposed to know to do an element wise multiply instead of matrix wise, just from those equations and figures above. Since In that equation, y and log don’t have some kind of * does that technically mean it’s element wise? Or is there something else here I’m not getting

    – Carson P
    Mar 25 at 4:38

















MCVE ? So that we can see the two codes actually in action?

– tryman
Mar 25 at 4:15






MCVE ? So that we can see the two codes actually in action?

– tryman
Mar 25 at 4:15














@tryman Do those two screen grabs help clear it up? The whole problem is just writing a cost function for a neural network in Octave

– Carson P
Mar 25 at 4:19





@tryman Do those two screen grabs help clear it up? The whole problem is just writing a cost function for a neural network in Octave

– Carson P
Mar 25 at 4:19













@tryman I added a lot more details. Hopefully that will clear everything up

– Carson P
Mar 25 at 4:24





@tryman I added a lot more details. Hopefully that will clear everything up

– Carson P
Mar 25 at 4:24













With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).

– tryman
Mar 25 at 4:36






With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).

– tryman
Mar 25 at 4:36














@tryman thanks! What I’m really trying to figure out is how I was supposed to know to do an element wise multiply instead of matrix wise, just from those equations and figures above. Since In that equation, y and log don’t have some kind of * does that technically mean it’s element wise? Or is there something else here I’m not getting

– Carson P
Mar 25 at 4:38






@tryman thanks! What I’m really trying to figure out is how I was supposed to know to do an element wise multiply instead of matrix wise, just from those equations and figures above. Since In that equation, y and log don’t have some kind of * does that technically mean it’s element wise? Or is there something else here I’m not getting

– Carson P
Mar 25 at 4:38













1 Answer
1






active

oldest

votes


















1














Transfer from the comments:

With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).



Update 1:

In regards to your question in the comment.
From the first screenshot, you represent each value yk(i) in the entry Y(i,k) of the Y matrix and each value h(x^(i))k as H(i,k). So basically, for each i,k you want to compute Y(i,k) log(H(i,k)) + (1-Y(i,k)) log(1-H(i,k)). You can do it for all the values together and store the result in matrix C. Then C = Y.*log(H) + (1-Y).*log(1-H) and each C(i,k) has the above mentioned value. This is an operation .* because you want to do the operation for each element (i,k) of each matrix (in contrast to multiplying the matrices which is totally different). Afterwards, to get the sum of all the values inside the 2D dimensional matrix C, you use the octave function sum twice: sum(sum(C)) to sum both columnwise and row-wise (or as @ Irreducible suggested, just sum(C(:))).



Note there may be other errors as well.






share|improve this answer

























  • To sum over all entries of a matrix C just write: sum(C(:))

    – Irreducible
    Mar 25 at 12:29












  • @Irreducible That too! Incorporated your suggestion.

    – tryman
    Apr 4 at 10:32











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Transfer from the comments:

With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).



Update 1:

In regards to your question in the comment.
From the first screenshot, you represent each value yk(i) in the entry Y(i,k) of the Y matrix and each value h(x^(i))k as H(i,k). So basically, for each i,k you want to compute Y(i,k) log(H(i,k)) + (1-Y(i,k)) log(1-H(i,k)). You can do it for all the values together and store the result in matrix C. Then C = Y.*log(H) + (1-Y).*log(1-H) and each C(i,k) has the above mentioned value. This is an operation .* because you want to do the operation for each element (i,k) of each matrix (in contrast to multiplying the matrices which is totally different). Afterwards, to get the sum of all the values inside the 2D dimensional matrix C, you use the octave function sum twice: sum(sum(C)) to sum both columnwise and row-wise (or as @ Irreducible suggested, just sum(C(:))).



Note there may be other errors as well.






share|improve this answer

























  • To sum over all entries of a matrix C just write: sum(C(:))

    – Irreducible
    Mar 25 at 12:29












  • @Irreducible That too! Incorporated your suggestion.

    – tryman
    Apr 4 at 10:32















1














Transfer from the comments:

With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).



Update 1:

In regards to your question in the comment.
From the first screenshot, you represent each value yk(i) in the entry Y(i,k) of the Y matrix and each value h(x^(i))k as H(i,k). So basically, for each i,k you want to compute Y(i,k) log(H(i,k)) + (1-Y(i,k)) log(1-H(i,k)). You can do it for all the values together and store the result in matrix C. Then C = Y.*log(H) + (1-Y).*log(1-H) and each C(i,k) has the above mentioned value. This is an operation .* because you want to do the operation for each element (i,k) of each matrix (in contrast to multiplying the matrices which is totally different). Afterwards, to get the sum of all the values inside the 2D dimensional matrix C, you use the octave function sum twice: sum(sum(C)) to sum both columnwise and row-wise (or as @ Irreducible suggested, just sum(C(:))).



Note there may be other errors as well.






share|improve this answer

























  • To sum over all entries of a matrix C just write: sum(C(:))

    – Irreducible
    Mar 25 at 12:29












  • @Irreducible That too! Incorporated your suggestion.

    – tryman
    Apr 4 at 10:32













1












1








1







Transfer from the comments:

With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).



Update 1:

In regards to your question in the comment.
From the first screenshot, you represent each value yk(i) in the entry Y(i,k) of the Y matrix and each value h(x^(i))k as H(i,k). So basically, for each i,k you want to compute Y(i,k) log(H(i,k)) + (1-Y(i,k)) log(1-H(i,k)). You can do it for all the values together and store the result in matrix C. Then C = Y.*log(H) + (1-Y).*log(1-H) and each C(i,k) has the above mentioned value. This is an operation .* because you want to do the operation for each element (i,k) of each matrix (in contrast to multiplying the matrices which is totally different). Afterwards, to get the sum of all the values inside the 2D dimensional matrix C, you use the octave function sum twice: sum(sum(C)) to sum both columnwise and row-wise (or as @ Irreducible suggested, just sum(C(:))).



Note there may be other errors as well.






share|improve this answer















Transfer from the comments:

With a quick look, in J = (1/m) * sum(sum((-Y*log(H]))' - (1-Y)*log(1-H]))'))) there is definetely something going on with the parenthesis, but probably on how you pasted it here, not with the original code as this would throw an error when you run it. If I understand correctly and Y, H are matrices, then in your 1st version Y*log(H) is matrix multiplication while in the 2nd version Y.*log(H) is an entrywise multiplication (not matrix-multiplication, just c(i,j)=a(i,j)*b(i,j) ).



Update 1:

In regards to your question in the comment.
From the first screenshot, you represent each value yk(i) in the entry Y(i,k) of the Y matrix and each value h(x^(i))k as H(i,k). So basically, for each i,k you want to compute Y(i,k) log(H(i,k)) + (1-Y(i,k)) log(1-H(i,k)). You can do it for all the values together and store the result in matrix C. Then C = Y.*log(H) + (1-Y).*log(1-H) and each C(i,k) has the above mentioned value. This is an operation .* because you want to do the operation for each element (i,k) of each matrix (in contrast to multiplying the matrices which is totally different). Afterwards, to get the sum of all the values inside the 2D dimensional matrix C, you use the octave function sum twice: sum(sum(C)) to sum both columnwise and row-wise (or as @ Irreducible suggested, just sum(C(:))).



Note there may be other errors as well.







share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 4 at 10:32

























answered Mar 25 at 5:00









trymantryman

1,120617




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  • To sum over all entries of a matrix C just write: sum(C(:))

    – Irreducible
    Mar 25 at 12:29












  • @Irreducible That too! Incorporated your suggestion.

    – tryman
    Apr 4 at 10:32

















  • To sum over all entries of a matrix C just write: sum(C(:))

    – Irreducible
    Mar 25 at 12:29












  • @Irreducible That too! Incorporated your suggestion.

    – tryman
    Apr 4 at 10:32
















To sum over all entries of a matrix C just write: sum(C(:))

– Irreducible
Mar 25 at 12:29






To sum over all entries of a matrix C just write: sum(C(:))

– Irreducible
Mar 25 at 12:29














@Irreducible That too! Incorporated your suggestion.

– tryman
Apr 4 at 10:32





@Irreducible That too! Incorporated your suggestion.

– tryman
Apr 4 at 10:32

















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