I was going through a recent computationaly effiecent deep-learning architecte - PVANet:
https://arxiv.org/abs/1608.08021
I'm trying to understand how they got the #params,#MAC figures in Table 1 of the paper. For instance the number of computations for conv1_1 is 397M but I get:
(528*320)*16*3*49 = 660M
the formula I used - (image dim)output channelsinput channels*kernel size
Related
I'm trying to create a new dataset of hidden state probabilities using a hidden Markov model. Everything works fine unless each time the output dataset comes up with different values (sometimes the same values) for hidden_states_train and hidden_states_test hence resulting a different column sizes in the columns stack/ a feature mismatch. e.g New dataset size (15261, 197) (5087, 194), New dataset size (15261, 197) (5087, 197) etc.
I can't figure out why this is happening each time I run the code. I tried to give same number of samples for both X_train_st and X_test_st but this keeps happening. If I set n_comp in range a smaller range e.g for n_comp in range(1,6) then often it results the same shapes.
Can someone shed some light to what's going on and a possible fix, please?
newX = X_train_st
newXtest = X_test_st
for n_comp in range(1,16):
print("fitting to HMM and decoding %d ..." % n_comp , end="")
modelHMM = GaussianHMM(n_components=n_comp, covariance_type="diag").fit(X_train_st)
hidden_states_train = to_categorical(modelHMM.predict(X_train_st))
hidden_states_test = to_categorical(modelHMM.predict(X_test_st))
print("done")
newX = np.column_stack((newX,hidden_states_train))
newXtest = np.column_stack((newXtest,hidden_states_test))
print('New dataset size',newX.shape,newXtest.shape)
for my thesis I have to calculate the number of workers at risk of substitution by machines. I have calculated the probability of substitution (X) and the number of employee at risk (Y) for each occupation category. I have a dataset like this:
X Y
1 0.1300 0
2 0.1000 0
3 0.0841 1513
4 0.0221 287
5 0.1175 3641
....
700 0.9875 4000
I tried to plot a histogram with this command:
hist(dataset1$X,dataset1$Y,xlim=c(0,1),ylim=c(0,30000),breaks=100,main="Distribution",xlab="Probability",ylab="Number of employee")
But I get this error:
In if (freq) x$counts else x$density
length > 1 and only the first element will be used
Can someone tell me what is the problem and write me the right command?
Thank you!
It is worth pointing out that the message displayed is a Warning message, and should not prevent the results being plotted. However, it does indicate there are some issues with the data.
Without the full dataset, it is not 100% obvious what may be the problem. I believe it is caused by the data not being in the correct format, with two potential issues. Firstly, some values have a value of 0, and these won't be plotted on the histogram. Secondly, the observations appear to be inconsistently spaced.
Histograms are best built from one of two datasets:
A dataframe which has been aggregated grouped into consistently sized bins.
A list of values X which in the data
I prefer the second technique. As originally shown here The expandRows() function in the package splitstackshape can be used to repeat the number of rows in the dataframe by the number of observations:
set.seed(123)
dataset1 <- data.frame(X = runif(900, 0, 1), Y = runif(900, 0, 1000))
library(splitstackshape)
dataset2 <- expandRows(dataset1, "Y")
hist(dataset2$X, xlim=c(0,1))
dataset1$bins <- cut(dataset1$X, breaks = seq(0,1,0.01), labels = FALSE)
I have a convolutional neural network whose output is a 4-channel 2D image. I want to apply sigmoid activation function to the first two channels and then use BCECriterion to computer the loss of the produced images with the ground truth ones. I want to apply squared loss function to the last two channels and finally computer the gradients and do backprop. I would also like to multiply the cost of the squared loss for each of the two last channels by a desired scalar.
So the cost has the following form:
cost = crossEntropyCh[{1, 2}] + l1 * squaredLossCh_3 + l2 * squaredLossCh_4
The way I'm thinking about doing this is as follow:
criterion1 = nn.BCECriterion()
criterion2 = nn.MSECriterion()
error = criterion1:forward(model.output[{{}, {1, 2}}], groundTruth1) + l1 * criterion2:forward(model.output[{{}, {3}}], groundTruth2) + l2 * criterion2:forward(model.output[{{}, {4}}], groundTruth3)
However, I don't think this is the correct way of doing it since I will have to do 3 separate backprop steps, one for each of the cost terms. So I wonder, can anyone give me a better solution to do this in Torch?
SplitTable and ParallelCriterion might be helpful for your problem.
Your current output layer is followed by nn.SplitTable that splits your output channels and converts your output tensor into a table. You can also combine different functions by using ParallelCriterion so that each criterion is applied on the corresponding entry of output table.
For details, I suggest you read documentation of Torch about tables.
After comments, I added the following code segment solving the original question.
M = 100
C = 4
H = 64
W = 64
dataIn = torch.rand(M, C, H, W)
layerOfTables = nn.Sequential()
-- Because SplitTable discards the dimension it is applied on, we insert
-- an additional dimension.
layerOfTables:add(nn.Reshape(M,C,1,H,W))
-- We want to split over the second dimension (i.e. channels).
layerOfTables:add(nn.SplitTable(2, 5))
-- We use ConcatTable in order to create paths accessing to the data for
-- numereous number of criterions. Each branch from the ConcatTable will
-- have access to the data (i.e. the output table).
criterionPath = nn.ConcatTable()
-- Starting from offset 1, NarrowTable will select 2 elements. Since you
-- want to use this portion as a 2 dimensional channel, we need to combine
-- then by using JoinTable. Without JoinTable, the output will be again a
-- table with 2 elements.
criterionPath:add(nn.Sequential():add(nn.NarrowTable(1, 2)):add(nn.JoinTable(2)))
-- SelectTable is simplified version of NarrowTable, and it fetches the desired element.
criterionPath:add(nn.SelectTable(3))
criterionPath:add(nn.SelectTable(4))
layerOfTables:add(criterionPath)
-- Here goes the criterion container. You can use this as if it is a regular
-- criterion function (Please see the examples on documentation page).
criterionContainer = nn.ParallelCriterion()
criterionContainer:add(nn.BCECriterion())
criterionContainer:add(nn.MSECriterion())
criterionContainer:add(nn.MSECriterion())
Since I used almost every possible table operation, it looks a little bit nasty. However, this is the only way I could solve this problem. I hope that it helps you and others suffering from the same problem. This is how the result looks like:
dataOut = layerOfTables:forward(dataIn)
print(dataOut)
{
1 : DoubleTensor - size: 100x2x64x64
2 : DoubleTensor - size: 100x1x64x64
3 : DoubleTensor - size: 100x1x64x64
}
This has become quite a frustrating question, but I've asked in the Coursera discussions and they won't help. Below is the question:
I've gotten it wrong 6 times now. How do I normalize the feature? Hints are all I'm asking for.
I'm assuming x_2^(2) is the value 5184, unless I am adding the x_0 column of 1's, which they don't mention but he certainly mentions in the lectures when talking about creating the design matrix X. In which case x_2^(2) would be the value 72. Assuming one or the other is right (I'm playing a guessing game), what should I use to normalize it? He talks about 3 different ways to normalize in the lectures: one using the maximum value, another with the range/difference between max and mins, and another the standard deviation -- they want an answer correct to the hundredths. Which one am I to use? This is so confusing.
...use both feature scaling (dividing by the
"max-min", or range, of a feature) and mean normalization.
So for any individual feature f:
f_norm = (f - f_mean) / (f_max - f_min)
e.g. for x2,(midterm exam)^2 = {7921, 5184, 8836, 4761}
> x2 <- c(7921, 5184, 8836, 4761)
> mean(x2)
6676
> max(x2) - min(x2)
4075
> (x2 - mean(x2)) / (max(x2) - min(x2))
0.306 -0.366 0.530 -0.470
Hence norm(5184) = 0.366
(using R language, which is great at vectorizing expressions like this)
I agree it's confusing they used the notation x2 (2) to mean x2 (norm) or x2'
EDIT: in practice everyone calls the builtin scale(...) function, which does the same thing.
It's asking to normalize the second feature under second column using both feature scaling and mean normalization. Therefore,
(5184 - 6675.5) / 4075 = -0.366
Usually we normalize all of them to have zero mean and go between [-1, 1].
You can do that easily by dividing by the maximum of the absolute value and then remove the mean of the samples.
"I'm assuming x_2^(2) is the value 5184" is this because it's the second item in the list and using the subscript _2? x_2 is just a variable identity in maths, it applies to all rows in the list. Note that the highest raw mid-term exam result (i.e. that which is not squared) goes down on the final test and the lowest raw mid-term result increases the most for the final exam result. Theta is a fixed value, a coefficient, so somewhere your normalisation of x_1 and x_2 values must become (EDIT: not negative, less than 1) in order to allow for this behaviour. That should hopefully give you a starting basis, by identifying where the pivot point is.
I had the same problem, in my case the thing was that I was using as average the maximum x2 value (8836) minus minimum x2 value (4761) divided by two, instead of the sum of each x2 value divided by the number of examples.
For the same training set, I got the question as
Q. What is the normalized feature x^(3)_1?
Thus, 3rd training ex and 1st feature makes out to 94 in above table.
Now, normalized form is
x = (x - mean(x's)) / range(x)
Values are :
x = 94
mean(89+72+94+69) / 4 = 81
range = 94 - 69 = 25
Normalized x = (94 - 81) / 25 = 0.52
I'm taking this course at the moment and a really trivial mistake I made first time I answered this question was using comma instead of dot in the answer, since I did by hand and in my country we use comma to denote decimals. Ex:(0,52 instead of 0.52)
So in the second time I tried I used dot and works fine.
I have the following problem. I have to compute dense SIFT interest points in a very high dimensional image (182MP). When I run the code in the full image Matlab always close suddently. So I decided to run the code in image patches.
the code
I tried to use blocproc in matlab to call the c++ function that performs the dense sift interest points detection this way:
fun = #(block_struct) denseSIFT(block_struct.data, options);
[dsift , infodsift] = blockproc(ndvi,[1000 1000],fun);
where dsift is the sift descriptors (vectors) and infodsift has the information of the interest points, such as the x and y coordinates.
the problem
The problem is the fact that blocproc just allow one output, but i want both outputs. The following error is given by matlab when i run the code.
Error using blockproc
Too many output arguments.
Is there a way for me doing this?
Would it be a problem for you to "hard code" a version of blockproc?
Assuming for a moment that you can divide your image into NxM smaller images, you could loop around as follows:
bigImage = someFunction();
sz = size(bigImage);
smallSize = sz ./ [N M];
dsift = cell(N,M);
infodsift = cell(N,M);
for ii = 1:N
for jj = 1:M
smallImage = bigImage((ii-1)*smallSize(1) + (1:smallSize(1)), (jj-1)*smallSize(2) + (1:smallSize(2));
[dsift{ii,jj} infodsift{ii,jj}] = denseSIFT(smallImage, options);
end
end
The results will then be in the two cell arrays. No real need to pre-allocate, but it's tidier if you do. If the individual matrices are the same size, you can convert into a single large matrix with
dsiftFull = cell2mat(dsift);
Almost magic. This won't work if your matrices are different sizes - but then, if they are, I'm not sure you would even want to put them all in a single one (unless you decide to horzcat them).
If you do decide you want a list of "all the colums as a giant matrix", then you can do
giantMatrix = [dsift{:}];
This will return a matrix with (in your example) 128 rows, and as many columns as there were "interest points" found. It's shorthand for
giantMatrix = [dsift{1,1} dsift{2,1} dsift{3,1} ... dsift{N,M}];