In order to prevent divisions by zero in TensorFlow, I want to add a tiny number to my dividend. A quick search did not yield any results. In particular, I am interested in using the scientific notation, e.g.
a = b/(c+1e-05)
How can this be achieved?
Assuming a, b and c are tensors. The formula you have written will work as expected. 1e-5 will be broadcasted and added on the tensor c. Tensorflow automatically typecasts the 1e-5 to tf.constant(1e-5).
Tensorflow however has some limitations with non-scalar broadcasts. Take a look at my other answer.
Related
I am using this function of tensorflow to get my function jacobian. Came across two problems:
The tensorflow documentation is contradicted to itself in the following two paragraph if I am not mistaken:
gradients() adds ops to the graph to output the partial derivatives of ys with respect to xs. It returns a list of Tensor of length len(xs) where each tensor is the sum(dy/dx) for y in ys.
Blockquote
Blockquote
Returns:
A list of sum(dy/dx) for each x in xs.
Blockquote
According to my test, it is, in fact, return a vector of len(ys) which is the sum(dy/dx) for each x in xs.
I do not understand why they designed it in a way that the return is the sum of the columns(or row, depending on how you define your Jacobian).
How can I really get the Jacobian?
4.In the loss, I need the partial derivative of my function with respect to input (x), but when I am optimizing with respect to the network weights, I define x as a placeholder whose value is fed later, and weights are variable, in this case, can I still define the symbolic derivative of function with respect to input (x)? and put it in the loss? ( which later when we optimize with respect to weights will bring second order derivative of the function.)
I think you are right and there is a typo there, it was probably meant to be "of length len(ys)".
For efficiency. I can't explain exactly the reasoning, but this seems to be a pretty fundamental characteristic of how TensorFlow handles automatic differentiation. See issue #675.
There is no straightforward way to get the Jacobian matrix in TensorFlow. Take a look at this answer and again issue #675. Basically, you need one call to tf.gradients per column/row.
Yes, of course. You can compute whatever gradients you want, there is no real difference between a placeholder and any other operation really. There are a few operations that do not have a gradient because it is not well defined or not implemented (in which case it will generally return 0), but that's all.
I've been reading the docs to learn TensorFlow and have been struggling on when to use the following functions and their purpose.
tf.split()
tf.reshape()
tf.transpose()
My guess so far is that:
tf.split() is used because inputs must be a sequence.
tf.reshape() is used to make the shapes compatible (Incorrect shapes tends to be a common problem / mistake for me). I used numpy for this before. I'll probably stick to tf.reshape() now. I am not sure if there is a difference between the two.
tf.transpose() swaps the rows and columns from my understanding. If I don't use tf.transpose() my loss doesn't go down. If the parameter values are incorrect the loss doesn't go down. So the purpose of me using tf.transpose() is so that my loss goes down and my predictions become more accurate.
This bothers me tremendously because I'm using tf.transpose() because I have to and have no understanding why it's such an important factor. I'm assuming if it's not used correctly the inputs and labels can be in the wrong position. Making it impossible for the model to learn. If this is true how can I go about using tf.transpose() so that I am not so reliant on figuring out the parameter values via trial and error?
Question
Why do I need tf.transpose()?
What is the purpose of tf.transpose()?
Answer
Why do I need tf.transpose()? I can't imagine why you would need it unless you coded your solution from the beginning to require it. For example, suppose I have 120 student records with 50 stats per student and I want to use that to try and make a linear association with their chance of taking 3 classes. I'd state it like so
c = r x m
r = records, a matrix with a shape if [120x50]
m = the induction matrix. it has a shape of [50x3]
c = the chance of all students taking one of three courses, a matrix with a shape of [120x3]
Now if instead of making m [50x3], we goofed and made m [3x50], then we'd have to transpose it before multiplication.
What is the purpose of tf.transpose()?
Sometimes you just need to swap rows and columns, like above. Wikipedia has a fantastic page on it. The transpose function has some excellent properties for matrix math function, like associativeness and associativeness with the inverse function.
Summary
I don't think I've ever used tf.transpose in any CNN I've written.
Quite simply, what I want to do is the following
A = np.ones((3,3)) #arbitrary matrix
B = np.ones((2,2)) #arbitrary matrix
A[1:,1:] = A[1:,1:] + B
except in Tensorflow (where the matrices can be arbitrarily complicated tensor expressions). Neither A nor B is a Tensorflow Variable, but just a run-of-the-mill tensor.
What I have gathered so far: tensors are immutable, so I cannot assign to a submatrix. tf.scatter_nd is the current option for sub-assignment, but does not appear to support sub-matrices, only slices.
Methods that should work, but are perhaps not ideal:
I could pad B with zeros, but I'm sure this leads to instantiation of
an unnecessarily large B - can it be made sparse, maybe?
I could use the padding idea, but write it as a low-rank decomposition, e.g. in Numpy: A+U.dot(B).U.T where U is a stacked zero and identity matrix. I'm not sure this is actually advantageous.
I could split A into submatrices, and stack them back together. Might be the most efficient, but sounds like the code would be convoluted.
Ideally, I want to do this operation N times for progressively smaller matrices, resulting in one large final result, but this is tangential.
I'll use one of the hacks for now, but I'm hoping someone can tell me what the idiomatic version is!
I have two tensors, a of rank 4 and b of rank 1. I'd like to produce aprime, of rank 3, by "contracting" the last axis of a away, by replacing it with its dot product against b. In numpy, this is as easy as np.tensordot(a, b, 1). However, I can't figure out a way to do this in Tensorflow.
How can I replace the last axis of a tensor with a value equal to that axis's dot product against another tensor (of course, of the same shape)?
UPDATE:
I see in Wikipedia that this is called the "Tensor Inner Product" https://en.wikipedia.org/wiki/Dot_product#Tensors aka tensor contraction. It seems like this is a common operation, I'm surprised that there's no explicit support for it in Tensorflow.
I believe that this may be possible via tf.einsum; however, I have not been able to find a generalized way to do this that works for tensors of any rank (this is probably because I do not understand einsum and have been reduced to trial and error)
Aren't you just using tensor in the sense of a multidimensional array? Or in some disciplines a tensor is 3d (vector 1d, matrix 2d, etc). I haven't used tensorflow but I don't think it has much to do with tensors in that linear algebra sensor. They talk about data flow graphs. I'm not sure where the tensor part of the name comes from.
I assume you are talking about an expression like:
In [293]: A=np.tensordot(np.ones((5,4,3,2)),np.arange(2),1)
resulting in a (5,4,3) shape array. The einsum equivalent is
In [294]: B=np.einsum('ijkl,l->ijk',np.ones((5,4,3,2)),np.arange(2))
np.einsum implements Einstine Notation, as discussed here: https://en.wikipedia.org/wiki/Einstein_notation. I got this link from https://en.wikipedia.org/wiki/Tensor_contraction
You seem to be talking about straight forward numpy operations, not something special in tensorflow.
I would first add 3 dimensions of size 1 to b so that it can be broadcast along the 4'th dimension of a.
b = tf.reshape(b, (1, 1, 1, -1))
Then you can multiply b and a and it will broadcast b along all of the other dimensions.
a_prime = a * b
Finally, reduce the sum along the 4'th dimension to get rid of that dimension and replace it with the dot product.
a_prime = tf.reduce_sum(a_prime, [3])
This seems like it would work (for the first tensor being of any rank):
tf.einsum('...i,i->...', x, y)
I have some data in a pandas dataframe (although pandas is not the point of this question). As an experiment I made column ZR as column Z divided by column R. As a first step using scikit learn I wanted to see if I could predict ZR from the other columns (which should be possible as I just made it from R and Z). My steps have been.
columns=['R','T', 'V', 'X', 'Z']
for c in columns:
results[c] = preprocessing.scale(results[c])
results['ZR'] = preprocessing.scale(results['ZR'])
labels = results["ZR"].values
features = results[columns].values
#print labels
#print features
regr = linear_model.LinearRegression()
regr.fit(features, labels)
print(regr.coef_)
print np.mean((regr.predict(features)-labels)**2)
This gives
[ 0.36472515 -0.79579885 -0.16316067 0.67995378 0.59256197]
0.458552051342
The preprocessing seems wrong as it destroys the Z/R relationship I think. What's the right way to preprocess in this situation?
Is there some way to get near 100% accuracy? Linear regression is the wrong tool as the relationship is not-linear.
The five features are highly correlated in my data. Is non-negative least squares implemented in scikit learn ? ( I can see it mentioned in the mailing list but not the docs.) My aim would be to get as many coefficients set to zero as possible.
You should easily be able to get a decent fit using random forest regression, without any preprocessing, since it is a nonlinear method:
model = RandomForestRegressor(n_estimators=10, max_features=2)
model.fit(features, labels)
You can play with the parameters to get better performance.
The solutions is not as easy and can be very influenced by your data.
If your variables R and Z are bounded (for ex 0<R<1 -3<Z<2) then you should be able to get a good estimation of the output variable using neural network.
Using neural network you should be able to estimate your output even without preprocessing the data and using all the variables as input.
(Of course here you will have to solve a minimization problem).
Sklearn do not implement neural network so you should use pybrain or fann.
If you want to preprocess the data in order to make the minimization problem easier you can try to extract the right features from the predictor matrix.
I do not think there are a lot of tools for non linear features selection. I would try to estimate the important variables from you dataset using in this order :
1-lasso
2- sparse PCA
3- decision tree (you can actually use them for features selection ) but I would avoid this as much as possible
If this is a toy problem I would sugges you to move towards something of more standard.
You can find a lot of examples on google.