I want to implement next formula: Si = (x1 - w1) + ... + (xn - wn) like a Conv=x1*w1 + ... xn*wn for some area under X (input tensor) and kernel W. Important that this operation repeat on all areas X with slice W with stride and padding params like a simple convolution.
How do this?
I found similar question a while ago on stackoverflow. But it finished on custom implementation on C++ and compilation or changes CUDA source or something like that.
Is there an easier way today?
If I get you correctly then you compute Si=(x1 + ... + xn)-(w1 + ... + wn)? The sum of the weights is a single number, so you don't have a "Kernel" anymore. The first sum you can compute via tf.nn.conv2d and a filter that is initialized with tf.ones. But I don't think that's what you meant to do, so could you maybe specify your question further?
Related
I am using Tensorflow to minimize a function. The function takes about 10 parameters. Every single parameter has bounds, e.g. a minimum and a maximum value the parameter is allowed to take. For example, the parameter x1 needs to be between 1 and 10.
I also have a pair of parameters that need to have the following constraint x2 > x3. In other words, x2 must always be bigger than x3. (In addition to this, x2 and x3 also have bounds, similarly to the example of x1 above.)
I know that tf.Variable has a "constraint" argument, however I can't really find any examples or documentation on how to use this to achieve the bounds and constraints as mentioned above.
Thank you!
It seems to me (I can be mistaken) that constrained optimization (you can google for it in tensorflow) is not exactly the case for which tensroflow was designed. You may want to take a look at this repo, it may satisfy your needs, but as far as I understand, it's still not solving arbitrary constrained optimization, just some classification problems with labels and features, compatible with precision/recall scores.
If you want to use constraints on the tensorflow variable (i.e. some function applied after gradient step - which you can do manually also - by taking variable values, doing manipulations, and reassigning then), it means that you will be cutting variables after each step done using gradient in general space. It's a question whether you will successfully reach the right optimization goal this way, or your variables will stuck at boundaries, because general gradient will point somewhere outside.
My approach 1
If your problem is simple enough. you can try to parametrize your x2 and x3 as x2 = x3 + t, and then try to do cutting in the graph:
x3 = tf.get_variable('x3',
dtype=tf.float32,
shape=(1,),
initializer=tf.random_uniform_initializer(minval=1., maxval=10.),
constraint=lambda z: tf.clip_by_value(z, 1, 10))
t = tf.get_variable('t',
dtype=tf.float32,
shape=(1,),
initializer=tf.random_uniform_initializer(minval=1., maxval=10.),
constraint=lambda z: tf.clip_by_value(z, 1, 10))
x2 = x3 + t
Then, on a separate call additionally do
sess.run(tf.assign(x2, tf.clip_by_value(x2, 1.0, 10.0)))
But my opinion is that it won't work well.
My approach 2
I would also try to invent some loss terms to keep variables within constraints, which is more likely to work. For example, constraint for x2 to be in the interval [1,10] will be:
loss += alpha*tf.abs(tf.math.tan(((x-5.5)/4.5)*pi/2))
Here the expression under tan is brought to -pi/2,pi/2 and then tan function is used to make it grow very rapidly when it reaches boundaries. In this case I think you're more likely to find your optimum, but again the loss weight alpha might be too big and training will stuck somewhere nearby, if required value of x2 lies near the boundary. In this case you can try to use smaller alpha.
In addition to the answer by Slowpoke, reparameterization is another option. E.g. let's say you have a param p which should be bounded in [lower_bound,upper_bound], you could write:
p_inner = tf.Variable(...) # unbounded
p = tf.sigmoid(p_inner) * (upper_bound - lower_bound) + lower_bound
However, this will change the behavior of gradient descent.
I am trying to understand the fisheye model by Scaramuzza, which is implemented in Matlab, see https://de.mathworks.com/help/vision/ug/fisheye-calibration-basics.html#mw_8aca38cc-44de-4a26-a5bc-10fb312ae3c5
The backprojection (uv to xyz) seems fairly straightforward according to the following equation:
, where rho=sqrt(u^2 +v^2)
However, how does the projection (from xyz to uv) work?! In my understanding we get a rather complex set of equations. Unfortunately, I don't find any details on that....
Okay, I believe I understand it now fully after analyzing the functions of the (windows) calibration toolbox by Scaramuzza, see https://sites.google.com/site/scarabotix/ocamcalib-toolbox/ocamcalib-toolbox-download-page
Method 1 found in file "world2cam.m"
For the projection, use the same equation above. In the projection case, the equation has three known (x,y,z) and three unknown variables (u,v and lambda). We first substitute lambda with rho by realizing that
u = x/lambda
v = y/lambda
rho=sqrt(u^2+v^2) = 1/lambda * sqrt(x^2+y^2) --> lambda = sqrt(x^2+y^2) / rho
After that, we have the unknown variables (u,v and rho)
u = x/lambda = x / sqrt(x^2+y^2) * rho
v = y/lambda = y / sqrt(x^2+y^2) * rho
z / lambda = z /sqrt(x^2+y^2) * rho = a0 + a2*rho^2 + a3*rho^3 + a4*rho^4
As you can see, the last equation now has only one unknown, namely rho. Thus, we can solve it easily using e.g. the roots function in matlab. However, the result does not always exist nor is it necessarily unique. After solving the unknown variable rho, calculating uv is very simple using the equation above.
This procedure needs to be performed for each point (x,y,z) separately and is thus rather computationally expensive for an image.
Method 2 found in file "world2cam_fast.m"
The last equation has the form rho(x,y,z). However, if we define m = z / sqrt(x^2+y^2) = tan(90°-theta), it only depends on one variable, namely rho(m).
Instead of solving this equation rho(m) for every new m, the authors "plot" the function for several values of m and fit an 8th order polynomial to these points. Using this polynomial they can calculate an approximate value for rho(m) much quicker in the following.
This becomes clear, because "world2cam_fast.m" makes use of ocam_model.pol, which is calculated in "undistort.m". "undistort.m" in turn makes use of "findinvpoly.m".
Say that I want to sample a matrix with each entry sampled from a distribution defined by an entry in another matrix. I unroll my matrix and apply map_fn to each element. With a relatively small matrix (128 x 128), the following gives me several PoolAllocator warnings (GTX TITAN Black) and does not train in any reasonable amount of time.
def sample(x):
samples = tf.map_fn(lambda z:
tf.random_normal([1], mean=z,
stddev=tf.sqrt(z * (1 - z))),
tf.reshape(x, [-1])) # apply to each element
return tf.cond(is_training, lambda: tf.reshape(samples, shape=tf.shape(x)),
lambda: tf.tanh(x))
Is there a better way to apply an elementwise operation like this?
Your code will run much faster if you can use Tensor-at-a-time operations instead of elementwise operations like tf.map_fn.
Here it looks like you want to sample from a normal distribution for each element, where the parameters of the distribution are different for each value in an input tensor. Try something like this:
def sample(x):
samples = tf.random_normal(shape=[128, 128]) * tf.sqrt(x * (1 - x)) + x
tf.random_normal() generates a normal distribution with mean 0.0 and standard deviation 1.0 by default. You can use point-wise tensor operations to fix up the standard deviation (by multiplying) and the mean (by adding) for each element. In fact, if you look at how tf.random_normal() is implemented, that's precisely what it does internally.
(You would probably also do better using a Python conditional to distinguish training from test time.)
If you plan to do this sort of thing a lot, you might file a feature request on github asking to generalize tf.random_normal to accept Tensors with more general shapes for mean and stddev. I see no reason why that shouldn't be supported.
Hope that helps!
See the tensorflow.contrib.distributions module, which has a Normal class with a sample method that does this for you.
I need to do something very similar to what is detailed in this post. But the way the stencils are done are not obvious to me... well the stencil for _flux is, but the ones for temp_bz & temp_bx are not.
I think the picture would get clearer with variables, instead of numbers (something like stencil = np.array([[a, b], [c, d]]) with a=0.5, b=...
As example, if the recurrence relation is
flux2[i,j] = a*flux2[i-1,j] + b*bz[i-1,j]*dx + c*flux2[i,j-1] - d*bx[i,j-1]*dz
how the code would be changed ?
Having flux2, bz and bx variables, and assuming they are numpy arrays (if they are not, they should), you could write that ecuation in a vectorized form as follows:
flux2[1:,1:] = a * flux2[:-1,1:] + b * bz[:-1,1:] * dx + c * flux2[1:,:-1] - d * bx[1:,:-1] * dz
Note that, since you didn't mention dz, I assumed it is a constant, if it is a matrix of the same shape as flux2, replace with dz[1:, 1:] (same applies to dx).
That line above will vectorize the operation to every i,j of the matrix, and thus, remove the for loop, giving a considerable speedup.
You would have to define the boundary conditions for row and column 0, as your equation doesn't define what to do in those special cases.
So, in short, as your stencil only uses one position for each variable, and only has 4 interactions, I would say is way faster to calculate it in its analytic form, rather than convolving 3 images with almost all-0 stencils (which would be quite a lot of overkill).
I am trying to fit a quadratic to some experimental data and using polyfit in numpy. I am looking to get a concave curve, and hence want to make sure that the coefficient of the quadratic term is negative, also the fit itself is weighted, as in there are some weights on the points. Is there an easy way to do that? Thanks.
The use of weights is described here (numpy.polyfit).
Basically, you need a weight vector with the same length as x and y.
To avoid the wrong sign in the coefficient, you could use a fit function definition like
def fitfunc(x,a,b,c):
return -1 * abs(a) * x**2 + b * x + c
This will give you a negative coefficient for x**2 at all times.
You can use curve_fit
.
Or you can run polyfit with rank 2 and if the last coefficient is bigger than 0. run again linear polyfit (polyfit with rank 1)