NoRM | MongoDB driver for .Net | SQL Database library
kandi X-RAY | NoRM Summary
kandi X-RAY | NoRM Summary
NoRM is a MongoDB driver for .Net designed to provide access to strongly/statically-typed documents and collections.
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Lines of Code : 209License : Non-SPDX (Apache License 2.0)
Copydef fold_batch_norms(input_graph_def):
"""Removes batch normalization ops by folding them into convolutions.
Batch normalization during training has multiple dynamic parameters that are
updated, but once the graph is finalized these become con def _BatchNormGrad(grad_y,
x,
scale,
pop_mean,
pop_var,
epsilon,
data_format,
is_training=True):
"""Returns the gra def benchmark_batch_norm(self):
print("Forward convolution (lower layers).")
shape = [8, 128, 128, 32]
axes = [0, 1, 2]
t1 = self._run_graph("cpu", shape, axes, 10, "op", True, False, 5)
t2 = self._run_graph("cpu", shape, axes, 10 Community Discussions
Trending Discussions on NoRM
QUESTION
I have 120 vectors in a matrix points (120 x 2). I calculate their squared norms:
ANSWER
Answered 2022-Mar-14 at 20:13Yes, table can round numeric input.
table() calls factor() which calls as.character(), and as.character() does some rounding:
QUESTION
I've written and optimized a Shiny app, and now I'm struggling with the IT section of the organization where I work to have it published on their servers. Currently, they are claiming that the app is not W3C compliant, which is true, according to the W3C validator.
The errors I'm trying to solve, with no success, are:
Bad value “complementary” for attribute “role” on element “form”.The value of the “for” attribute of the “label” element must be the ID of a non-hidden form control.
Such errors can be seen also in very minimal shiny apps, like:
...ANSWER
Answered 2022-Mar-04 at 08:05The following only deals with the first of the errors you mention (as this one is pretty clear thanks to @BenBolkers comment), but hopefully it points you to the right tools to use.
I'd use htmltools::tagQuery to make the needed modifications - please check the following:
QUESTION
Imagine you have a segmentation map, where each object is identified by a unique index, e.g. looking similar to this:
For each object, I would like to save which pixels it covers, but I could only come up with the standard for loop so far. Unfortunately, for larger images with thousands of individual objects, this turns out to be very slow--for my real data at least. Can I somehow speed things up?
ANSWER
Answered 2022-Feb-23 at 17:27If I understand the question correctly, You would like to see where any object is located, right? So if we start with one matrix (that is, all shapes are in one array, where empty spaces are zeros and object one consists of 1s, object 2 of 2s etc.) then You can create a mask, showing which pixels (or values in a matrix) are non-zero like this:
my_array != 0
Does that answer Your question?
Edit for clarification
QUESTION
I am currently having issue with the implementation of the Metropolis-Hastings algorithm.
I am trying to use the algorithm to calculate integrals of the form
In using this algorithm, we can obtain a long chain of configurations ( in this case, each configuration is just a single numbers) such that in the tail-end of the chain the probability of having a particular configuration follows (or rather tends to) a gaussian distribution.
My code seems to be messing up with obtaining the said gaussian distributions. There is a strange dependence on the transition probablity (the probablity of picking a new candidate configuration depending on the previous configuration in the chain). However, if this transition probability is symmetric, there should be no dependence on this function at all (it only affects speed at which phase space [space of potential configurations] is explored and how quickly the chain converges to the desired distribution)!
In my case I am using a normal distribution transition function (which satisfies the need to be symmetric), with width d. For each d I use I do indeed get a gaussian distribution however the standard deviation, sigma, depends on my choice of d. The resulting gaussian should have a sigma of roughly 0.701 but I find that the value I actually get depends on the parameter d, when it shouldn't.
I am not sure where the error in this code is, any help would be greatly appreciated!
...ANSWER
Answered 2022-Feb-02 at 20:28You need to save x even when it doesn't change. Otherwise the center values are under-counted, and more so as d increases, which increases the variance.
QUESTION
I have a set of data values for a scalar 3D function, arranged as inputs x,y,z in an array of shape (n,3) and the function values f(x,y,z) in an array of shape (n,).
EDIT: For instance, consider the following simple function
...ANSWER
Answered 2021-Nov-16 at 17:10All you need is just reshape F[:, 3] (only f(x, y, z)) into a grid. Hard to be more precise without sample data:
If the data is not sorted, you need to sort it:
QUESTION
I have this basic VBA SQL-statement. I searches an external database and returns all the records where the field [LabNumberPrimary] = [labnummer] in the external database.
My VBA code repeats itself with some minor adjustments. How do I combine the 2 statements so my VBA code gets smaller and more user friendly?
1st statement:
...ANSWER
Answered 2021-Dec-14 at 13:44Try this:
QUESTION
I have the following MATLAB snippet:
...ANSWER
Answered 2021-Nov-22 at 16:12I invite you to read the documentation for norm. It is a good idea to always read the documentation to a function and not make assumptions about what it does. In short, with a matrix input, norm computes the matrix norm:
norm(R,1)is the maximum absolute column sum ofR.norm(R,Inf)is the maximum absolute row sum ofR.norm(R,2)is approximatelymax(svd(R)).
The 1-norm and infinity-norm of the matrix are computed in a similar way, and are therefore expected to be similar in cost. Computing the sum over rows or columns, and the max of the result, is quite cheap.
The 2-norm of the matrix in contrast requires a singular value decomposition, which is significantly more expensive.
In Julia, norm computes the vector norm. To compute a matrix norm, use opnorm.
To compute the vector norm of rows or columns of a matrix in MATLAB, use vecnorm (since R2017b). To compute the norm of the vectorized matrix, use norm(R(:)).
PS: The real question is why is the infinity-norm in Julia so slow? It should be cheaper than the 1-norm and much cheaper than the 2-norm!
QUESTION
Anti-closing preamble: I have read the question "difference between penalty and loss parameters in Sklearn LinearSVC library" but I find the answer there not to be specific enough. Therefore, I’m reformulating the question:
I am familiar with SVM theory and I’m experimenting with LinearSVC class in Python. However, the documentation is not quite clear regarding the meaning of penalty and loss parameters. I recon that loss refers to the penalty for points violating the margin (usually denoted by the Greek letter xi or zeta in the objective function), while penalty is the norm of the vector determining the class boundary, usually denoted by w. Can anyone confirm or deny this?
If my guess is right, then penalty = 'l1' would lead to minimisation of the L1-norm of the vector w, like in LASSO regression. How does this relate to the maximum-margin idea of the SVM? Can anyone point me to a publication regarding this question? In the original paper describing LIBLINEAR I could not find any reference to L1 penalty.
Also, if my guess is right, why doesn't LinearSVC support the combination of penalty='l2' and loss='hinge' (the standard combination in SVC) when dual=False? When trying it, I get the
...ValueError: Unsupported set of arguments
ANSWER
Answered 2021-Nov-18 at 18:08Though very late, I'll try to give my answer. According to the doc, here's the considered primal optimization problem for LinearSVC:
,phi being the Identity matrix, given that LinearSVC only solves linear problems.
Effectively, this is just one of the possible problems that LinearSVC admits (it is the L2-regularized, L1-loss in the terms of the LIBLINEAR paper) and not the default one (which is the L2-regularized, L2-loss).
The LIBLINEAR paper gives a more general formulation for what concerns what's referred to as loss in Chapter 2, then it further elaborates also on what's referred to as penalty within the Appendix (A2+A4).
Basically, it states that LIBLINEAR is meant to solve the following unconstrained optimization pb with different loss functions xi(w;x,y) (which are hinge and squared_hinge); the default setting of the model in LIBLINEAR does not consider the bias term, that's why you won't see any reference to b from now on (there are many posts on SO on this).
- ,
hingeor L1-loss - ,
squared_hingeor L2-loss.
For what concerns the penalty, basically this represents the norm of the vector w used. The appendix elaborates on the different problems:
- L2-regularized, L1-loss (
penalty='l2',loss='hinge'): - L2-regularized, L2-loss (
penalty='l2',loss='squared_hinge'), default inLinearSVC: - L1-regularized, L2-loss (
penalty='l1',loss='squared_hinge'):
Instead, as stated within the documentation, LinearSVC does not support the combination of penalty='l1' and loss='hinge'. As far as I see the paper does not specify why, but I found a possible answer here (within the answer by Arun Iyer).
Eventually, effectively the combination of penalty='l2', loss='hinge', dual=False is not supported as specified in here (it is just not implemented in LIBLINEAR) or here; not sure whether that's the case, but within the LIBLINEAR paper from Appendix B onwards it is specified the optimization pb that's solved (which in the case of L2-regularized, L1-loss seems to be the dual).
For a theoretical discussion on SVC pbs in general, I found that chapter really useful; it shows how the minimization of the norm of w relates to the idea of the maximum-margin.
QUESTION
I was trying to an example of the book -"Dynamical Systems with Applications using Python" and I was asked to plot the phase portrait of Verhulst equation, then I came across this post: How to plot a phase portrait of Verhulst equation with SciPy (or SymPy) and Matplotlib?
I'm getting the same plot as the user on the previous post. Whenever, I try to use the accepted solution I get a "division by zero" error. Why doesn't the accepted solution in How to plot a phase portrait of Verhulst equation with SciPy (or SymPy) and Matplotlib? works?
Thank you very much for you help!
Edit:
Using the code from the previous post and the correction given by @Lutz Lehmann
...ANSWER
Answered 2021-Nov-11 at 18:31With the help of @Lutz Lehmann I could rewrite the code to get want I needed.
The solutions is something like this:
QUESTION
When trying to initialize a Vector using the result of some operation in Eigen, the result seems to be different depending on what syntax is used, i.e., on my machine, the assertion at the end of the following code fails:
...ANSWER
Answered 2021-Oct-05 at 15:31The condition number of the matrix that defines the linear system you are solving is at the order of 10⁷. Roughly speaking, this means that after solving this system numerically the last 7 digits will be incorrect. Thus, leaving you with roughly 9 correct digits or an error of around 10⁻⁹. It seems like
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