laplace | source project to enable screen sharing directly via browser | Websocket library
kandi X-RAY | laplace Summary
kandi X-RAY | laplace Summary
Laplace is an open-source project to enable screen sharing directly via browser. Made possible using WebRTC for low latency peer-to-peer connections, and WebSocket implemented in golang for WebRTC signaling.
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Top functions reviewed by kandi - BETA
- create a new session stream
- new a session
- init ui
- Start the streaming media
- join the room
- Update the room of the room
- called from server
- Retrieve the location based on the current URL .
- Update status indicator
- Gets the websocket URL .
laplace Key Features
laplace Examples and Code Snippets
Lines of Code : 46License : Non-SPDX (Apache License 2.0)
Copydef log_cdf_laplace(x, name="log_cdf_laplace"):
"""Log Laplace distribution function.
This function calculates `Log[L(x)]`, where `L(x)` is the cumulative
distribution function of the Laplace distribution, i.e.
```L(x) := 0.5 * int_{-infty} def determinant(self) -> float:
"""
returns the determinant of an nxn matrix using Laplace expansion
"""
if self.__height != self.__width:
raise Exception("Matrix is not square")
if self.__height def __init__(self,
loc,
scale,
validate_args=False,
allow_nan_stats=True,
name="LaplaceWithSoftplusScale"):
parameters = dict(locals())
with ops.name_scope(name, value Community Discussions
Trending Discussions on laplace
QUESTION
I want to add a value to each non-zero element in my sparse matrix. Can someone give me a method to do that.
...ANSWER
Answered 2022-Apr-16 at 15:01Offered without comment:
QUESTION
Maple is returning a limit in my answer and will not apply the limit. I am specifically trying to evaluate the Laplace Integral of the function f(t) = cos(omega*t)^2
I use the laplace() command to confirm my answer. If Maple would apply the limit, then I would get the expected answer.
What is happening? How can I force Maple to evaluate the limit?
...ANSWER
Answered 2022-Feb-02 at 14:34inttrans[laplace](cos(w*t)^2,t,s);
(s^2+2*w^2)/(s^2+4*w^2)/s
int(exp(-s*t)*cos(w*t)^2,t=0..infinity)
assuming s>0;
(s^2+2*w^2)/(s^2+4*w^2)/s
int(exp(-s*t)*cos(w*t)^2,t=0..infinity)
assuming s<0;
undefined
QUESTION
I want to train a model with a Naive Bayes classifier using the tidymodels framework.
Tidymodels uses the discrim packages, which itself uses the klaR package to estimate Naive Bayes models.
Specifying a NB model in the tidymodels framework can be done with e.g.:
...ANSWER
Answered 2022-Jan-26 at 23:11You can set this model parameter as an engine argument:
QUESTION
I want to use quadratic terms to fit my general linear mixed model with id as a random effect, using the lme4 package. It's about how the distance to settlements influences the probability of occurrence of an animal. I use the following code (I hope it is correct):
...ANSWER
Answered 2022-Jan-23 at 15:16A couple of points.
- Coefficients of non-linear model terms do not have a straightforward interpretation and you should make effect plots to be able to communicate the results from your analyses. You may use
effectPlotData()from theGLMMadaptivepackage to do this. Refer to this page for more information. - To be able to appraise whether including a quadratic effect of
dist_settlementsimproves the model fit, you should fit a model without the squared term (i.e. only the linear effect ofdist_settlements) and a model with the squared term. Then perform a likelihood ratio test to appraise whether inclusion of complex terms improves the model fit. In case of LMMs, make sure to fit both models using maximum likelihood, not REML. For GLMMs, you don't have to borther about (RE)ML. - The variance of the random intercepts is rather close to 0, which may require your attention. Refer to this answer and this section of Ben Bolker's github for more information on this topic.
You may want to take a look at this great lecture series by Dimitris Rizopoulos for more information on (G)LMMs.
QUESTION
I am hoping to move my custom camera video pipeline to use video memory with a combination of numba and cupy and avoid passing data back to the host memory if at all possible. As part of doing this I need to port my sharpness detection routine to use cuda. The easiest way to do this seemed to be to use cupy as essential all I do is compute the variance of a laplace transform of each image. The trouble I am hitting is the cupy variance computation appears to be ~ 8x slower than numpy. This includes the time it takes to copy the device ndarray to the host and perform the variance computation on the cpu using numpy. I am hoping to gain a better understanding of why the variance computation ReductionKernel employed by cupy on the GPU is so much slower. I'll start by including the test I ran below.
...ANSWER
Answered 2022-Jan-14 at 21:58I have a partial hypothesis about the problem (not a full explanation) and a work-around. Perhaps someone can fill in the gaps. I've used a quicker-and-dirtier benchmark, for brevity's sake.
The work-around: reduce one axis at a timeCupy is much faster when reduction is performed on one axis at a time. In stead of:
x.sum()
prefer this:
x.sum(-1).sum(-1).sum(-1)...
Note that the results of these computations may differ due to rounding error.
Here are faster mean and var functions:
QUESTION
Is there a way to numerically solve the following PDE in Python?
The second term on the RHS has a derivative with respect to time as well as space.
I tried using Py-PDE package in Python, it solves only the form dy(x,t)/dt = f(y(x,t)) so I tried to use a root finding algorithm similar to scipy fzero to get the solution to dy(x,t)/dt - f(y(x,t),dy(x,t)/dt) = 0 (solving for dy(x,t)/dt).
...ANSWER
Answered 2021-Dec-13 at 07:24Since no one has posted an answer yet, I managed to get a minimal working example by using scipy odeint with a method of lines to solve the PDE, that is, by discretizing the Laplace operator, and then wrapping the differential equation inside fsolve to get dydt:
QUESTION
I'm traiying to code, as exercise, a PL/pgSQL function that implements Laplace's Theorem to calculate determinant then print the solution when called. I thingI'm new in this language and can't find what's wrong with the code. I think the idea is OK, maybe there's a sintax error! Help!
...ANSWER
Answered 2021-Nov-17 at 17:43You never change k. On the other hand, you increment l too much.
One of the l = l + 1 (there are two of them) should be k = k + 1.
QUESTION
Random coefficient Poisson models are rather difficult to fit, there tends to be some variability in parameter estimates between lme4 and glmmADMB. But in my case:
...ANSWER
Answered 2021-Nov-14 at 00:46I got a little bit carried away. tl;dr as pointed out in comments, it's hard to get glmmADMB to work with a Poisson model, but a model with overdispersion (e.g. negative binomial) is clearly a lot better. Furthermore, you should probably incorporate some aspect of random slopes in the model ...
Packages (colorblindr is optional):
QUESTION
I'm trying to plot the decision boundary of the SVM classifier using a precomputed Laplace kernel (code below) on the similar lines of this scikit-learn post. I'm taking test points as mesh grid values (xx, yy) just like as mentioned in the post and train points as X and y. I'm able to fit the pre-computed kernel using train points.
ANSWER
Answered 2021-Nov-05 at 11:23The issue is getting your meshgrid into the same dimensions as the training matrix, before applying the laplacian. So if we run the code below to fit the svm :
QUESTION
I'm trying to calculate the accuracy score, of a SVM using Laplacian kernel (as a pre-computed kernel). However, I'm getting the error as below when I try to calculate the accuracy score.
My code :
...ANSWER
Answered 2021-Oct-20 at 03:43You calculated pred_y using your train inputs which has 105 elements and y_test has 45 elements.
You need to add a step:
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