3DN | 3DN : 3D Deformation Network , CVPR | Machine Learning library
kandi X-RAY | 3DN Summary
kandi X-RAY | 3DN Summary
3DN: 3D Deformation Network, CVPR 2019
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
- Train the model
- Create a tensorflow variable on the CPU
- Write OFF file
- Return the nn distance between two points
- Get a trained model
- Group points by index
- Calculate loss for endpoints
- 2D convolutional convolution layer
- Computes the knn_point of the kn
- Get a batch of data
- Normalize the center of the given point
- Get the vertices and tras
- Get a texture
- Reset the queue
- Processes the worker
- Compute approx_match between two coordinates
- Group the points at the given index
- Calculate match cost between two matches
- Return the nn_distance between two coordinates
- Calculate the mesh lattice
- Write out an OFF file
- Read a label file
- Calculates the nn - distance correlation between two points
- Get the repulsion loss
- Calculate the KNN algorithm
- Get decoder
- Reads off file
- Example example
- 3D convolutional layer
- Transpose input tensors
- Calculate loss for end_points
3DN Key Features
3DN Examples and Code Snippets
Community Discussions
Trending Discussions on 3DN
QUESTION
I am trying to add new empty folder into existing project. In SConscript.3dn
file I use:
ANSWER
Answered 2021-Apr-30 at 07:50solution is: e.Execute(Mkdir('$RUNTIMEDIR/resources'))
QUESTION
I have a react app running, and I wanted to add route based code-splitting using React.lazy to it.
Currently my code is,
...ANSWER
Answered 2020-Jun-06 at 16:07Like I mentioned in the question, the babel-minify-webpack-plugin
was causing the issue for some reason. My guess is, they were saving function definitions as strings to save space, and using eval somewhere inside its logic. But that's just my guess.
Anyway, the Github page for babel-minify-webpack-plugin says that it is deprecated, so I ended up removing that from my project, and using the terser-webpack-plugin instead. Everything seems to work now, and the build time is also significantly reduced. My advice is, avoid using the babel-minify-webpack-plugin
and use some other minification plugin instead
QUESTION
I have two applications angularjs. I had to take some pages from one to migrate to the second. Since I activate the bundle optimization with System.Web.Optimization, the application no longer runs. Here is the state of the console:
...
ANSWER
Answered 2019-Oct-31 at 14:24With Implicit Annotation, code will break when minified.
From the Docs:
Implicit AnnotationCareful: If you plan to minify your code, your service names will get renamed and break your app.
You can add an ng-strict-di
directive on the same element as ng-app
to opt into strict DI mode.
QUESTION
I have mapped and filtered two objects. I've hit a wall in my thinking. I cannot see how to add the key/value from one object to the other when the values match.
This type of sort is fairly new to me and wearing my logic thinking out, so any help will be appreciated.
I want to add the 'lat' and 'lon' values, with keys from the 'todaysDataPostcode' object, to the 'data' object.
This should be done when the postcodes match.
I'm using an approach I have done before, but the final stage is different and it's proving difficult to figure out
here's the code:
...ANSWER
Answered 2019-Sep-17 at 14:08Is that what you've been after?
QUESTION
Currently matching values from two different objects. When match is found, add value from one object to the other.
Recursively do this and return one new object.
I am a little lost. I have tried .map within .map and I have tried filter.
I will post the code below and any help will be massively appreciated. It is my last hurdle.
i want to match on either location
or postcode
Then match the years.
Then add the price
to the nested Object year
Here is my current function:
...ANSWER
Answered 2019-Sep-11 at 12:25After getting a match:
Say for example: Year 95, store the year and a price in a NewObject. Then while searching for the year in the second data set try to use year to search I hope that is what you have done recently.
Now use this syntax in the loop while mapping data for second object
SecondObject.Year.k = NewObject.price
QUESTION
Hi I am trying to click this link using selenium but it does not get clicked
This is the HTML from the website
...ANSWER
Answered 2019-Apr-10 at 09:08Delete [@href]
from your Xpath, you need a Xpath to the webElement not to the atribute of the webElement.
QUESTION
So that might be silly, but I'm stuck with this recursion T(n) = 5T(n/2) + O(nlogn)
. I know from Master Theorem that it's supposed to be
, but I can't really get there.
So far I got to a point of
I just wanted to know if I'm going the right direction with that
...ANSWER
Answered 2018-Jun-20 at 01:01You're definitely on the right track here! Let's see if we can simplify that summation.
First, notice that you can pull out the log n term from the summation, since it's independent of the sum. That gives us
(n log n) (sum from k = 0 to lg n (5/2)k)
That sum is the sum of a geometric series, so it solves to
((5/2)log n + 1 - 1) / (5/2 - 1)
= O((5/2)lg n)
Here, we can use the (lovely) identity that alogb c = clogb a to rewrite
O((5/2)lg n) = O(nlg 5/2)
= O(nlg 5 - 1)
And plugging that back into our original formula gives us
n log n · O(nlg 5 - 1) = O(nlg 5 log n).
Hmmm, that didn't quite work. We're really, really close to having something that works here, though! A good question to ask is why this didn't work, and for that, we have to go back to how you got that original summation in the first place.
Let's try expanding out a few terms of the recurrence T(n) using the recursion method. The first expansion gives us
T(n) = 5T(n / 2) + n log n.
The next one is where things get interesting:
T(n) = 5T(n / 2) + n log n
= 5(5T(n / 4) + (n / 2) log (n / 2)) + n log n
= 25T(n / 4) + (5/2) log (n / 2) + n log n
Then we get
T(n) = 25T(n / 4) + (5/2) log (n/2) + n log n
= 25(5T(n / 8) + (n / 4) log (n / 4)) + (5/2) log (n/2) + n log n
= 125T(n / 8) + (25/4)n log (n / 4) + (5/2) log (n/2) + n log n
The general pattern here seems to be the following sum:
T(n) = sum from i = 0 to lg n (5/2)kn lg(n/2k)
= n sum from i = 0 to lg n (5/2)k lg(n/2k)
And notice that this is not your original sum! In particular, notice that the log term isn't log n, but rather a function that grows much more slowly than that. In fact, as k gets bigger, that logarithmic term gets much, much smaller. In fact, if you think about it, the only time we're really paying the full lg n cost here is when k = 0.
Here's a cute little trick we can use to make this sum easier to work with. The log function grows very, very slowly, so slowly, in fact, that we can say that log n = o(nε) for any ε > 0. So what what happens if we try upper-bounding this summation by replacing lg (n / 2k) with (n / 2k)ε for some very small but positive ε? Well, then we'd get
T(n) = n sum from i = 0 to lg n (5/2)k lg(n/2k)
= O(n sum from i = 0 to lg n (5/2)k (n / 2k)ε)
= O(n sum from i = 0 to lg n (5/2)k nε (1 / 2ε)k)
= O(n1+ε sum from i = 0 to lg n (5 / 21+ε))
This might have seemed like some sort of sorcery, but this technique - replacing logs with tiny, tiny polynomials - is a nice one to keep in your back pocket. It tends to come up in lots of contexts!
The expression we have here might look a heck of a lot worse than the one we started with, but it's about to get a lot better. Let's imagine that we pick ε to be sufficiently small - say, so that that 5 / 21+ε is greater than one. Then that inner summation is, once again, the sum of a geometric series, and so we simplify it to
((5/21+ε)lg n + 1 - 1) / (5/21+ε - 1)
= O((5/21+ε)lg n)
= O(nlg (5/21+ε)) (using our trick from before)
= O(nlg 5 - 1 - ε)
And that's great, because our overall runtime is then
T(n) = O(n1+ε nlg 5 - 1 - ε)
= O(nlg 5),
and you've got your upper bound!
To summarize:
Your original summation can be simplified using the formula for the sum of a geometric series, along with the weird identity that alogb c = clogb a.
However, that won't give you a tight upper bound, because your original summation was slightly off from what you'd get from the recursion method.
By repeating the analysis using the recursion method, you get a tighter sum, but one that's harder to evaluate.
We can simplify that summation by using the fact that log n = o(nε) for any ε > 0, and use that to rejigger the sum to make it easier to manipulate.
With that simplification in place, we basically redo the analysis using the same techniques as before - sums of geometric series, swapping terms in exponents and logs - to arrive at the result.
Hope this helps!
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install 3DN
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page