mx3 | sample project showcasing/collecting cross platform | Application Framework library
kandi X-RAY | mx3 Summary
kandi X-RAY | mx3 Summary
Cross platform has been well studied on desktop, but this is an exploration in doing that on mobile and an open request for comments and improvements.
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QUESTION
I would like to efficiently calculate all pairwise cross products of the rows of two matrices, A and B, which are nx3 and mx3 in size. And would ideally like to achieve this in einsum notation.
i.e. the output Matrix C, would be (n X m x 3),
where
C[0][0] = cross(n[0],m[0])
C[0][1] = cross(n[0],m[1])
...
C[1][0] = cross(n[1],m[0])
...
Due to the approach I am taking, using for loops aren't an option.
Any help would be much appreciated.
...ANSWER
Answered 2021-Apr-22 at 04:40Looks like cross broadcasts the leading dimensions.
QUESTION
I sent a reply to an email and this is what I got in return, can you translate it for me? and what should I do? Mail Delivery System MAILER-DAEMON@mx3.imadiff.net sent :
I'm sorry to have to inform you that your message could not be delivered to one or more recipients. It's attached below.For further assistance, please send mail to If you do so, please include this problem report. You can delete your own text from the attached returned message. The mail system europeanchampion@pfls.fr: host barracuda03.imadiff.net[194.69.195.26] said:
554 rejecting banned content (in reply to end of DATA command) Reporting-MTA: dns; mx3.imadiff.net X-Postfix-Queue-ID: 6B26E1A50420 X-Postfix-Sender: rfc822; vermontjf@gmail.com Arrival-Date: Sun, 4 Apr 2021 10:45:16 +0200 (CEST)
Final-Recipient: rfc822; europeanchampion@pfls.fr Original-Recipient: rfc822;europeanchampion@pfls.fr Action: failed Status: 5.0.0 Remote-MTA: dns; barracuda03.imadiff.net Diagnostic-Code: smtp; 554 rejecting banned content
...ANSWER
Answered 2021-Apr-15 at 10:18Seems pretty clear in the response you received. The message transfer agent, who's Internet address is mx3.imadiff.net flagged your email as having content that it determined was contraband for their system. You need to look at what you sent and figure it out ... maybe it had curse words in it, or it had an attachment that the filter didn't like. You could try contacting the recipient and asking them what their mail filter system looks for as contraband and then look at the message you sent to see if you can figure out why it got flagged. Then remove whatever that is and re-send it.
QUESTION
I want to implement the following routine without using loops, just using Numpy functions or ndarray methods. Here is the code:
...ANSWER
Answered 2021-Mar-30 at 08:38If I understand correctly, np.reshape
should work. While there are different options for the order
argument, it sounds like you want to reshape the last two axes of length 3M
in the original array both to an array of shape (M,3)
in the 'C'
order (the default). This can be achieved with
QUESTION
I want to calculate the camera calibration parameters from the picture coordinates and object coordinates of some coded markers.
...ANSWER
Answered 2021-Mar-19 at 15:02Well the answer is quite simple:
QUESTION
So a bit of context, i was trying to make a code which reads in the number of jobs and then the jobs themselves goes something like this 4 1 2 3 4
Heres the thing if i enter the number of jobs as 4 then enter 3 jobs, then for some reason it reads in the third job again .
is there any way to check and raise an error if this happens? i would compare two values to check if they were the same, but two values can be same.
also i would be using a file lets say file.txt file.txt would have the number of jobs and jobs in the format 4 1 2 3 4 and i would use ./mx3 < file.txt where mx3 is the program compiled using gcc c99
...ANSWER
Answered 2020-Oct-08 at 10:27- Can you post some code for better understanding ?
I didn't understand your problem at all but i think that could help you :
Return Value of scanf
: copy and past from here
These functions return the number of input items successfully matched and assigned, which can be fewer than provided for, or even zero in the event of an early matching failure.
The value EOF is returned if the end of input is reached before either the first successful conversion or a matching failure occurs. EOF is also returned if a read error occurs, in which case the error indicator for the stream (see ferror(3)) is set, and errno is set indicate the error.
(I'm new on stack forgive me if i'm clumsy and explain me if possible for my self improvement ^^)
QUESTION
Let's say I have a mx3
2D array, what I want is:
- First sort the array according to its last column (the column indexed 2).
- Find all the array slices whose values in the last column are the same, sort each array slice according to the column in the middle.
- Find all the array slices in which the two values in the last two columns are the same among rows, sort each array slice according to the first column.
An example of this sorting is like
Below is the solution I came up with, which essentially implements the quick-sort algorithm:
...ANSWER
Answered 2020-Sep-30 at 06:27You can use lexsort:
QUESTION
I have an mx3 array that is used to create a 3d model. Is there is a fast way to extract all the points that belong to a given plane using numpy or other python functions? The plane will take the Ax+By+Cz+D=0 form. I'm currently looping through all the points in the array to find the points that satisfy this equation.
...ANSWER
Answered 2020-Sep-21 at 06:51Vectorization will be much faster. In the example below, all points below lie on integer values in the region -100 < x,y,z < 100. The matrix p contains one million points; we calculate all points that lie on a given plane (almost instantaneously):
QUESTION
I have turned a double for loop
into a single for loop
using vectorization
. I would like to now get rid of the last loop
.
I want to slice
an Nx3 array
of coordinates and calculate distances between the sliced portion and the remaining portion without using a for loop.
(1) the slice is always 3x3
.
(2) the slice is variable i.e., Mx3
where M is always significantly smaller than N
Vectorizing the interaction of 1 row of the slice interacting with the remainder is straightforward. However, I am stuck using a for loop to do (in the case of the slice of size 3) 3 loops, to calculate all distances.
Context:The Nx3 array is atom coordinates, the slice is all atoms in a specific molecule. I want to calculate the energy of a given molecule interacting with the rest of the system. The first step is calculating the distances between each atom in the molecule, with all other atoms. The second part is to use those distances in a function to calculate energy, and that is outside the scope of this question.
Here is what I have for a working minimal example (I have vectorized
the inner loop, but, need to (would really like to...) vectorize
the outer loop
. That loop won't always be of only size 3, and python
is slow at for loops.
ANSWER
Answered 2020-Mar-28 at 23:11Here how you can do it:
QUESTION
I have a set of points stored in a variable called pts
, which is a Mx3
matrix.
Where pts(:,1)
are the x coordinates on the image, pts(:,2)
are the y coordinates on the image, and pts(:,3)
are the probabilities.
How can I visualize the points on the image as a heatmap. Points with higher probabilities shown in dense red color. Similar to the image attached.
Any help would be much appreciated!
...ANSWER
Answered 2019-Dec-01 at 10:32Due to the fact you didn't supply the data I cant use your example, but a simple solution to your problem is to use imagesc
on top of your image with the z
coordinate value as the color. you obviously should play with the transparency in order to get the desired result.
some simple example:
QUESTION
The point set A
is a Nx3
matrix, and from two point sets B
and C
with the same size of Mx3
we could get the lines BC
betwen them. Now I want to compute the distance from each point in A
to each line in BC
. B
is Mx3
and C
is Mx3
, then the lines are from the points with correspoinding rows, so BC
is a Mx3
matrix. The basic method is computed as follows:
ANSWER
Answered 2019-Nov-01 at 15:11You can remove the for
loops by doing this (it should speed-up at the cost of memory, unless M
and N
are small):
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Install mx3
Optionally gem install xcpretty to make the output of xcodebuild nice
Run make play. You should see "Hello, #{your login name}" printed to the console, if so, you seem to already have all the requirements met for building on iOS.
Run make android to build the example android application.
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