Let us say I have two shapes. One is a polygon and the other a square. Each shape consists of lines which two points (a latitude/logitude pair). I would like to determine the degree to which the square is within the polygon (percentages 0 ... 100). Is this possible? Any pointers to relevant algorithms would be very much appreciated. Thanks.
Christian
In C++ you could try CGAL. or GEOS
Found CGAL in the intersection detection section of the Stonybrook Algorithm Repository maintained by Steven Skiena and GEOS on OpenGeo
If you know the projection, add the data to a Postgis/postgres table, compute the area, and then do a matrix between all your boxes such that the data is area_box_1 / area_box_2
Boxes | 1 | 2 | 3 | 4 | 5
1 20% 0% 0% 5% 0%
----------------------------------
2
----------------------------------
3
----------------------------------
4
----------------------------------
5
Related
I have a dataset in csv format. one columns is images and the other one are it's corresponding labels. I have done with single labels and multi label classification problems but this data looks impossible for me to convert to a multi class that's why I am asking here. Please first check the data format below
| uid | labels |
| --- | --- |
| 1.jpg | The cardiac silhouette and mediastinum size are within normal limits. There is no pulmonary edema. There is no focal consolidation. There are no XXXX of a pleural effusion. There is no evidence of pneumothorax. |
| 2.jpg | Borderline cardiomegaly. Midline sternotomy XXXX. Enlarged pulmonary arteries. Clear lungs. Inferior XXXX XXXX XXXX.
| 3.jpg | There are diffuse bilateral interstitial and alveolar opacities consistent with chronic obstructive lung disease and bullous emphysema. There are irregular opacities in the left lung apex, that could represent a cavitary lesion in the left lung apex.There are streaky opacities in the right upper lobe, XXXX scarring. The cardiomediastinal silhouette is normal in size and contour. There is no pneumothorax or large pleural effusion. |
I want to split the labels with (.) and (,) so that they will act as multi labels (each image will have many labels from 5-7) and then I want to convert these into numpy array or any sort which I can feed to my image classifier models.
Problems is to split and convert this data in such a form that can be used by image classifier models.
Thank You
We've signed up to the Pro plan and now we need to create a report using Map Image REST API to generate heatmaps using multiple colors (more than 4 colors).
I saw on the documentation that there is a limit of 4 levels and colors, I'm wondering if it's possible to use more colors in order to reach our requirements.
Do you have plans to increase the limits or beta version that doesn't have those limits?
For instance, we need to create 6 areas each one with different colors and 6 levels on the same map as shown on the following image, I should be able to use 6 different colors but only shows up 4 colors.
Map image example with 6 areas
Here is the request
GET https://image.maps.ls.hereapi.com/mia/1.6/heat
?apiKey={{API_KEY}}
# Area 1 - Yellow
&a0=49.27,-123.48
&rad0=1900
&l0=0
# Area 2 - Red
&a1=49.25,-123.38
&rad1=1500
&l1=1
# Area 3 - Blue
&a2=49.18,-123.342144
&rad2=1500
&l2=2
# Area 4 - Green
&a3=49.28,-123.35
&rad3=1000
&l3=3
# Area 5 - Orange
&a4=49.21,-123.55
&rad4=1800
&l4=4
# Area 6 - White
&a5=49.30,-123.60
&rad5=1000
&l5=5
#
&z=11
&w=900
&h=900
&plt=FCFF00,EB2501,001EFF,1FE80C,FF8C0D,FFFFFF
Thanks!
I can't speak to our plans for this API, I can, however, raise a ticket internally asking that this be considered. My guess is that it's for performance as well as "length of URL" concerns, but at minimum I can ask.
A bag contains 16 balls of following colors: 8 red, 4 blue, 2 green, 1 black and 1 white. Anisha picks a ball randomly from the bag and messages Babu its color using a string of zeros and ones. She replaces the ball in the bag and repeats this experiment many times. What is the minimum expected length of the message she has to convey to Babu per experiment?
(a)3/2
(b)log 5
(c)15/8
(d)31/16
(e)2
According to me, since the ball is taken out with replacement. At any time, there are 16 balls of 5 different colors in the bag. To encode 5 colors, ceiling of log5 (base 2) i.e. 3 bits should be needed but the answer given is (15/8). Can someone point out my mistake and provide some hint for the correct solution?
using static huffman compression you can encode the more common colours in fewer bits than the rare colours, that being the case on can expect that common colours will usually be chosen.
eg:
red 1
blue 01
green 001
white 0001
black 0000
on average from 16 draws there will be
8 reds = 8 bits
4 blues = 8 bits
2 greens = 6 bits
1 white = 4 bits
1 black = 4 bits
for a total of 30/16 bits on average
Your answer is right as the maximum value needed for encoding. But consider the following coding scheme 1 - red (1/2 prob), 01 - blue (1/4 prob), 00 - green (1/8 prob), 001 - black (1/16 prob), 000 - white (1/16 prob) multiply message length by probability and you should have 1 + 5/8 ( not 15/8 ... though)
Is there a semantically correct way of showing a tooltip in a table header cell that explains something about the column's data?
My table looks like this:
---------------------
| URL | Pageviews |
---------------------
| url/1 | 5 / 20 |
---------------------
| url/2 | 2 / 14 |
---------------------
I want the "Pageviews" text to have a tooltip that says "Week / Total" (explaining why the data looks like that).
Is abbr the right element for me to use? (it would look like this: <th><abbr title="Week / Total">Pageviews</abbr></th> I can't seem to find a better one. The definition says it's supposed to be used for abbreviations, which is not my case.
HTML5 allows you to use the "title" attribute on any element.
http://dev.w3.org/html5/alt-techniques/ is a good summary of such techniques.
<th title="Week"> Pageviews </th>
I'd highly recommend adding a <caption/> under the table with such data for usability and UX purposes !
I'm on the FIRST robotics team at my high school, and we are working on developing a kiwi drive robot, where there are three omni wheels mounted in a equilateral triangle configuration, like this:
The problem is programming the robot to drive the motors such that that the robot moves in the direction of a given joystick input. For example, to move "up", motors 1 and 2 would be powered equally, while motor 3 would be off. The joystick position is given as a vector, and I was thinking that if the motors were expressed as vectors too, vector projection might be what I need. However, I'm not sure if this is right, and if it is, how I would apply it. I also have a feeling that there may be multiple solutions to one joystick position. Any help would be greatly appreciated.
I've built 9 robots during my time at school (1 FIRST, 8 RoboCup). We used the same omnidrive layout as you do. Beta's answer looks correct but add rotation to all wheels afterwards:
W1 = -1/2 X - sqrt(3)/2 Y + R
W2 = -1/2 X + sqrt(3)/2 Y + R
W3 = X + R
[This is Beta's formula with some added Rotation]
You need to think about the available ranges for your motors. I am guessing it can take a PWM signal of +/-255, so either the input or the output has to be adjusted somewhat. (It's not that hard...)
A good paper with details
To answer your specific questions: Vector projection is essentially what you are doing here. You apply it by having a matrix M, your input from the joystick I and your output to the motors O. Thus O = M * I;
M = [(-0.5 -sqrt(3)/2 +1)
(-0.5 +sqrt(3)/2 +1)
(1 0 +1)]
First let's define some terms. In keeping with the usual convention, the X axis will point to the right and the y axis will point up (so that the thrust of wheel 3 is along the X axis). We'll call the motion of the wheels W1, W2 and W3, each defined so that Wi > 0 means that the wheel rotates in the clockwise direction. In your example, if W1 < 0, W2 = W1 and W3 = 0, the robot will move in the +Y direction.
If all three wheels rotated at the same rate (W1 = W2 = W3) the robot would rotate in place. I'm guessing you don't want that, so the sum of the rotations must be zero: W1 + W2 + W3 = 0.
The motion of each wheel contributes to the motion of the robot; they add as vectors:
W1 = -1/2 X - sqrt(3)/2 Y
W2 = -1/2 X + sqrt(3)/2 Y
W3 = X
So if you know the desired X and Y from the joystick, you have W1, W2 and W3. As we've already seen, the difference between W1 and W2 is what drives Y motion. Their sum drives motion in X.
Though this system can be solved mathematically, in 2002, FIRST Team 857 chose to solve it mechanically. Our control system used three joysticks mounted with their X-axes forming an equilateral triangle, and handles replaced with ball-socket arms connected with a Y-shaped yoke. Map the X-axis of each stick directly to a motor speed, and the control system has been solved. As an advantage, this system is very intuitive for laypeople to run--push the yoke in the direction you want to go, rotate it to turn.
As you have recognized, the first part of this will be finding an appropriate equation to represent the resultant motion for any motor settings. Depending on the level of control and feedback you have on your motor speeds, I would suggest the process you go thorough should start with writing a vector equation: (define positive X as straight ahead)
-M1Cos(30)+M2Cos(30)=X (the negative is because 1 and 2 must be powered the same magnitude, but opposite polarities for forward motion)
M1Sin(30)+M2Sin(30)-M3 = Y (as anticlockwise motion on 1 and 2 will result in the robot moving left in the Y and anticlockwise motion on 3 will result in the robot moving to the right)
The other input that you need to add into this is the desired rotation of the robot, thankfully, M1+M2+M3 = W (Rotational velocity)
Your joystick input will give you X,Y and W, so you have 3 equations with 3 unknowns.
From here it is simultaneous equations, so you may end up with multiple solutions, but these can generally be restricted based on possible motor speeds and the like.
An example of this is the rec::robotino::com::OmniDrive Class - the source code for this method is available too...