I'm relatively new to Pine and was having trouble with a couple of things.
So I've got a indicator that I want to look back for the last 5 full days of activity, however not each stock trades in the premarket/afterhours so I want to be able to look back say 5 days and then calculate the number of bars on the time frame (i.e. for a 5 min or hourly resolution). That way I can calculate the number of bars present on the current ticker for that time frame and adjust my indicator length (instead of just having a constant figure of 80 bars or whatever).
The second thing I want to do is only have the indicator update on the completion of a full bar (for example the hourly resolution). So I want it to only calculate up to x-1 bars. Right now I'm finding that it's updating on each new tick. ( I was able to figure this one out, just changed it from i= 0 to length -1 to i=1 to length).
Any help is greatly appreciated.
Thanks,
A
According to JPEG2000 specs, Number of tiles in X and Y directions is calculated by following formula:
numXtiles = (Xsiz − XTOsiz)/ XTsiz
&
numYtiles = (Ysiz − YTOsiz)/ YTsiz
But it is not mentioned about the range of numXtiles or numYtiles.
Can we have numXtiles=0 while numYtiles=250 (or any other value) ?
In short, no. You will always need at least one row and one column of tiles to place your image in the canvas.
In particular, the SIZ marker of the JPEG 2000 stream syntax does not directly define the number of tiles, but rather the size of each tile. Since the tile width and height are defined to be larger than 0 (see page 453 of "JPEG 2000 Image compression fundamentals, standards and practice", by David Taubman and Michael Marcellin), you will always have at least one tile.
That said, depending on the particular implementation that you are using, there may be a parameter numXtiles that you can set to 0 without crashing your program. In that case, the parameter is most likely being ignored or interpreted differently.
SCENARIO
I am working on an application that keeps track of blood glucose levels into a graph. On the graph there are "markings" (ex: -200mg) going in vertical order along the y axis on the right side of the screen and "hours" (ex: -12:00 PM) will be along the x axis on the bottom of the graph. I have to plot out little 'dots' to display what the blood glucose level was throughout the way.
ISSUE
I am trying to calculate how to position the 'dots' in the correct time and mg level and I'm having difficulty calculating the positions. I can access the "markings" and retrieve it's marking.center.x to indicate which 'Time Slot' (x axis) and the marking.center.y to indicate which 'MG Level' the 'dot' needs to go into. Problem is it isn't always exactly 12:00 PM or 200mg where it will need to be placed. In fact that would be very rare.
WHAT I NEED
Based on the following variables:
dot.mgLevel
The dot will already know where it needs to go based on the information retrieved from the medical device. It will know the time and mgLevel to assign itself.
marking.mgLevel
The markings will each have evenly distributed values that such as -100mg, -200mg, -300mg ect...
timemarking.timeslot
Each time marking on the bottom will each have evenly distributed times allocated every 30 min. Such as -12:00PM, -12:30PM, -1:00PM ect...
If the dot has a mg Level of 330mg and the closest marking on the mg Level is 300mg, then I need to be able to calculate how much further up the dot needs to move from 300 going towards the 400mg marker.
SO...
If the distance between the markings are 100pt and the dot's mgLevel is 330mg, then I know that I need to move the dot from the 300mg marking toward the 400mg marking by exactly 30pt. That's because it's simple math because the distance between the markings is 100. But in real life it isn't 100, so I need to be able to calculate this.
MY ULTIMATE QUESTION
Say distance between markings is 241 and each marking represents multiples of a hundred. Say my dot has a mgLevel of 412. How do I calculate how far I need to move the dot so that it will be in the correct place?
I THINK?
I think I need to make 241 equal 100%. But I need help.
Distance between markings is 241pt
Markings are multiples of 100mg
1mg will occupy 2.41pt. So 412mg will occupy (2.41 * 412) pt. To know how much to move for the next dot, take the difference in mg and multiply by 2.41.
In general, if distance between 2 markings in points is d, markings are multiples of m, and desired accuracy is k decimal places, 1mg will occupy g:
let divisor = pow(10.0, Double(k))
let g = round((d/m)*divisor) / divisor
I want to have a different scale using a slider for the Min and Max. 0 should be in the middle, but I want to have a scale 0-50 for max and -200-0 for min.
Is that possible? How could I achieve that?
I don't think that this is possible out of the box.
But what you can do is, e.g. setup the slider in a way that for min it has 0 and for max 1000. Now you will have to check in which part of the part, user left the slider. If it's in the lower half, you will have to normalize those values to the desired range, the same for the upper range. The exact middle could be a bit tricky because you will have more points in the slider than in the ranges you are providing, but I guess with some fiddling around you can do this. :)
Hope that helped :)
I believe this is quite possible.
Here is my idea of how you can do it
Take a slider with values -200 to 200
For values > 0. Divide value by 4. (i.e. 200 / 4 = 50)
This will make your positive range from 0 to 50 but negative range from -200 to 0
Similarly while setting value to slider in positive manner you can multiply it by 4.
I use zeros to initialize my matrix like this:
height = 352
width = 288
nFrames = 120
imgYuv=zeros([height,width,3,nFrames]);
However, when I set the value of nFrames larger than 120, MATLAB gives me an error message saying out of memory.
The original function is
[imgYuv, S, A]= changeYuv(fileName, width, height, idxFrame, nFrames)
my command is
[imgYuv,S,A]=changeYuv('tilt.yuv',352,288,1:120,120);
Can anyone please tell me what's going on here?
PS: one of the purposes of the function is to load a yuv video which consists more than 2000 frames. Is there any possibility to implement that?
There are three ways to avoid the error
Process a limited number of
frames at any given time.
Work
with integer arrays. Most movies are
in 8-bit format, while Matlab
normally works with doubles.
uint8 takes 1 byte per element,
while double takes 8 bytes. Thus,
if you create your array as B =
zeros(height,width,3,nFrames,'uint8)`,
it only uses 1/8th of the memory.
This might work for 120 frames,
though for 2000 frames, you'll run
again into trouble. Note that not
all Matlab functions work for
integer arrays; you may have to
reimplement those that require
double.
Buy more RAM.
Yes, you (or rather, your Matlab session) are running out of memory.
Get out your calculator and find the product height x width x 3 x nFrames x 8 which will tell you how much memory you have tried to get in your call to zeros. That will be a number either close to or in excess of the RAM available to Matlab on your computer.
Your command is:
[imgYuv,S,A]=changeYuv('tilt.yuv',352,288,1:120,120);
That is:
352*288*120*120 = 1459814400
That is 1.4 * 10^9. If one object has 4 bytes, then you need 6GB. That is a lot of memory...
Referencing the code I've seen in your withdrawn post, your calculating the difference between adjacent frame histograms. One option to avoid massive memory allocation might be to just hold two frames in memory, instead of reading all the frames at once.
The function B = zeros([d1 d2 d3...]) creates an multi-dimensional array with dimensions d1*d2*d3*...
Depending on width and height, given the 3rd dimension of 3 and the 4th dimension of 120 (which effectively results in width*height*360), may result in a very huge array. There are certain memory limits on every machine, maybe you reached these... ;)