When loading a CSV with PHPSpreadsheet I am losing float precision. Seems to be a maximum precision of 13. Is this a PHPSpreadheet issue or PHP issue?
The cell value 4.02020325142409 is loaded as 4.0202032514241 (13th character rounded up)
The cell value 3.90812005382548 is loaded as 3.9081200538255 (13th character rounded up)
The cell value 4.55605765112764 i s loaded as 4.5560576511276 (13th character same)
The cell value 4.4730378939229 is loaded as 4.4730378939229 (Only 13 characters)
By referring to PHP float point numbers documentation, I think it should be the PHP issue rather than PhpSpreadsheet.
The size of a float is platform-dependent, although a maximum of
approximately 1.8e308 with a precision of roughly 14 decimal digits is
a common value (the 64 bit IEEE format).
Not sure about your usage on those float number but a warm reminder from doc stated that :
Warning...So never trust floating number results to the last digit,
and do not compare floating point numbers directly for equality. If
higher precision is necessary, the arbitrary precision math functions
and gmp functions are available....
Related
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.
I am processing images, which are long, usually a few hundred thousand pixel in length. The height is usually in the 500-1000 pixel range. The process involves modifying the images on a column by column basis. So, for example, I have a column of constant values that needs to be subtracted from each column in the image.
Currently I split the image into smaller blocks, put them into linearized 2d arrays. Then I make a linearized 2d array from the column of constant values that is the same size as the smaller block. Then a (image array - constant array) operation is done until the full image is processed.
Should I copy the constant column to the device, and then just operate column by column? Or should I try to make as large of a "constant array" as possible, and then perform the subtraction. I am not looking for 100% optimization or even close to that, but an idea about what the right approach to take is.
How can I optimize this process? Any resources to learn more about this type of processing would be appreciated.
Constant memory is up to 64KB, so assuming your pixels are 4 bytes or less, then you should be able to handle an image height up to about 16K pixels, and still put the entire "constant column" in constant memory.
After that, you don't need to process things "column by column". Constant memory is optimized for access when every thread is requesting the same value from constant memory, which perfectly describes your case.
Therefore, your thread code can be trivially simple:
#define MAX_COL_SIZE 1024
__constant__ float const_column[MAX_COL_SIZE];
__global__ void img_col_kernel(float *in, float *out, int num_cols, int col_size){
int idx = threadIdx.x + blockDim.x*blockIdx.x;
if (idx < num_cols)
for (int i = 0; i < col_size; i++)
out[idx+i*num_cols] = in[idx+i*num_cols] - const_column[i];
}
(coded in browser, not tested)
Set up const_column in your host code using cudaMemcpyToSymbol prior to calling img_col_kernel. Call the kernel with a 1D grid including a total number of threads equal to or greater than your image width (num_cols). Pass the "linearized 2D" pointers to your input and output images to the kernel (in and out). The above kernel should run pretty fast, and essentially be bound by memory bandwidth for images of width 1000 or more. For small images, you may want to increase the number of threads by dividing your image vertically into say, 4 pieces, and operate with 4 times as many threads (and 4 regions of constant memory).
How can I display fractional numbers in an UITextField, e.g. 5 and 1/5 should be displayed as the number "5" followed by the numerator "1" on top, a horizontal line, and the denominator "5" at the bottom. Should I subclass should I subclass a UITextField?
Is there any sample code somewhere that shows how can I achieve this?
If it is a commonly found fraction, you can find the unicode character (of 1/5 character) and then you can assign your textfield's text property something like this "\u{2155}" where 2155 is the unicode value of 1/5 character.
For finding out unicode values, you can use this website http://www.fileformat.info/info/unicode/
For more reading on displaying unicode character, you may use Apple Documentation
However, if the fraction is not a commonly found one, then for them there will be no Unicode characters. In such cases, you have to explicitly convert it into a string like "\(myNumeratorIntegerVariable)/\(myDenominatorIntegerVariable)"
This question already has answers here:
Objective-C - How to increase the precision of a float number
(3 answers)
Closed 10 years ago.
In my application I am just dividing 50 by 3 I want to store the exact result value of this.
If I use float it gives 16.666666 and if i use double then it gives 16.666667.
Actually,I am creating three labels inside a frame by dividing the height of the frame I am deciding the height of each label. so I f i do not get exact value it creates a gap between labels.if if i pass 60 then it works fine because 60/3 results 20 but if I pass 50 then there is a gap.
If you want to make your frame divide into three equal-height areas, then the height of your frame in pixels needs to be divisible by three. You can't display fractional pixels, they are not divisible; each height as measured in pixels needs to be an integer number.
The only way to store an "exact" value would be to create a class called "Rational" (or similar) and store the numerator and denominator of the fraction as separate ivars. Floats and doubles (or any literal computer representation for that matter) cannot store rational numbers with an infinite number of decimal places or transcendental real numbers.
The way to use the "Rational" class would be to store the numerator and denominator, and then apply the appropriate maths to these values (if you wish to propagate "exactness" through the program). The slightly easier way would be to display the rational number as numerator and denominator but use the float/double approximation for the underlying mathematics.
float t= (float)50/3;
long double t1= (long double)50/3;
NSLog(#"%.30f %.30LF",t, t1);
produced output "16.666666030883789062500000000000 16.666666666666666666088425508008".
I would suggest you to go with long double which is not exact but precise enough.
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... ;)