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What's the shortest way to combine say 20 256x256 thumbnails into a single large image of 4 rows by 5 columns using RMagick?
Assuming all the images are in the current directory and named from 1.jpg to n.jpg and row * col = n.
include Magick
row = NUM_ROWS
col = NUM_COLS
ilg = ImageList.new
1.upto(col) {|x| il = ImageList.new
1.upto(row) {|y| il.push(Image.read((y + (x-1)*col).to_s + ".jpg").first)}
ilg.push(il.append(false))}
ilg.append(true).write("out.jpg")
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Only if it's possible; for example one panel for FPS and another one for MS
Thanks, best regards
Yes, you can have multiple:
var stats1 = new Stats();
stats1.showPanel(0); // Panel 0 = fps
stats1.domElement.style.cssText = 'position:absolute;top:0px;left:0px;';
document.body.appendChild(stats1.domElement);
var stats2 = new Stats();
stats2.showPanel(1); // Panel 1 = ms
stats2.domElement.style.cssText = 'position:absolute;top:0px;left:80px;';
document.body.appendChild(stats2.domElement);
And then in your main loop
stats1.update();
stats2.update();
stats.js r17
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I am trying to transfer some part of my image to another matrix with this code:
p1 = zeros(512,512,3);
p1(1:128, 1:128, 1:3) = image(1:128, 129:256, 1:3);
And when this code did not work I tried with 3 for loop
and after that I tried to transfer R G B layers separately:
p1(1:128, 1:128, 1) = image(1:128, 129:256, 1);
But none of these codes work. Could anyone tell me what to do?
In Matlab you can always assign one matrix to another with the same size. For example:
A = ones (4,2,3,4,5);
B = zeros(4,2,3,4,5);
A = B
will run with no errors.
It is possible that the type of the image you are using is uint8 (8-bit unsigned integer), while zeros creates a double matrix. This will result in improper behavior of operations or built-in functions, if you do not cast (change the type of) your matrices properly. Use whos to check the type of your image:
whos image
Try to cast the zeros to uint8:
p1 = uint8(zeros(512,512,3));
...
Here is an example:
image = imread('peppers.png');
partail_im = uint8(zeros(size(image)));
partail_im(1:128,1:128,:) = image(1:128,129:256,:);
imshow(partail_im);
Note: It's better not to use image as the name of any variable since it's name of a built-in function image.
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When I type "dispatch_after" in XCode, it auto-completes GCD code as below:
int64_t delayInSeconds = 2.0;
dispatch_time_t popTime = dispatch_time(DISPATCH_TIME_NOW, delayInSeconds * NSEC_PER_SEC);
dispatch_after(popTime, dispatch_get_main_queue(), ^(void){
<#code to be executed on the main queue after delay#>
});
let's look this statement :
int64_t delayInSeconds = 2.0;
int64_t is the synonyms of "long long",but here we assign a double value "2.0" to this kind of variable delayInSeconds.So the fractional part of the value will be discarded.If you assign it with value "0.3",the variable's value will be 0. I think this template of code will mislead programmers and the right code should be float delayInSeconds = 2.0;Am I right? thanks in advance.
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The source file is named "ReadOutputFile.m".
function ReadOutputFile(inputFile, outputFile)
SPIT = 2;
R_REGULARITY = 8;
%read input, threshold, and level
spitclip = wavread(inputFile);
spitclip = spitclip*(2^15);
fid = fopen('threshBinary.bin', 'rb', 'native');
thresh = fread(fid, 'int16');
fclose(fid);
fid = fopen('signalLevelBinary.bin', 'rb', 'native');
lvl = fread(fid, 'int16');
fclose(fid);
%get start and stop information from output file
outputNumeric = csvread(outputFile);
starts = outputNumeric(:,1);
stops = outputNumeric(:,2);
types = outputNumeric(:,8);
reasons = outputNumeric(:,9);
regularity = outputNumeric(:,7);
indicesOfSpits = find(types==SPIT);
numEvents = length(starts);
eventGate = zeros(1,length(spitclip));
firstSpits = zeros(1,length(spitclip));
chainedSpits = zeros(1,length(spitclip));
maxVal = 800;
%loop through events and plot spits only
%color code by reason
for n = 1:length(indicesOfSpits)
i = indicesOfSpits(n);
if(reasons(i) == R_REGULARITY)
chainedSpits(starts(i):stops(i)) = regularity(i)*800;
else
firstSpits(starts(i):stops(i)) = maxVal;
end
end
plot(spitclip);
hold on;
grid on;
plot(lvl,'r');
plot(thresh, 'y');
plot(firstSpits, 'm');
plot(chainedSpits, 'k');
That looks like Matlab (or the free equivalent octave)
http://en.wikipedia.org/wiki/M_(programming_language)
You would wonder, how easy it is to find the programming language.
By just knowing the suffix.
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Could you give me a hint on how to solve this problem?
How can I find out whether most cars have the same color if the only function I can use is to check whether two cars have the same color?
With equality, you can first build equivalence classes and then count the largest one.
You can create a "list" where the color of the car is the key value and it counts how many cars of that color you have and stores a car of that color.
It goes something like this:
you will have a struct of with this fields:
struct color
car *car_with_color = NULL
int count = 0
so you go to the first car store it's color and add 1:
color[1].car_with_color=&cars[1];
color[1].count=1;
then for the next car you do:
if (cars[2].color == color[1].car_with_color->color)
color[1].count++;
else
color[2].car_with_color=&cars[2];
color[2].count=1;
and so on:
for (i=1;i<=NUM_OF_CARS;i++){
j=1;
while (color[j].car_with_color != NULL){
if (cars[i].color == color[j].car_with_color->color){
color[i].count++;
break;
}else
j++;
}
if (color[j].car_with_color == NULL){
color[j].car_with_color=&cars[i];
color[j].count=1;
}
}
it's written in a C like pseudo-code.