The approach you are using is a valid approach, but a little outdated. From HoneyComb, there is a new way to fix all of this. Your resources folder should now look like this: Please refer to the link I have posted and familiarize yourself with Smallest Width concept. Hope this helps 🙂 EDIT: Adding to … Read more
Yes, it does behave like -1 in numpy.reshape(), i.e. the actual value for this dimension will be inferred so that the number of elements in the view matches the original number of elements. For instance: import torch x = torch.arange(6) print(x.view(3, -1)) # inferred size will be 2 as 6 / 3 = 2 # … Read more
You are right that one can get image dimensions before it’s fully loaded. Here’s a solution (demo): var img = document.createElement(‘img’); img.src=”https://stackoverflow.com/questions/6575159/some-image.jpg”; var poll = setInterval(function () { if (img.naturalWidth) { clearInterval(poll); console.log(img.naturalWidth, img.naturalHeight); } }, 10); img.onload = function () { console.log(‘Fully loaded’); }
Just tried this out and it works for me: int finalHeight, finalWidth; final ImageView iv = (ImageView)findViewById(R.id.scaled_image); final TextView tv = (TextView)findViewById(R.id.size_label); ViewTreeObserver vto = iv.getViewTreeObserver(); vto.addOnPreDrawListener(new ViewTreeObserver.OnPreDrawListener() { public boolean onPreDraw() { // Remove after the first run so it doesn’t fire forever iv.getViewTreeObserver().removeOnPreDrawListener(this); finalHeight = iv.getMeasuredHeight(); finalWidth = iv.getMeasuredWidth(); tv.setText(“Height: ” + finalHeight … Read more
From the documentation of width(): This method is also able to find the width of the window and document. $(window).width(); // returns width of browser viewport $(document).width(); // returns width of HTML document Simple jsFiddle Demo In the demo, I have set html { width: 1000px; }, which is bigger than the viewport. The width … Read more
You can use the getimagesize function like this: list($width, $height) = getimagesize(‘path to image’); echo “width: ” . $width . “<br />”; echo “height: ” . $height;
The canonical way of doing this in numpy would be to use np.transpose‘s optional permutation argument. In your case, to go from ijkl to klij, the permutation is (2, 3, 0, 1), e.g.: In [16]: a = np.empty((2, 3, 4, 5)) In [17]: b = np.transpose(a, (2, 3, 0, 1)) In [18]: b.shape Out[18]: (4, … Read more
try(ImageInputStream in = ImageIO.createImageInputStream(resourceFile)){ final Iterator<ImageReader> readers = ImageIO.getImageReaders(in); if (readers.hasNext()) { ImageReader reader = readers.next(); try { reader.setInput(in); return new Dimension(reader.getWidth(0), reader.getHeight(0)); } finally { reader.dispose(); } } } Thanks to sfussenegger for the suggestion
In general you want to size your blocks/grid to match your data and simultaneously maximize occupancy, that is, how many threads are active at one time. The major factors influencing occupancy are shared memory usage, register usage, and thread block size. A CUDA enabled GPU has its processing capability split up into SMs (streaming multiprocessors), … Read more
System.Drawing.Image img = System.Drawing.Image.FromFile(@”c:\ggs\ggs Access\images\members\1.jpg”); MessageBox.Show(“Width: ” + img.Width + “, Height: ” + img.Height);