gradient

I need to create black iphone like gradient in my android application. Please view black gradient at the top in the image below. How to do it? Thanks Something like this, perhaps? <shape xmlns:android="http://schemas.android.com/apk/res/android" android:shape="rectangle"> <gradient android:startColor="#FF000000" android:endColor="#FF8E8E8E" android:angle="270"/> </shape> I've posted a blog about an Iphone look an Android: http://www.dibbus.com/2011/06/android-gradient-toolbar/ You can find some examples of gradients and what is possible using xml and/or 9-pacth images. You can create gradients

This question already has answers here : Multi-color linear gradient in WinForms (2 answers) Closed 3 years ago . I am trying to fill a rectangle with a multi-color diagonal gradient in winforms that looks like the following example: diagonal gradient I know this is a WPF example, but is it possible to get similar results in winforms? punker76 Here is a little example for you void MainFormPaint(object sender, PaintEventArgs e) { LinearGradientBrush br = new LinearGradientBrush(this.ClientRectangle, Color.Black, Color.Black, 0 , false); ColorBlend cb = new ColorBlend(); cb.Positions = new[] {0,

I'm playing around with createRadialGradient() on HTML5 canvas. It works like a charm, except when I'm trying to implement (semi-)transparency. I made this jsFiddle to make things hopefully clearer: http://jsfiddle.net/rfLf6/1/ . function circle(x, y, r, c) { ctx.beginPath(); var rad = ctx.createRadialGradient(x, y, 1, x, y, r); rad.addColorStop(0, c); rad.addColorStop(1, 'transparent'); ctx.fillStyle = rad; ctx.arc(x, y, r, 0, Math.PI*2, false); ctx.fill(); } ctx.fillRect(0, 0, 256, 256); circle(128, 128, 200, 'red'); circle(64, 64, 30, 'green'); What is happening is a circle (radial gradient

In tensorflow, if one has two tensors x and y and one want to have the gradients of y with respect to x using tf.gradients(y,x) . Then what one actually gets is : gradient[n,m] = sum_ij d y[i,j]/ d x[n,m] There is a sum over the indices of y, is there a way to avoid this implicit sum? To get the whole gradient tensor gradient[i,j,n,m] ? Here is my work around just taking the derivative of each component (as also mentionned by @Yaroslav) and then packing them all together again in the case of rank 2 tensors (Matrices): import tensorflow as tf def twodtensor2list(tensor,m,n): s = [[tf.slice

I want to create a "glow" effect on top of an image. On dark it's almost working, but on bright images you can see a gray rectangle on image, as seen in below image: Here is my code: public static Bitmap drawModifiedImage(Bitmap bitmap) { Canvas bigCanvas = new Canvas(bitmap); Bitmap b = Bitmap.createBitmap(bitmap.getWidth(), 20, Bitmap.Config.ARGB_8888); Canvas c = new Canvas(b); c.drawColor(Color.TRANSPARENT); LinearGradient grad = new LinearGradient(bitmap.getWidth()/2, 0, bitmap.getWidth()/2, 20, Color.WHITE, Color.TRANSPARENT, Shader.TileMode.CLAMP); Paint p = new Paint(); p.setStyle

java: I want to use the Gradient style Transparent Background to JFrame. On the top the transparency should be 100% but when going down it should go on decreasing and at the bottom it should be 20% I know i can use the images already having such effect but i want to provide the themes facility and allowing user to use their favorite images but allow transparency at the run time. Sun added support for translucent backgrounds to java in 6u10 but it is not formally described in the API. In Java 7 the functionality was formally added to the API via the setBackground(), setOpacity(), and setShape()