gradient

问题 In Android, I used the following code to generate a gradient background that I need: <gradient android:angle="90" android:startColor="#40000000" android:endColor="#00000000" android:type="linear" /> The background goes from light to relatively dark from top to bottom. I wonder how to do the same in Python with PIL, since I need the same effect on another program written in Python. 回答1: Here's something that shows ways to draw either multicolor rectangular horizontal and vertical gradients.

问题 This maybe a simple question, but I didn't find a clear explanation for this. I know in css3, the syntax for setting up a radial gradient is as following: radial-gradient(shape size at position, start-color, ..., last-color); If I set sth like this: radial-gradient(circle at 50% 50%, #f00 50%, transparent 50%); I thought the red circle will be positioned in the center, and the radius of the red circle would be 50%*containerWidth (or containerHeight). i.e. the circle will just touch the edge

问题 How can I go about specifying absolute Offset s for the GradientStop s in my LinearGradientBrush ? I have a GridView with a LinearGradientBrush as the background: <Grid.Background> <LinearGradientBrush StartPoint="0,0" EndPoint="0,1"> <LinearGradientBrush.GradientStops> <GradientStop Offset="0" Color="White" /> <GradientStop Offset="0.25" Color="White" /> <GradientStop Offset="0.4" Color="WhiteSmoke" /> </LinearGradientBrush.GradientStops> </LinearGradientBrush> </Grid.Background> When the

问题 I have a list of values from 0-1. I want to convert this list to an image by using a gradient that converts these floating point values to RGB values. Are there any tools in Java that provide you with this functionality? 回答1: 0 should be mapped 0 1 should be mapped 255 keep in mind that you need 3 of them to make a color so multiply by 255 the floating number and cast it to int. 回答2: Perhaps GradientPaint can do what you want. It's unclear how you want a list of floating point values to be

问题 I want to change the background color from my combobox. But I would like to retain the color gradients. I've tried using this code but still does not get the effect. <Setter Property="Background" Value="White"/> <!-- It's only white :( --> 回答1: <ComboBox> <ComboBox.Background> <LinearGradientBrush EndPoint="0,1"> <GradientStopCollection> <GradientStop Color="Blue" Offset="0.5" /> <GradientStop Color="White" Offset="0.5" /> </GradientStopCollection> </LinearGradientBrush> </ComboBox.Background

问题 I have been trying to call filter css method from jquery for IE, but I am aint b able to do so? What I using ? $('.gtob').mouseover(function(){ $(this).css("background-image","-moz-linear-gradient(100% 100% 90deg, #373737, #000000)"); $(this).css("background-image","-webkit-gradient(linear, 0% 0%, 0% 100%, from(#373737), to(#000000))"); $(this).css("filter","progid:DXImageTransform.Microsoft.gradient( startColorstr='#373737', endColorstr='#000000',GradientType=0)"); }); The first two lines

问题 I want to plot a filled circle in Gnuplot, but with transparency changing with the distance to the center of the circle, i.e. a transparency gradient (solid near to the center, and transparent near to the radius of circle). Is there any way of doing this in Gnuplot? Thank you in advance! 回答1: One option would be to define a custom function describing the gradient (e.g., a Gaussian) and then use pm3d map to visualize it: set terminal pngcairo size 400,400 set output 'fig.png' unset border

问题 can anyone explain why I get such dramatically different results for the Laplace operator in Matlab when I use laplacian = del2(image); versus [x, y] = gradient(image); [xx, xy] = gradient(x); [yx, yy] = gradient(y); laplacian = xx + yy; Shouldn't these come to the same thing? They get particularly divergent when one includes a dx term. Putting my example up here in case it helps: I have a test field consisting of [5; 2.5+2.5i; 5i; -2.5+2.5i; -5; -2.5-2.5i; -5i; 2.5-2.5i] times its transpose

问题 Here's the current code I am using. <? header("Content-type: image/png"); // example: <img src="gradient.php?height=600&width=100&start=00FF00&end=ff0000" /> $height=100; $width=1; $start='000000'; $end='FFFFFF'; extract($_REQUEST); // overwrite using vars from url $start_r = hexdec(substr($start,0,2)); $start_g = hexdec(substr($start,2,2)); $start_b = hexdec(substr($start,4,2)); $end_r = hexdec(substr($end,0,2)); $end_g = hexdec(substr($end,2,2)); $end_b = hexdec(substr($end,4,2)); $image =

问题 x = Conv2D(768, (3, 3), padding='same', activation='relu', kernel_initializer='normal', name='rpn_conv1',trainable=trainable)(base_layers) x_class = Conv2D(num_anchors, (1, 1), activation='sigmoid', kernel_initializer='uniform', name='rpn_out_class',trainable=trainable)(x) # stop gradient backflow through regression layer x_regr = Conv2D(num_anchors * 4, (1, 1), activation='linear', kernel_initializer='zero', name='rpn_out_regress',trainable=trainable)(x) How to use K.stop_gradient() to stop