invert

问题 How do I invert the text-color of an element using jQuery? <div style="color: rgb(0, 0, 0)">Invert me</div> 回答1: A bit late but better late than never: function invert(rgb) { rgb = Array.prototype.join.call(arguments).match(/(-?[0-9\.]+)/g); for (var i = 0; i < rgb.length; i++) { rgb[i] = (i === 3 ? 1 : 255) - rgb[i]; } return rgb; } console.log( invert('rgba(255, 0, 0, 0.3)'), // 0, 255, 255, 0.7 invert('rgb(255, 0, 0)'), // 0, 255, 255 invert('255, 0, 0'), // 0, 255, 255 invert(255, 0, 0) /

问题 How do I invert the text-color of an element using jQuery? <div style="color: rgb(0, 0, 0)">Invert me</div> 回答1: A bit late but better late than never: function invert(rgb) { rgb = Array.prototype.join.call(arguments).match(/(-?[0-9\.]+)/g); for (var i = 0; i < rgb.length; i++) { rgb[i] = (i === 3 ? 1 : 255) - rgb[i]; } return rgb; } console.log( invert('rgba(255, 0, 0, 0.3)'), // 0, 255, 255, 0.7 invert('rgb(255, 0, 0)'), // 0, 255, 255 invert('255, 0, 0'), // 0, 255, 255 invert(255, 0, 0) /

问题 I have a string A and want to test if another string B is not part of it. This is a very simple regex whose result can be inverted afterwards. I could do: /foobar/.test('foobar@bar.de') and invert it afterwards, like this: !(/foobar/).test('foobar@bar.de') The problem I have is, that I need to do it within the regular expression and not with their result. Something like: /!foobar/.test('foobar@bar.de') (which does not work) In other words: the regular expression should test for a non

问题 I'm using WinForms. I have a picture-box in my form. When I open a picture in the picture-box I am able to invert the colors back and forth on a click of a button, but my code is extremely slow. How can I increase the performance. private void Button1_Click(object sender, System.EventArgs e) { Bitmap pic = new Bitmap(PictureBox1.Image); for (int y = 0; (y <= (pic.Height - 1)); y++) { for (int x = 0; (x <= (pic.Width - 1)); x++) { Color inv = pic.GetPixel(x, y); inv = Color.FromArgb(255, (255

问题 Lets say I have a function that is nlogn in space requirements, I want to work out the maximum size of input for that function for a given available space. i.e. I want to find n where nlogn=c. I followed an approach to calculate n, that looks like this in R: step = function(R, z) { log(log(R)-z)} guess = function(R) log(log(R)) inverse_nlogn = function(R, accuracy=1e-10) { zi_1 = 0 z = guess(R) while(abs(z - zi_1)>accuracy) { zi_1 = z z = step(R, z) } exp(exp(z)) } But I can't get understand

问题 I am creating an application that will take a URL as input, retrieve the page's html content off the web and extract everything that isn't contained in a tag . In other words, the textual content of the page, as seen by the visitor to that page. That includes 'masking' out everything encapsuled in <script></script> , <style></style> and <!-- --> , since these portions contain text that is not enveloped within a tag (but is best left alone). I have constructed this regex: (?:<(?P<tag>script

问题 I'd like to comb off the n largest extremes from a timeseries. heapq works perfectly for the nlargest def nlargest(series, n): count = 0 heap = [] for e in series: if count < n: count+=1 hp.heappush(heap, e) else: # keeps heap size fixed hp.heappushpop(heap,e) ''' note: heap[0] is smallest ''' return heap but what about the n smallest? Note that i want a subset of the original series, so heapify and reversing the order won't work. What I'd like is essentially to overload the comparison

问题 I am trying to make a function that will invert a Three.js Geometry across the xz plane. Scaling the geometry by (1, -1, 1) will effectively do just this, but now the planes that were facing outwards are facing inwards, so I can see through the geometry. I know there is a way to render both sides of the faces, but I'd rather not so I can speed up render times. Is there an easy way I can toggle which face of a plane is shown? 回答1: material.side = THREE.BackSide or material.side = THREE

问题 In ILNumerics, I want to inverting a matrix A. The result is matrix B. Here is my code: ILArray<double> data = ILSpecialData.sinc(50, 50); ILArray<double> data2 = ILMath.zeros(50, 50); ILMath.invert(data, data2); The method is: ILMath.invert Method (ILInArray<Double>, ILOutArray<Double>) public static void invert( ILInArray<double> A, ILOutArray<double> outArray ) but I get the zero matrix in matrix B (data2) instead of an inversion of matrix A (data). What should I do to fix this? If it

问题 Okay, I've learned a bit more and have rephrased my question. I've got a need to flip or invert text 180 degrees (so it appears upside-down) on a RDLC report. I have some custom VB code that takes the text, converts it to a bitmap, then flips the rotates the canvas 180 degrees. The effect of this makes the text look a bit.. dithered... or fuzzy. It's not a sharp font anymore. The problem I'm experiencing is I'm using a special TTF Barcode font that creates something a scanner can read. When I