smooth

问题 I want to rotate a bitmap image based on user click by 10 deg. Following numerous stackoverflow and google answers, I tried various combinations of Matrix rotation. However the image doesn't really rotate as expected and gives a jittery view of rotation + oscillation about canvas center. To test I am increasing rotation angle by 10 deg (instead of clicks) each time object's draw method is called. The image is a symmetrical circle [64x64 enclosing rectangle] and I expect it to rotate at center

问题 I'm trying to make a live video streaming with Azure but i realy don't know what i should try first. First, a have a web site with a latest version of Flowplayer on it. Now i need to know how to make a stream video from my web camera to Azure (what kind of software do i need)? I'm trying to use Expression Encoder 4 Pro (but it dosen't want to Stream video to Azure Blobs). Next step i need to know what i should to do on Azure? I mean some settings, maybe i should make a VM with IIS role on it,

问题 Matlab's smooth function, by default, smooths data using a 5-point moving average. What would be the best way to do the same in python? For example, if this is my data 0 0.823529411764706 0.852941176470588 0.705882352941177 0.705882352941177 0.676470588235294 0.676470588235294 0.500000000000000 0.558823529411765 0.647058823529412 0.705882352941177 0.705882352941177 0.617647058823529 0.705882352941177 0.735294117647059 0.735294117647059 0.588235294117647 0.588235294117647 1 0.647058823529412 0

问题 I know that the smoothing parameter(lambda) is quite important for fitting a smoothing spline, but I did not see any post here regarding how to select a reasonable lambda (spar=?), I was told that spar normally ranges from 0 to 1. Could anyone share your experience when use smooth.spline()? Thanks. smooth.spline(x, y = NULL, w = NULL, df, spar = NULL, cv = FALSE, all.knots = FALSE, nknots = NULL, keep.data = TRUE, df.offset = 0, penalty = 1, control.spar = list(), tol = 1e-6 * IQR(x)) 回答1:

问题 I am looking for a solution to convert a freehand, user drawn SVG path, consisting of lots auf LineTo segments, into a smoother one. Preferred language would be JavaScript, but any advice is welcome. 回答1: first of all, I would recommend using a good graphics library, such as raphael. It will simplify the process of actually using javascript to perform the drawing. A very simple method of smoothing is to convert all lineto commands with equivalent curveto commands and calculate some control

问题 I am trying to resize a borderless form but when I increase the size using the right/bottom side, I get a gap between the border and the old client area that depends of the speed you move the mouse. The effect is more noticeable when you resize from the left border or even from the bottomleft corner, it's horrible everywhere (I tried with other commercial apps and it happens as well). This effect happens as well when I change to sizable border, but it's not as awful as when I remove form

问题 I am trying to resize a borderless form but when I increase the size using the right/bottom side, I get a gap between the border and the old client area that depends of the speed you move the mouse. The effect is more noticeable when you resize from the left border or even from the bottomleft corner, it's horrible everywhere (I tried with other commercial apps and it happens as well). This effect happens as well when I change to sizable border, but it's not as awful as when I remove form

问题 I already know the jQuery plugin ScrollTo, but I didn't find any way up to now to realize the following: The users gets to my site (by typing, NOT by clicking a link on my page) domain.com/bla.php#foo and the anchor "#foo" exists. Now I want that the browser of the user does NOT automatically scroll to "#foo", instead I want to smoothly scroll so that the element '#foo' is about in the middle of the view and NOT on the absolute top position of the users view. thanks so far! 回答1: If you don't

问题 I am just testing the following code to round corners on UIImage: - (UIImage *)makeRoundedImage:(UIImage *)image radius:(float)radius; { CALayer *imageLayer = [CALayer layer]; imageLayer.frame = CGRectMake(0, 0, image.size.width, image.size.height); imageLayer.contents = (id) image.CGImage; imageLayer.masksToBounds = YES; imageLayer.cornerRadius = radius; UIGraphicsBeginImageContext(image.size); [imageLayer renderInContext:UIGraphicsGetCurrentContext()]; UIImage *roundedImage =

问题 I'd like to sloothly move an image (or an element) from its actual X, Y location to X1, Y1. When the distance between X and X1 is equal to that between Y and Y1 its easy. But what if the X difference is say 100px and Y diff is 273px? Being new to Javascript, I don't want to re-invent the wheel! Besides, since I'm learning, I do NOT want to use jQuery or the likes. I want pure javascript. Please supply with simple script :-) 回答1: One solution: function translate( elem, x, y ) { var left =