Skip to main content

graphics - Rectangle with rounded edges



Inspired from :


Fit an image within a Rectangle [] in Graphics


I would like now to fit an image within a Rectangle[] with rounded edges as shown in the example below :


enter image description here


Is it possible ?



Answer



Using pieces from the linked answers, and Heike's code:


Text piece:


 txt1 = Take[
ExampleData[{"Text", "LoremIpsum"}, "Lines"], {1, -1, 2}][[1]] //

StringTake[#, 330] &;

Image piece:


 pic = Import["http://dailytechgadgets.files.wordpress.com/2011/02/old-ferrari.jpg"];

And Heike's code for rectangles:


 rec[ll_, ur_, pic_] := 
Module[{crop, boxrat},
boxrat = #2/#1 & @@ MapThread[Abs[#2 - #1] &, {ll, ur}];
crop = ImageCrop[pic, Transpose[{ImageDimensions[pic]}],

AspectRatio -> boxrat];
Inset[crop, Min /@ Transpose[{ll, ur}], {Left, Bottom},
Abs[ur - ll]]]

and Heike's code again for putting all together -- just adding RoundingRadius to rectangle objects and commenting out lines that produce lines--:


 Graphics[{EdgeForm[{Thickness[0.005`], Black}], FaceForm[White], 
Rectangle[{0, 0}, {160, 90}, RoundingRadius -> 4],
FaceForm[Darker[Gray]],
Rectangle[{0, 0}, {80, 63},
RoundingRadius -> 4],(*code for picture*){rec[{80, 0}, {160, 63},

pic], FaceForm[Opacity[0]],
Rectangle[{80, 0}, {160, 63},
RoundingRadius -> 4]},(*code for text*)
Inset[Pane[
Style[txt1, 12, TextAlignment -> Left], {Scaled[1],
Scaled[0.75`]}, Alignment -> Center,
ImageSizeAction -> "Scrollable"], {0, 8}, {Left, Bottom}, {78,67}],
Flatten[Transpose[{Flatten[(Table[
RandomChoice[{GrayLevel[0.15`], c0[[#1]]}], {3}] &) /@
Range[2, 4, 1]],

MapThread[
Function[{Xs, Ys},
Rectangle[{Xs, Ys}, {Xs + 16, Ys + 9},
RoundingRadius -> 4]], {Flatten[Table[Range[0, 32, 16], {3}]],
Flatten[(Table[#1, {3}] &) /@ Range[63, 81, 9]]}]}]](*,{Black,
Thickness[0.005`],Line[{{0,63},{159,63}}]}*)},
PlotRange -> {{0, 160}, {0, 90}}, Method -> {"ShrinkWrap" -> True},
ImagePadding -> 2, ImageMargins -> 0, ImageSize -> 500]

produces:



image output


You need further refinements to clean up. In particular, the image needs to be masked with a rectangle with rounded corners.


EDIT: For masking an image with a round-cornered reactangle, play with the parameters of Rectangle in


 pic2 = ImageAdd[ pic, 
Graphics[{Black, Rectangle[{2, 1}, {4, 2}, RoundingRadius -> .2]}]]

Update: For better masking using SetAlphaChannel and better handling of image size (Thanks: @ssch and @Jens)


roundImage[img_, r_] := Module[{dim = ImageDimensions[img], sr}, 
sr = Max[dim]*r;
SetAlphaChannel[img,

Graphics[{White, Rectangle[Scaled[{0, 0}], Scaled[{1, 1}], RoundingRadius -> sr]},
Background -> Black, PlotRangePadding -> 0,
PlotRange -> {{0, 0}, dim}\[Transpose], AspectRatio -> Automatic] ]]

Comments

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...