Skip to main content

output formatting - How to get this grid normal height?


When I execute this


Grid[{{x, x}, {SpanFromAbove, x}, {SpanFromAbove, x}}, 

 ItemSize -> {{Automatic}, 
  {Automatic, {1 -> Scaled[0.2], 2 -> Scaled[0.4], 3 -> Scaled[0.4]}}}, 
 Frame -> All, 
 Alignment -> {Center, Center}, Spacings -> {{2}, {1}}]

I get an extremely tall grid instead of a normal height grid. Is this a bug or am I doing something wrong? Version 10.0.1 Win8.1 64bit



This just keeps getting worse. I am have used a Pane as some have suggested that the grid needs a sized container for it to work with Scaled. I get back a truncated grid. This is really baffling. Any ideas how to get this to work.


Pane[
 Grid[{{x, x}, {SpanFromAbove, x}, {SpanFromAbove, x}}, 

  ItemSize -> {{{Scaled[0.5]}}, 
   {Automatic, {1 -> Scaled[0.2], 2 -> Scaled[0.4], 3 -> Scaled[0.4]}}}, 
  Frame -> All, Alignment -> {Center, Center}, Spacings -> {{1}, {1}}], 
 ImageSize -> {432, 216}]

Mathematica graphics


It gives a cut off grid. ??? I can just make out the top of the "x" in the first column. It shouldn't be this difficult to have a grid a prescribed size and scale some rows within it. I must be doing something wrong.



Answer



My advice, unless it is really basic problem, do not use SpanFrom~. Usually you will face "issues" like that. Next time try with nested Grids.


Like I've said, the whole layout management is broken and everyone who tried do something more complex than simple grid with automatic options will agree.



This is just another example, your case works if vertical heights sum up to: .5...


Framed[
 Grid[{
       {x, x},
       {SpanFromAbove, x},
       {SpanFromAbove, x}},
  ItemSize -> {
               {{Scaled@.499}},                  (*specX*)
               {Scaled@.1, Scaled@.2, Scaled@.2} (*specY*)
    }, Frame -> All, Alignment -> {Center, Center}, Spacings -> {0, 0}]

 , 
     ImageSize -> {432, 216}, FrameMargins -> 0, Alignment -> {Left, Top}]


enter image description here



Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

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...
Post a Comment
Read more