Skip to main content

plotting - ArrayPlot and non-integer PlotRange


There seems to weird bug regarding non-integer PlotRange in ArrayPlot. For example...


ArrayPlot[
Table[Sin[x/50 π] Tanh[y/50 π], {y, 100}, {x, 100}],
DataRange -> {{0, 1}, {0, 1}},
PlotRange -> {{0, 1}, {0.5, .9}}
]


gives an error:



Value of option PlotRange -> {{0,1},{0.5,0.9}} is not All, Full, Automatic, a positive machine number, or an appropriate list of range specifications.



The same code works, if one of the plot range parameters for y-axes is turned into an integer.


Did I do smth wrong? Is there a workaround?


I need to use array plot as ListContourPlot is too slow for large datasets and MatrixPlot has the same problem.


I am using Mathematica 8 64 bit version on linux.


Edit


After some more testing I see, that it works whenever PlotRange includes at least one integer. For example PlotRange -> {{0, 1}, {0.5, 1.2}} works as it includes 1. So one possible workaround is just to scale the range, such that max-min would be larger than 1.



But I am still looking forward, if anyone finds a way to have shorter than 1 range on the axes. Dirty way would be just using manual 'Ticks'.


Edit 2


Possible workaround.


As I wrote in the previous edit, it works, then the range includes at least one integer. So one could just scale the range. Here is a naive hard-coded example how it might be done.


ArrayPlot[
Table[Sin[x/50 π] Tanh[y/50 π], {y, 100}, {x, 100}],
DataRange -> {{0, 1}, {0, 1} 10},
PlotRange -> {{0, 1}, {0.1, 0.8} 10},
FrameTicks -> {Table[{y, ToString[y/10 // N]}, {y, 1, 8, 1}],
Table[{x, ToString[x // N]}, {x, 0, 1, 0.25}]},

AspectRatio -> 1
]

This means that I have bypassed the only side-effect of scaling the range, which is messing up the tick labels.


I would submit it as a solution, when I have waited my 8 hours and there is no nicer one proposed.



Answer



Here is a workaround that works in V7,8,9,10 (the underlying problem persists):


Show[
ArrayPlot[
Table[Sin[x/50 π] Tanh[y/50 π], {y, 100}, {x, 100}],

DataRange -> {{0, 1}, {0, 1}}],
PlotRange -> {{0, 1}, {0.5, .9}}, FrameTicks -> All]

Mathematica graphics


Comments

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

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...